What's the physicochemical difference between a chromophore and a fluorophore? Let's take Hematoxylin, which is a pH indicator and a stain for histology, and therefore a chromophore, but does not fluoresce:

Now let's take Eosin, which stains tissue pinkish, but is also fluorescent:

I would like to know which features of the molecules decide if it is a chromophore or a fluorophore. As far as I can see it (I'm a biologist with some training in organic chemistry), both have conjugated systems, although the system in Eosin spans a lot of more bonds. Is it really just the size of the conjugated system that decides this, e.g., if Hematoxylin had a larger conjugated system, would it also be fluorescent?

The physical chemistry of fluorescence is a little bit complicated, and so you should consult a text book for "the reason" that things fluoresce.

One of the things that I have noticed is that fluorescent compounds often have several "heavy halogens" Br and I attached to some poly-aromatic hydrocarbon.

Disclaimer: this is not to say that the halogens are the fluorescing portion of the molecule (it's probably almost the whole molecule, conjugation + halogen), and it does not mean that all compounds that contain Br and I will be fluorescent. It also does not mean that Br and I are required for fluorescence.

I think that a lot of fluorescent molecules were discovered by chance, or even brute-force changing substituents until the molecule exhibits the desired fluorescence.  (It's really too hard to predict with qualitative theory, and electronic structure theory descriptions of fluorescence are also very difficult.)


Why is acetylsalicylic acid still more acid than benzoic acid? In this question, it is explained why salicylic acid is a stronger acid than benzoic acid. So, why acetylsalicylic acid (aspirin), whose conjugate base cannot hydrogen bond since there is no hydrogen, is still more acid than benzoic acid?

aspirin: $\mathrm{p}K_\text{a}$ = 3.5
benzoic acid: $\mathrm{p}K_\text{a}$ = 4.2

Acetoxyl is a moderate activating group for the aromatic electrophilic substitution, so it should mean that the resonance-driven electron-donating effect prevails over the inductive electron-withdrawing effect, therefore it should raise the $\mathrm{p}K_\text{a}$ of the acid.

Is it different between electrophilic substitution reactivity and acidity? Does the inductive effect prevail in this case?

The acetyl moiety in acetylsalicylic acid has a $(-I)$-effect, i.e. compared to a proton attached to the aromatic ring (as in the case of benzoic acid), it withdraws electron density out of the aromatic ring.  This provides an additional stabilization of the conjugated base, that is equivalent with rendering acetylsalicylic acid a stronger Brønsted acid than benzoic acid.  Hence, the $pK_a$ of acetylsalicylic acid is lower than the one of benzoic acid.

Regarding the second paragraph, please notice $pK_a$ is a thermodynamic property -- where time does not play a role.  In contrast, the substitution reactivity you likely refer to (Hammett parameters) refers to reaction kinetics and rate constants.  Here, time does play a role.


Reason for enthalpy being non-zero at constant temperature Suppose a reaction is occuring at constant temperature. Then $\mathrm dH=nC_{p}\mathrm dT = nC_{p} × 0=0$.

Please explain what the mistake is that I have done. Clearly, there are many reactions which take place at constant temperature and have a definite enthalpy change. But according to the equation $\mathrm dH=nC_{p}\mathrm dT$, if temperature $T$ is constant then $\mathrm dT =0$ and $\mathrm dH = 0$. But this is clearly not true. There are reactions with constant temperature and $\mathrm dH≠0$. Please explain what is wrong?

The only way that an exothermic reaction can occur with "constant temperature" is if the heat generated is constantly removed in some way. Even if the heat removal is essentially instantaneous, it is incorrect to say that there is no temperature change. Rather there is an infinite number of infinitely small temperature changes as the system heats up by an amount dT and then is returned to the set T. If you were able to measure these infinitely small changes and also to know the instantaneous Cp (which changes as the reaction composition changes), you could sum them up and get a non-zero value for the enthalpy change. Obviously, that's not a practical approach.
Instead, to determine the enthalpy change experimentally, it is easiest to perform the reaction within an insulated apparatus such that the temperature is not constant and the change in temperature can be used to determine the enthalpy change.


What is the cost of 1.00 g of calcium ions as provided by this brand of dry milk? A student determined the $\ce{Ca^{2+}}$ ion content in a sample of nonfat dry milk. For complete titration of the $\ce{Ca^{2+}}$ ion present, a sample weighting $1.483 ~\mathrm{g}$ required $41.33 ~\mathrm{mL}$ of $1.183 \times 10^{-2} ~\mathrm{M}$ EDTA solution. Calculate the mass of the $\ce{Ca^{2+}}$ ions in the titrated sample.

So I did:

$41.33 ~\mathrm{mL} = 0.04133 \mathrm{L}$

$(1.183 \times 10^{-2}~\mathrm{mol\, L^{-1}})\times (0.04133 \mathrm{L}) = 4.89 \times 10^{-4} ~\mathrm{mol}\, \ce{Ca^{2+}}$

$(4.89 \times 10^{-4} ~\mathrm{mol}\, \ce{Ca^{2+}}) \times 40.078 ~\mathrm{g\, (mol\, \ce{Ca^{2+}})^{-1}} = 0.0196 ~\mathrm{g}\, \ce{Ca^{2+}}$

My question is:

A $25.06$ oz box of the powdered milk analyzed above costs $5.46$.

  Calculate the cost of $\pu{1.00 g}$ of $\ce{Ca^{2+}}$ ion as provided by this brand of dry milk.

How do I do that?

First you calculate the value $v$ of the powder per gram:
$$ 25.06~\mathrm{oz} = 710.44~\mathrm{g} \; \hat{=} \; 5.46~\$ \Rightarrow v=0.007685~\mathrm{\$\, (g\, Powder)^{-1}}$$

Then you calculate the mass of powder ($m$) that contains one gram of calcium ions:
$$ m=\frac{1.483~\mathrm{g\, Powder}}{0.0196~\mathrm{g\, \ce{Ca^{2+}}}} = 75.66~\mathrm{\frac{g\, Powder}{g\, \ce{Ca^{2+}}}} $$

Finally, you multiply the two results for the price $p$:
$$ p = v\times m = 0.58~\mathrm{\$\, (g\, \ce{Ca^{2+}})^{-1}}$$


Reaction between Iron (III) Oxide and CO? I'm a science tutor and I came across a stoichiometry problem asking how much Iron is produced in a reaction between $\ce{Fe2O3}$ and $\ce{CO}$:

$$\ce{Fe2O3 + CO -> Fe + ?}$$

I know how to do the stoichiometry part, but I'm stumped about how to figure out the products. My guess is $\ce{Fe}$ and $\ce{CO2}$ since the $\ce{CO}$ would want to change to the more stable $\ce{CO2}$, thus deoxidizing the $\ce{Fe}$? All the student's problems so far have just been between simple ionic salts, and I never remember coming across one like this when I took AP Chen myself in highschool, since $\ce{CO}$ is covalent.

The reaction products depends on stoichiometric ratio of the reactants as well as reaction conditions (i.e temperature, pressure etc.)

$$\ce{Fe2O3 + CO ->[500-600 C] 2FeO + CO2}$$

  Iron(III) oxide and carbon monoxide to produce iron(II) oxide and
  carbon dioxide. This reaction takes place at a temperature of
  500-600°C. (source)

$$\ce{3Fe2O3 + CO ->[400 C] 2Fe3O4 + CO2}$$

  Iron(III) oxide react with carbon monoxide to produce iron(II,III)
  oxide and carbon dioxide. This reaction takes place at a temperature
  near 400°C. (source)

$$\ce{Fe2O3 + 3CO ->[700 C] 2Fe + 3CO2}$$

  Iron(III) oxide react with carbon monoxide to produce iron and carbon
  dioxide. This reaction takes place at a temperature near 700°C. (source)


Properties of plasmas In chemistry one can recognize that the four states of matter are solid, liquid, gas and plasma. The first is rigid, and has a definite shape and volume. The second doesn't have a shape, and assumes the shape of its container, but it has a fixed volume. The third doesn't have either a shape or a fixed volume and assumes the volume and shape of its container. What about the fourth one (plasma)?

Plasma is made up of ionized gas, so molecules have a positive electric charge and valency electrons are totally or partially separated by their nuclei.
Plasma is different from a gas since it has a high temperature and is radiation emitting; think of the Sun and other stars, which are made up of plasma and show both these properties.
Plasma hasn't got a proper volume, like gases; e.g. stars can expand or contract under the opposite effects of gravity and nuclear fusion. For example, this property is important to comprehend the formation of white dwarfs and neutron stars, process caused by high pression due to gravity.
Little trivia: there is also a fifth state of matter, whose name is Bose-Einstein Condensate (BEC).


Why is freezing point depression greater than boiling point elevation for equimolar substances? I know that the freezing point depression and boiling point elevation are given by the formulas $\Delta\mathrm{T}_{b}=ib\mathrm{K}_{eb}$ and $\Delta\mathrm{T}_{f}=ib\mathrm{K}_{F}.$ It seems that when this formula is applied for equimolar concentrations, the magnitude of freezing point depression is greater than boiling point elevation. Is there a qualitative reason for this? If so, what is it?

To find an answer, let us take a look at how the cryoscopic constant (for decrease in melting point) and the ebullioscopic constant (for increase of boiling point) are defined:

Please note that the Kelvin-scale is used for temperature here. The boiling point is always above the melting point (we keep pressure constant to keep the example simple).

For most substances, the enthalpy of vaporization is greater than the enthalpy of fusion. It follows that the denominator for K_eb is greater, resulting in a smaller K_eb.

This means that the effect of greater enthalpy of vaporization compared with enthalpy of fusion outweighs the effect of boiling temperature being greater than the melting temperature (even though it's the square of absolute temperature).


Question about haloalkanes Are haloalkanes considered a homologous series? For example, when assigning chloromethane to a homologous series, do we say that it belongs to the alkanes or haloalkanes?

Edit: I think that it would be a haloalkane. However, I'm unsure if haloalkanes are considered a homologous series.

While homologous series are a very nice thing for teachers to quiz students about, they practically don’t matter to modern lab chemists. Sure, we all know that methanol, ethanol and propan-1-ol form a series and that a $\ce{CH2}$ unit is added when going from one to the next but in terms of actual lab properties the three are different enough to be remembered individually. As for most synthetic applications, linear molecules (the only ones for which a homologous series makes some sense) are not really any target of relevance so they won’t be thought of.

That said, I see nothing against applying a term that works well for many functional groups (e.g. $\ce{OH}$ or $\ce{NH2}$) to another functional group ($\ce{Cl}$) so the only argument against calling them one is ‘my teacher said so and it’s important that I know that in the upcoming exam’.


Why would the 2+ molecular ion of butane have a mass spectrum peak at 29? Question 7 of Edexcel's June 2014 Unit 2 exam paper reads:

  In a mass spectrum of butane, C4H10, where would a peak be seen for
  the molecular ion if it had a charge of 2+?

The answer is 29.

My question is: why?

The mass spectrometer observes the mass to charge ratio (m/z). Most of the time, small organic molecules like butane will be singly charged by the ionizer, which means that the mass can be read off directly since z =1.

If butane was somehow ionized to the +2 state, then the mass spec would observe 58/2=29, which is the mass of butane (58 g/mol) divided by the charge (+2).

This becomes particularly important for larger compounds with multiple easily ionizable groups, such as peptides. A series of peaks corresponding to different charge states can be used to identify the compound of interest.


Concentration of Amino acid calculation? I found the $\mathrm{pH}$ of alanine solution, and I need its concentration, I'm given the formula below:
$$\mathrm{pH}=\frac 12 (\mathrm{p}K_\mathrm{a}- \log c)$$
My try:
I have calculated $\mathrm{pH}$, to find $c$ I need $\mathrm{p}K_\mathrm{a}$. Can someone explain that formula, what is $\mathrm{p}K_\mathrm{a}$?
I know that this amino acid has $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$, which do I use?

To clarify:
$\mathrm{p}K_\mathrm{a1}: \ce{NH3^{+}-R-COOH <=> NH3^{+}-R-COO- + H+}$
$\mathrm{p}K_\mathrm{a2}: \ce{NH3^{+}-R-COO- <=> NH2-R-COO- + H+}$
For aminoacids, the the pH calculation is the same as for acidic salts of diprotic weak acids $\ce{MHA}$.
Unfortunately for you, $\mathrm{pH}=\dfrac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$ as the very good approximation, so it is almost independent on the concentration, so you cannot determine the concentration from $\mathrm{pH}$.
More exactly. you could, but by a much more complex formula as you listed above, as it would be very inacurate because of high $\dfrac{\mathrm{d}c}{\mathrm{d(pH)}}$
By other words, aminoacid solutions act as low capacity pH buffers ( because of quite a big difference of their $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$.


Vinegar reacting with metal pan? I was following a recipe for a seasoning containing rice vinegar. The directions noted that a metal saucepan should NOT be to used to heat the seasoning because it would react with the vinegar.

What would happen, chemically, if I used a metal (stainless steel) pan?

My thought is that the acetic acid and Iron could react to form Iron(III) acetate. Would this complex be in a concentration high enough to warrant the, "do not put in a metal pan," warning or would the reaction be something else?

Vinegar was used as one of the main household cleaning and odor removal agents for a good reason. I think in general you can get numerous iron (III) and chromium (III) (which are less pleasant) salts and complexes aside from acetate, e.g. chloride/iodide if you use iodised salt or hypochlorite from chlorinated water.
Diluted acetic acid (vinegar) etches majority of steels quicker than the glacial one, dissolving its passivating $\ce{Cr2O3}$ layer. Aside from concentration, reactivity also greatly depends on the steel type and acetic acid purity [1]:

Carbon and Alloy Steels. Steel is attacked quite rapidly by all concentrations of acetic acid, even at room temperature. Glacial $\ce{CH3COOH}$ at room temperature is less aggressive than aqueous solutions of the acid, but still gives a rate of attack of $\pu{0.8 to 1.3 mm/yr}$ ... Therefore, steel is normally unacceptable for use in $\ce{CH3COOH}$ service.
Stainless Steels. The chromium stainless steels of the 400 series occasionally exhibit low corrosion rates in laboratory tests in dilute acetic
acid. However, because field experience with these materials indicates high corrosion rates and pitting attack, they are rarely used for $\ce{CH3COOH}$ production equipment.
Impurities present in the manufacture of acetic acid, such as acetaldehyde, formic acid, chlorides, and propionic acid, are expected to increase the attack of stainless steels. Contamination with chloride can cause pitting,rapid stress-corrosion cracking, and accelerated corrosion of type 316 stainless steel. Up to $\pu{20 ppm}$ of chloride can be tolerated, but higher concentrations are likely to cause rapid equipment failure.
Transferring heat through a metal wall, as in heat exchangers, can drastically alter the corrosion characteristics of the metal.

If you know the exact steel type used, you can check its stability against various acidic media in several corrosion handbooks, e.g. [1] contains cumulative tables for corrosion behavior of manufactured steels tested under various conditions.

Craig, B. D.; Anderson, D. S. Handbook of Corrosion Data; ASM International, 1994. ISBN 978-0-87170-518-1.


Why use a salt bridge at all in an electrochemical cell? I was reading about electrochemical cells and I really don't understand why a salt bridge is needed. Can't you just combine the solutions? I've found a lot of webpages explaining electrochemical cells but strangely enough none of them seem to find this important enough to explain. All thoughts appreciated :)

The use of a salt bridge is to maintain the electrical neutrality throughout the cell.

A salt bridge will allow the movement of ions.
Electrons flow from anode to cathode.Due to this a negative charge will be developed in the cathode and a positive charge in the anode.This charge developed will oppose the electron flow (current from anode to cathode) and at one level the electron flow will be ceased.So the cell's action will be stopped.
So in order to prevent this, salt bridges are used.

There is a similar question in this site.Please refer that.

Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?


Can water be produced by mixing hydrogen gas and oxygen? Could there be some kind of process, a kind of "retrograde electrolysis" where hydrogen gas (or in some other form) and oxygen gas (or in some there form) are mixed with required energy to combine to form water as a byproduct?

Indeed, it's called combustion! When you burn hydrogen and oxygen gas, the reaction is:

$$\ce{2H2_{(g)} + O2_{(g)} -> 2H2O_{(g)} + energy}$$

Basically, you could see the former U.S. space shuttle as a large machine that turns hydrogen and oxygen into water. (Note: the space shuttle used liquid hydrogen and oxygen, but otherwise the principle and formula is exactly the same)

Since the Gibbs free energy of the forward reaction is negative ($-242\textrm{ kJ/mol}$), you don't need to add energy to the system, the reaction is spontaneous (mind you, you generally need a spark to ignite the combustion because there's a kinetic barrier to the reaction called the energy of activation).

The reverse reaction, water to hydrogen and oxygen, has a positive Gibbs free energy, so you need to add energy in the form of electricity to the system to make the reaction happen.


Coulomb force and potential energy in water vs vacuum If you have 2 ions of equal but opposite charge, will the force between them be larger in a vacuum and smaller in water? Would this be because the relative permittivity of water is greater than 1 (around 80)? Also, what does this mean for the interaction energy? Would an interaction be larger in vacuum? I am trying to figure out whether a reaction is more likely to take place in the two different media between the two ions...

Yes, much larger in vacuum. The interaction is proportional to the reciprocal of the dielectric constant (relative permittivity) $1/\epsilon$ so is less in water. Think of it as an attenuation of the electric field around an ion, the larger $\epsilon$ is the more the field is attenuated and is true whatever the charges on the ions are making them attractive or repulsive.  The force at distance r between charges $q_1, q_2$ is

$$F=\frac{q_1q_2}{(4\pi\epsilon_0)\cdot\epsilon r^2}$$

in SI units; $\epsilon_0 $ is the permittivity of free space. The charges are $q=ze$ where e is the charge on the electron and z the ionic valency, $\pm 1, 2 \cdots $ etc.

The interaction energy is

$$U=\frac{q_1q_2}{(4\pi\epsilon_0)\cdot\epsilon r} \qquad \mathrm{Joule}$$

The electric field around charge $q_1$ is

$$E=\frac{q_1}{(4\pi\epsilon_0)\cdot\epsilon r^2}$$

in V/m. The force acting on  a second charge $q_2$ is $F=Eq_2$

Its worth plugging in some numbers for the energy for different $\epsilon$ and comparing this to thermal energy $k_BT$.


What is the difference between a mixture and a composed substance? In my textbook it says, that a composed substance is a substance made of more than 1 type of atom. And It also says, in another chapter it says that a mixture is the grouping of more than one substance without any chemical reaction.

What is the difference between a mixture and a composed substance?

From the definitions in your textbook, it seems that a mixture is a combination of two or more substances that--though combined--maintain their chemical identifiability. Examples of mixtures might be:

Water/Alcohol Mixtures
Salt/Pepper Mixture
In pharmacy, pills, pastes/creams, and suppositories are mixtures

(there are many example of mixtures)

The consituent compounds might be separated with somewhat mild techniques, including dissolving in certain solvents, distillation, and crystallization.

Composed Substances (according to your textbook) are mixtures of two or more species that under go some chemical reaction. The species involved could lose most of their identifiability, becoming completely different compound(s).  Loosely speaking, chemical reactions can not be reversed by simple separation techniques.

(Of course, there is some arbitrariness here: particularly in what comprises "most" of their identifiability and what defines "simple" separation techniques.  Also, mixtures are actually held together with bonds, just like in composed substances.  They may be weaker, but even "weakness" of the interaction is a subjective concept.)


Length of NaH bond from the edge length of unit cell If the edge length of $\ce{NaH}$ unit cell is $\pu{488 pm}$, what is the length in $\pu{pm}$ of an $\ce{NaH}$ bond? The answer is $\pu{244 pm}$, I thought that $d = 2 \sqrt{2}\times r$ , do you not need to know the radius for this?

$\ce{NaH}$ has a rock salt type structure meaning each edge has two $\ce{Na-H}$ bonds. When working out bond lengths in a unit cell, you assume the atoms are point charges (i.e. they take up no volume) and so you do not need to know the radius, only the unit cell parameter which as you have said is $\pu{488 pm}$. Since there are two bonds along each edge, the bond length is simply half of the unit cell parameter, therefore $\pu{244 pm}$.

Finally it dawned on me, why you may have been concerned about knowing atomic radii, yes, you need to know the radii to determine whether $T_\mathrm{d}$ or $O_\mathrm{h}$ sites are preferred. However, given the context of the question, there must have been something previously alluding to the octahedral sites being occupied (and hence rock salt structure) rather than the tetrahedral sites (which would give you a ccp structure with half of $T_\mathrm{d}$ sites occupied). However, you still wouldn't need to use the ionic radii in any calculations pertaining to bond length. I have also included a drawn picture of the unit cell of Sodium Hydride to help you visualise how to calculate the bond length.


What is the conjugate acid of NaOH using the Brønsted-Lowry definition of acids? To the best of my knowledge, a conjugate acid of a base is the base after it has accepted a proton, or a $\ce{H+}$ ion. In this case:

$$\ce{NaOH + H+ -> Na+ + H2O}$$

Is the conjugate acid of $\ce{NaOH}$ the sodium ion, or the water? Common sense tells me it can't be the $\ce{Na+}$ ion, because it has no protons to donate, so how could it ever be an acid? So I am thinking that the conjugate acid is $\ce{H2O}$. However, wouldn't that mean that the conjugate acid of any base of the form

$$\ce{(something)OH + H+ -> (something)+ + H2O}$$
would be water, and that seems unsettling to me. I also believe that since $\ce{NaOH}$ undergoes the following reaction:

$$\ce{NaOH -> Na+ + OH-}$$

the $\ce{Na+}$ is something of a 'spectator ion' (not sure if that's the correct term), this seems to imply that $\ce{H2O}$ should be the conjugate acid. Basically, I'm really confused, and could use a little help sorting all this out.

Your first equation is more properly written as

$$\ce{NaOH + H3O+ -> Na+ + 2H2O}$$

in aqueous media. However, we can do better if we explicitly show the dissociation of $\ce{NaOH}$ as

$$\ce{NaOH -> Na+ + OH-}$$

and substitute that into the first expression (note that I write $\ce{2H2O}$ as $\ce{H2O + H2O}$) to get

$$\ce{Na+ + \underbrace{OH^{-}}_{base} + \underbrace{H3O^{+}}_{acid} -> Na+ + \underbrace{H2O}_{conjugate\;acid} + \underbrace{H2O}_{conjugate\;base}}$$

where we see that $\ce{H2O}$ is the conjugate acid of $\ce{OH-}$ as well as the conjugate base of $\ce{H3O+}$. The last bit - where water plays 2 roles - is due to water being amphoteric, or able to act as an acid or a base.


Why is it advised to put a Styrofoam plate under an Erlenmeyer flask? At school today, we were doing experiments with ammonium nitrate and sodium hydroxide, and mixing them with water into an Erlenmeyer flask (separate flasks, of course). My instructor told us that we had to keep a Styrofoam plate under the Erlenmeyer flask at all times.
After the experiment, she told us to find out why we put a Styrofoam plate under the flask by tomorrow or we would get a 0 on that assignment.
Does anyone know why? I'm thinking it's to protect the table from the heat or coldness of the flask, or to protect the table in case of spills. We have to get the question exactly as she wants it, so I am looking for help.

You provided the additional information in the comments that your experiment was about measuring the change in temperature that occurs upon mixing. This should give you a hint as to whether the substances mix with water in an endo or exothermic manner.

When measuring temperature changes it is crucial not to influence the temperature of the reaction mixture in any way. Naturally, that would for example mean not to keep the flask in your hand (body temperature is typically more than room temperature). But some error sources are not as easily visible. In your case, the table top was an error source. Whatever the material, it likely conducts heat better than air — meaning that if your flask was on the table a significant portion of the heat generated (or used) would be conducted away (or onto site) by the table resulting in a lower change in temperature. That is why these experiments are typically performed under maximally insulating conditions. One of the best and widely available insulators is polystyrene because of its high air content (air is a good insulator). Hence, keeping polystyrene under a flask reduces the error caused by the table conducting heat.


Is it possible create crystalline solvate of electrons? Okay, I know that this reads ridiculous insane, but one can obtain solvated electrons by adding sodium metal to liquid ammonia solution.

The chemical reaction as written in my inorganic chemistry textbook is this:

$$\ce{Na + x NH3 -> Na+ + e(NH3)_x-}$$

A metallic sheen is observed, and it supposed to have an absorption band at about $1500\ \mathrm{nm}$ that expands into the range of visible light corresponding to the solvated electrons.

These electrons can be react with alkynes as very powerful reducing agents to to remove an H from a C joined by a triple bond, also liberating hydrogen gas as such. This occurs in the solution made by an alkali metal, not from pure extracted electrons

$$\ce{RC#CH + e- -> RC#C- + 1/2 H2}$$

So I was wondering if pure electrons can be solvated in ammonia, if anyone could think of any means by which these electrons could be extracted from the layer, still solvated by ammonia, and some reaction chemistry could be used to trap electrons in a carbon cage in the form of a solid? It is already known that potassium can be electrostatically trapped in valinomycin between the carbonyl groups of valinomycin.

While it may be different small molecules have been trapped in fullerenes. If electrons could be solidified via extracting them from the metallic layer in the proposed reaction, and interacting them with a "host molecule" in host-guest chemistry, do you think that these materials would be highly conductive. It seems to me that captured electrons would have a much different chemistry than simply metals like copper which shed electrons when a current runs through them. Perhaps if these electrons could be absorbed into carbon nanotubes, then the conductivity of nanotubes would increase significantly and these electron complexes may be significantly more powerful computationally than silicon chips.

Here is an experiment by the University of Nottingham scientists showing the solvation of electrons: https://www.youtube.com/watch?v=tYjQXjUUvwY

What do you think about this? Any ideas?

At first I would like to advise you reading about electrides, which were mentioned in the comments. Secondly I won't write about electrides because the links given in the comments are much better at explaining them. So I am going to be a little bit abstract with this question. The first problem which you will come through is Coulomb's law:
$$\mathbf F=\frac{1}{4\pi \epsilon_{0}}\frac{Q_{1}Q_{2}}{|\mathbf r|^{2}}\frac{\mathbf r}{|\mathbf r|}$$
If you could somehow prepare even $0.01$ moles of electrons, they would be attracted to the ions which they were separated from very strongly even from kilometers away. The second problem is that electrons can't be simply trapped into any substance because of their wavelike nature and they would tend to "stay" in a localised or in a delocalised state in the molecoules. And here comes the two main things which are preventing such things to exist:
1. Charge distribution
If a conducting body is charged, the charge tends to build up on the edges, and in your situation the repulsive force between the electrons would be so huge, they would "kick" each other off on the edges of the pieces of the substance so it would loose it's electrons.An interesting phenomena such substance would have: Let's suppose the substance containing the electrons is a liquid. All the electrons would be located on the surface of the liquid because of the faraday effect.When the surface of the liquid is disturbed waves appear on it's surface. Let's suppose the electrons somehow can't be kicked off from the surface.
Then there would always be electric current in the liquid because the elecrons would always try to get to the most sharp edge on the surface and the liquid would generate extremely complex magnetic fields because of the currents flowing on the surface of the liquid (I am so excited about the idea of such a liquid, I am so sad I have no idea how it would be possible to make it)
2. Quantum mechanics
The band gap determines if a substance is an insulator or a semiconductor or a conductor. And adding more electrons to something doesn't modify the band gaps. Like if I charge a plastic rod with electrons it won't become a conductor. But there is an alternative solution. If you dope the substance it can modify the band gap energy and it can be more conductive but you can't simply trap electrons in something (F-centers are a completely different story). Imagine irradiating carbon nanotubes with electrons. I wouldn't suspect it becoming more conducting .
Anyway the best source for pure electrons is an electron gun
If you want to deepen your knowledge about the nature of electrons and conductivity I advise you learning some solid-state physics, quantum mechanics and electromagnetism.(None of them is a light weekend program)


Rate of nitration of C6H6, C6D6 and C6T6 Why the reaction rate (nitration) of $\ce{C6H6}$, $\ce{C6D6}$ and $\ce{C6T6}$ are equal? Can anyone explain this using simple words?

The rate-determining step is the attack of the benzene aromatic ring onto the electrophile, and the loss of aromaticity is what causes the instability of the transition state, resulting in a higher activation energy and thus a slower rate of reaction. The subsequent deprotonation is after the rate-determining step and thus will not affect the rate of reaction, despite C-T > C-D > C-H bond strength.


What does K° signify in this reaction? What does the K° signify in the above reaction. Is it just a typo, or is it supposed to symbolize something?
EDIT: This screenshot is from the solutions manual. In the original textbook, the author has used $K ^ 0$.

K° here means elemental potassium (zero oxidation state). The author wanted to avoid the confusion of K with the equilibrium constant. There are more quantities than we have symbols, so letters and symbols get re-used. It is called a pigeon hole effect.
Now, that the OP has posted the reference, Solomons, Organic Chemistry, Ed 11, Problem 11.14 Solution manual, it is clear that the circle is a typo for zero. In the 11 and the 12th edition, the symbol stands corrected to proper numerical "zero" in the main textbook (pg 510). It is not unusual to find typos in solution manuals and textbooks. We are all humans-after all.


Woodward Fieser rules for dienes - non conjugated double bond as exocyclic? When looking for exocyclic double bonds in a diene, should a double bond not connected to the conjugated system be considered?

For example, in this compound, should the marked double bond considered as an exocyclic double bond?

Woodward's rules are an empirical method for predicting the maximum absorption of a molecule based on the functional groups present (that is, conjugated functional groups which possess a distinctive maximum absorbance such as a diene or an unsaturated carbonyl).

  Should the marked double bond considered as an exocyclic double bond?

Exocyclic is defined as being outside of the ring system (or in more complex systems, outside of the ring system being considered), the bond highlighted is indeed inside the ring system and is as such endocyclic.

If we start with the alkene at the bottom left, then it is in conjugation with an endocyclic alkene (to the immediate right) and an exocyclic alkene (in the ring to the right hand side). These double bonds are all in conjugation with one another.

The bond you've pointed at isn't in conjugation with the 'triene'

The Woodward rules are specifically related to the effect of groups in conjugation, a simple test of which is to see if you can push curly arrows between the pi bonds. If you try this for your molecule you'll quickly see that the bond you've highlighted is too distant from the conjugated system to have any involvement.


How should I name this branched alkyl substituent? The parent chain is a decane (in black), but for the alkyl substituent (in red) I'm trying to decide between:

5-(1,2-dimethylpropyl)decane
5-(2-methylbutan-3-yl)decane
5-(3-methylbutan-2-yl)decane

Which of these is correct?

According to Section P-46 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), compound acyclic substituents consisting of a principal chain and one or more acyclic substituents can be named in two ways:

by using alkyl substituents according to method (1) of Subsection P-29.2

  (1) The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ replace the ending ‘ane’ of the parent hydride name. The atom with the free valence terminates a chain and always has the locant ‘1’, which is omitted from the name. This method is recommended primarily for saturated acyclic and monocyclic hydrocarbon substituent groups and for the mononuclear hydrides of silicon, germanium, tin, and lead. Substituent groups formed by this method are referred to as ‘alkyl-type substituent groups’;

by using alkanyl substituents according to method (2) of Subsection P-29.2

  (2) The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ are added to the name of the parent hydride with elision of the terminal letter ‘e’, if present, when followed immediately by the letter ‘y’. The locants for the atoms of free valences are as low as is consistent with any established numbering of the parent hydride and, except for mononuclear parent hydrides or the suffix ‘ylidyne’, the locant ‘1’ must be cited. This method is used to generate names of ‘alkanyl-type substituent groups’ that are simple substituent groups with free valences at positions other than ‘1’.

For both methods, the principal substituent chain has the greater number of skeletal atoms, i.e. the longest chain. Low locants are given first to the free valence; for method (1), the free valence always has the locant ‘1’.

The preferred names for simple substituent prefixes derived from saturated parent hydrides with a single free valence that is not at the end of the longest chain are formed according to method (2).

According to method (1), the alkyl-type name of compound substituent group given in the question is ‘1,2-dimethylpropyl’. The preferred alkanyl-type prefix according to method (2) is ‘3-methylbutan-2-yl’.

Therefore, the preferred name for the complete compound given in the question is ‘5-(3-methylbutan-2-yl)decane’.

Note that, according to current IUPAC nomenclature, locants are placed immediately before that part of the name to which they relate. Therefore, correct prefix is ‘3-methylbutan-2-yl’ (not ‘3-methyl-2-butanyl’).


Why is sulfuric acid used in the Jones oxidation of alcohols? When oxidising primary alcohols, $\ce{K2Cr2O7}$ is used because it's a useful & readily available oxidising agent.

Why does the sulfuric acid need to be added?
Does it have to do with the acidity of the resulting dichromate solution? How would this facilitate the oxidation?

It protonates the chromic acid to form an oxonium ion, increasing positive charge at the chromium center.

Using $\ce{H^+}$ typically protonates the electrophile, making it more electrophilic (more electron deficient = more electrophilic), just like with acid-catalyzed esterification. In fact, you're making a chromate ester in this reaction.

See the first step here in this image (and the rest is related, also):
Image source


How can I draw metallocenes in ChemDraw? How can I properly draw the structure of metallocenes (e.g. ferrocene) and similar sandwich compounds in ChemDraw?

In particular, two problems cannot be solved with the usual workflow when building a molecule in ChemDraw:

How can I draw a bond starting from the centre of a ring structure (e.g. cylopentadienyl or benzene) if there is no atom?
How can I connect the centre of another ring structure to an existing molecule with the correct bond length and angle?

This question came up among the editorial staff of the chemistry section of the German Wikipedia. I have answered this question there myself, and I think it might be useful to share the solution to this problem here.

In the first step, separately draw the individual ring structures. ChemDraw comes with an template library of predrawn structures, including some suitable cyclopentadienyl rings.

You can place the ring structures anywhere in your document.

Select one ring structure. On the Structure menu, click Add Multi-Center Attachment. This adds a multi-center node as attachment point to the structure.

Repeat this process for the other ring structure. Asterisks indicate the created multi-center nodes.

You can use the multi-center attachment points to draw the coordination bonds. Note that bonds connecting to delocalized systems should be drawn as plain bonds; it is not acceptable to use dashed bonds. When one bond crosses another, if necessary, you can click Bring to Front or Send to Back on the Object menu to indicate which bond is in front. The bond which is behind will have a small section cut out.

Click to select the end (i.e. the terminal atom) of the coordination bond in one structure, and then shift+click to select the end of the coordination bond in the second structure. On the Object menu, click Join.

The two structures are joined so that they share an atom. Note that the second structure is moved accordingly.

Finally, you can add any substituents as required.

(Also note that for unsubstituted ferrocene in the gas phase, the cyclopentadienyl rings are actually eclipsed.)


What type of analytical chemistry (such as spectroscopy) is suitable for high resolution air analysis I want to find some type of analytical chemistry method, such as spectroscopy, that is capable of analysis the chemical composition of air. If it can be used for high resolution analysis would be preferable.

Headspace GC-MS is the way to go. It's used to detect vapors coming off a solution. For example, it is used in the fragrance industry to test the time-dependent development of a perfume, since the lighter molecules will evaporate first and form a higher concentration in the headspace initially.

Unfortunately, for detection of noble gases in an extremely low concentration this is difficult. The technology has just been created and involves concentrating the rare gases via adsorption over high surface area carbon before analysis. Each gas adsorbs in a specific temperature range that is different.

From an article on the International Noble Gas Experiment, a modern method to detect trace amounts of Xenon from nuclear explosions:

  “Xenon is isolated from the air by a charcoal-containing purification device resulting in higher concentrations of the gas in its stable and unstable, i.e. radioactive, forms.  The isolated xenon is then measured for its level of radioactivity which helps to quantify the radioactive xenon contained in the sample.” Source

I know that a company called Air Products liquifies air before distilling Nitrogen and Oxygen, and then purifies Xenon for medical use. An in-depth process is described here.


Why do the units of rate constants change, and what does that physically mean? Just to be clear, I do understand that the units of the rate constant k is selected to make the equation dimensionally consistent. That is not what I am asking.
It should be the case that k 'means' something physically, which is implied by its dimensions/units. And I would also assume that what k 'means' should not be dependent on the order of the reaction, since I cannot see any obvious reason why it should. But we know that the overall order of a reaction changes the units of the rate constant, so I must be wrong.
My question is as stated: Why do the dimensions of the rate constant change depending on different reaction orders, and what does that physically mean?
As an extension to this, I would also like to understand how the same thing works for the equilibrium constant, where the stoichiometric coefficients of the species or the number of products/reactants change the units of the constant. Again, I cannot see the reason why these would change what the constant means dimensionally.
Sorry if this an obvious question. I am not a chemist.
Edit:
I wanted to clear up a few things.
First of all, as I understand I was expecting there to be a deeper meaning to the dimensions of the rate constant even though there isn't always an obvious one, since it is an empirical equation which merely simplifies a process rather than describing the fundamental workings of it. But I still want to understand intuitively (if there is indeed an intuitive reason) why the units are different for each overall order. As in the examples Maurice gives in his answer, a reaction of order zero produces dimensions of amount per time whereas a first order reaction produces [ratio] per time and so on for other real number orders. It just seemed strange to me that a constant in the equation, which seems to be doing the same job regardless of order, express something in completely different dimensions depending on order.
After reading your answers, this is what I thought: In a zeroth order reaction where the rate is constant, amount per time seems to be the only reasonable way to describe the rate since ratio per time inevitably produces a curve. Similarly for a first order reaction, amount per time wouldn't work because this value changes constantly. So, the constant is expressed in terms of the units which can describe the reaction given its nature. This was stated/suggested in some replies but thinking of it over an example is what cleared it up (if it is correct) and thank you for your responses. Perhaps this was obvious but a constant with varying units was not something I was familiar with.

Maybe it would make more sense if rate constants would have different symbols depending on whether it is a zero, first or second order reaction.
As an analogy, we might be asking "how big" an object is depending on the length of it. Depending whether the object is, say, a square, a cube or a hypercube, the answer would be in terms of its area, volume or 4D-volume, and the units (and dimensions) would be different in each case.
For the equilibrium constant, we avoid the issue by switching from concentrations to either concentrations divided by a standard concentration (this is a simplification) or to activities (the former multiplied by an activity coefficient that accounts for non-ideality).

Why do the dimensions of the rate constant change depending on different reaction orders, and what does that physically mean?

They have to change with the chosen system of units so that the equations are consistent. We could choose instead to have a dimensionless measure of concentration when we set up the kinetic equations. This would describe the reality equally well, but the kinetic constants would always have the same dimensions (1/time).
According to this argument, there is no physical meaning in rate constants having distinct dimensions depending on the order of reaction.


Can we determine the presence of carbon by simply heating or charring the substance? Char is made of Carbon, but just because a substance char, does it immediately mean it contains carbon?

Also,
Why is charring not a good test for detecting carbon in organic compounds?
I searched that because the test can't differentiate an elemental carbon from an organic carbon. Can someone elaborate or give an example?

Not all carbon-containing chemical char. For example, sodium carbonate, $\ce{Na2CO3}$ just produces $\ce{CO2}$. Sodium carbonate might be viewed as inorganic, though.

An organic chemical that behaves in a similar way is hexamethylenetetramine, or hexamine, $\ce{(CH2)6N4}$. On heating, it decomposes primarily into formaldehyde, $\ce{H-CHO}$, and ammonia, $\ce{NH3}$, though there are traces of other gases such as $\ce{HCN}$ and $\ce{CO}$. For that reason, hexamine is used as a smokeless fuel (no char left behind, as the saying goes).


What is the phase transformation temperature anatase to rutile phase in tin dioxide(SnO2)? Usually people discuss only pressure dependent phase transformations in case of either tin dioxide or titanium dioxide. It would be nice if someone gave the phase diagram for tin dioxide.

Anatase and rutile phases form in $\ce{TiO2},$ the presence of cationic dopants or impurities significantly affect the transformation kinetics and phase stability [1].
It should be noted that anatase is metastable at all temperatures and pressures.

References

Hanaor, D. A. H.; Sorrell, C. C. Review of the Anatase to Rutile Phase Transformation. J Mater Sci 2011, 46 (4), 855–874. https://doi.org/10/ct79p8.


How common are automated pipetting systems It's clear that automated pipetting is faster than doing it manually, but how common is it for a lab to have an automated pipetting system, and what usually are the circumstances that a lab have before they purchase such a system?

Speed isn't everything. Reproducibility might be much more important!

Imagine that you have to transfer different volumes from five different stock solutions to a 1536 well plate and adjust the total volume by adding water or the common solvent of your choice. A programmable pipetting automat might be a blessing here.


Can an ion isotope exist? Can an isotope have a charge, i.e. the number of neutrons and electrons differ in that atom?
If it does exists, is it correct to call it an ion isotope or isotope ion?

Short answer: Yes.
Longer answer:
Isotopes are atoms of the same element that have different numbers of neutrons in their nuclei. Everything else about them is the same. The numbers of protons and electrons in a neutral atom is governed by the atomic number.
So, for example, chlorine has two naturally occurring isotopes:
$$\begin{array}{|c|c|c|c|c|c|c|}\hline
\mathrm{isotope} & Z & A & p & n & e & \mathrm{abundance}\\ \hline
\ce{^35_17Cl} & 17 & 35 & 17 & 18 & 17 & 75.78\%\\
\ce{^37_17Cl} & 17 & 37 & 17 & 20 & 17 & 24.22\%\\ \hline
\end{array}$$
Where:
$$\begin{aligned}
&Z=\mathrm{atomic\ number}\\
&A = \mathrm{mass\ number}\\
&p = \mathrm{number\ of\ protons}\\
&n = \mathrm{number\ of\ neutrons}\\
&e = \mathrm{number\ of\ electrons}\\
\end{aligned}$$
Isotopes generally behave the same chemically1 since almost all chemistry is governed by the number of electrons ($Z$). Thus, anything that contains chloride ions contains a roughly 3:1 ratio of $\ce{^35Cl-}$ and $\ce{^37Cl-}$. We could call these ion isotopes or isotope ions, but in most all cases, we do not distinguish between the two in reactions or compounds involving chloride.
1 There is one caveat in the "isotopes behave the same chemically" rule of thumb that is worth mentioning. Since isotopes have different masses, they can produce kinetic isotope effects. Given the $KE=\frac{1}{2}mv^2$, two isotopes with different masses will move at different velocities and have the same kinetic energy. Heavier isotopes generally move more slowly.
This effect is most pronounced in hydrogen, where the second most common isotope $\ce{^2H}$ ($\ce{D}$) is twice as massive at the most common isotope $\ce{^1H}$ ($\ce{H}$). If two atoms have the same kinetic energy, then:
$$\begin{aligned}
m_1 v_1^2 &=m_2 v_2^2\\
\dfrac{m_1}{m_2} &=\dfrac{v_2^2}{v_1^2}\\
\sqrt{\dfrac{m_1}{m_2}} &=\dfrac{v_2}{v_1}
\end{aligned}$$
For the isotopes of hydrogen, the heavier isotope of hydrogen moves at only 70% of the velocity of the lighter isotope. This kinetic isotope effect can be used to probe reaction mechanisms. If the rate decreases dramatically when $\ce{H}$ is swapped for $\ce{D}$, then that hydrogen atom is involved in the rate determining step.
$$\begin{array}{|c|c|c|c|}\hline
\mathrm{isotope} & A & \mathrm{rel.\ mass} & \mathrm{rel.\ velocity}\\ \hline
\ce{^1H} & 1 & 1 & 1 \\
\ce{^2H} & 2 & 2 & \dfrac{1}{\sqrt{2}}\approx 0.707\\ \hline
\end{array}$$
For chlorine, the outcome is a little different. Since the ratio of the masses of the two isotopes is close to 1, the ratio of the velocities is also close to 1.
$$\begin{array}{|c|c|c|c|}\hline
\mathrm{isotope} & A & \mathrm{rel.\ mass} & \mathrm{rel.\ velocity}\\ \hline
\ce{^35Cl} & 35 & 1 & 1 \\
\ce{^37Cl} & 37 & 1.06 & 0.97\\ \hline
\end{array}$$


ABC arrangement of FCC/CCP I know that FCC shows ABC type but I am confused about it. For simple cubic and body centered, it is pretty easy to work out the packing arrangement but FCC is giving me a headache.

What is this Wikipedia article trying to tell in the diagram and how did choose A, B and C? What's up with the different orientations?

Can somebody tell me these in a simple way as I can't solve FCC problems based on structure.

Perhaps the simple way to look at this is think of the sphere structures as three layers. Think of the (x,y) coordinates of the spheres, neglecting the z coordinate. The first (white balls) and second layers (black balls) are the same in HCP and FCC. So let's label the first layer the A layer and the second layer the B layer.

Now we add the third layer.

In HCP, which is shown on the left in the above image, the center of spheres in the third layer would be placed directly over the center of a sphere in the A (first) layer. So the third layer (white ball)is another A layer.
In FCC, which is shown on the right in the above image, the spheres in the third layer (green ball) would be offset by 1/2 the radius of the sphere, above a sphere in the first layer. (Note that the dashed circle denotes a position directly above the A level circle.) So the third layer is neither an A layer nor a B layer. Let's call it a C layer. Now the fourth layer would again have spheres whose center are directly above the center of the spheres in the first layer. So it would be another A layer.

So HCP has (AB)(AB)(AB)... layers but FCC has (ABC)(ABC)(ABC)... layers. Where A, B and C denote different sets of (x,y) coordinates. Again the z coordinate being the layer of the ball coming out of the 2D image.


How does repeatedly freezing and thawing antibiotics decrease their antimicrobial effectiveness? This is something I have read on forums repeatedly, but I don't understand the mechanism behind how antibiotics lose their effectiveness. I understand how repeatedly thawing and freezing cells would damage them, but not how repeatedly thawing and freezing antibiotics make them lose effectiveness.

As indicated in the comments, freeze-thaw cycles should not affect molecules of size of typical antibiotics. Freeeze-thawing may affect colloids by disrupting their structure. Also, the effect would depend on how fast a sample was frozen/thawed etc (slow freezing would cause rearrangement of molecules and crystallization of water whereas rapid freezing would result in vitrification of water.)

The popular lab practice, in my opinion, would have perhaps arisen because it is better to keep the stock stored at the appropriate temperature (frozen) for longer life. Repeated thawing of the stock would definitely increase its chance of degradation because of mishandling and even otherwise. Therefore, it is better to aliquot smaller volumes; even if you have forgotten to keep it back into the freezer you'll end up losing only a small amount.

Not related to your question but this is an interesting fact: freeze-thaw cycles promote molecular assembly of RNA into higher order structures (ref).


Neutralizing sulfur-like smells in car A full bottle of jewelry cleaning (links product and material safety data sheet below) broke in my car and went unnoticed for a few days.  Now the car has a sulfur-like smell whenever I go into the car.  It dissipates within a few minutes, but then returns very quickly once I leave the car.

The primary chemical components are alcohol ethoxylates, and can decompose into carbon monoxide, carbon dioxide nitrogen Oxides, sulfur oxides and other toxic gases.

I have cleaned the area and had the carpet (rear driver) shampooed from a car detailer, but still the smell remains.

What chemical compound can I use to neutralize the smell permanently (enter equilibrium)?

Product Link (Amazon)
Product Link (Manufacturer)
Material data safety sheet

I guess that it actually was a silver cleaner, not one for gold or precious stones.

The latter mostly contain nonionic surfactants, which are made by the reaction of water-insoluble long chain alkanols (C12 - C15, from natural fatty acids) with oxirane. These washing agents dissolve well in water and help to remove grease and fingerprints from the jewelery. Their action isn't much different from that of a dishwasher for manual cleaning.

These cleaners don't smell the way you described and the car retailer probably used similar cleaners for the car.

Silver cleaners are a different league. The do contain the same surfactants (washing agents), but in order to actually remove the tarnish (silver sulfide) from the silverware, they do also contain a bit of acid (typically sulfuric acid) and thiourea, which probably is responsible for the smell.

Repeated shampoeing probably is the best method to get rid of the thiourea without ruining the carpet.


What are the limits of analytical chemistry? As I understand it, the branch of chemistry dealing with identifying substances present in a sample is called "analytical chemistry". This branch looks vast and complicated, with lots and lots of individual techniques applicable in specific circumstances. I'm hardly even a beginner in chemistry, so it's very difficult to get a good grasp of what can be done with current technology.

So, in general, how well can one identify the substances in a completely unknown sample? Does one need to have at least a vague idea of what's in it?

Differently put, let's say I give you a container with a mixture of ten randomly chosen chemicals from the entire CAS Registry, and suppose they didn't react with each other. How difficult would it be to identify all ten of them?

The trouble with your question is the use of the phrase in general. Most analytical chemistry is directed towards specific goals and isn't about trying to find what is in a sample in general.

This means that the range of techniques applied are highly dependent on the problem you want to solve. If you want to detect, for example, pesticide residues in foodstuffs you probably already know the list of things you want to detect and the range of techniques appropriate to identifying them. If you seek the active ingredient of a herbal remedy, a different range of techniques apply. Sometimes you want to verify that a pure substance you have made is what you think it is, this involves a different range of techniques again.

One common thread in much analytical chemistry is separation: if you seek to identify something in a mixture you want to separate the mixture into its different components. So a variety of types of chromatography may be used in the bulk or on a micro scale. Identifying very small amounts of contaminants in food (e.g. pesticides) probably involves some form of extraction into a solvent followed by separation using analytical chromatography (gas or high pressure liquid chromatography (HPLC)). Where the range of compounds is not known in advance this can be supplemented by mass spectroscopy (the chromatography separates the components, the mass spectrograph helps identify the fragments of the molecules in each component).

If you don't know what you are looking for in advance, it isn't easy to elect the right separation technique and a great deal of experimentation is likely to be required.

Another consideration is that some things are much easier to detect than others. Chromatographic and detection tools like the electron capture detector can identify organohalides to below parts per trillion levels. Atomic emission spectroscopy can identify some metals easily and at low levels. Other techniques can help identify structure; NMR is (sometimes) good at identifying structures, but usually needs pure compounds in relatively large quantities. X-ray techniques can be useful for minerals. Some compounds have very distinct features in their IR spectra or their UV absorption. And so on.

The bottom line is there are no general techniques. There are a wide range of tools to apply depending on the circumstances and the problem you are trying to solve. Almost everything starts with separation. Then, for simple problems, you may be able to apply a single known technique to find out what you want. For harder problems, a rage of tools may be required but the tools will be fairly specific to the problem.


What is "heating under reflux"? When dealing with organic reactions, heating under reflux is often required, such as the oxidation of Toluene using acidified $\ce{KMnO4}$ and dilute $\ce{H2SO4}$ to Benzoic acid($\ce{C6H6O2})$

What does this mean and how does it differ from "normal" heating?

Many organic reactions are unreasonably slow and can take an extended period of time to achieve any noticeable effect so heating is often used to increase the rate of reaction. However, many organic compounds have low boiling points and will vaporise upon exposure to such high heat, preventing the reaction from proceeding in full.

To address this, heating under reflux is used. This refers to heating a solution with an attached condenser to prevent reagents from escaping.

As seen above, any vapor will condense on the cool surface of the attached condenser and flow back into the flask.

The hot water bath pictured is an optional component of heating under reflux and is usually only used for particularly sensitive reactions. Also, using it limits the reaction temperature to 100 degrees Celsius.


How do I decide the electromeric effect on pent-2-ene? There are two possible case
π electrons shifting to C2
In this case, nucleophile attacks C3
It is stabilized by +I effect of two ethyl groups
π electrons shifting to C3
In this case, nucleophile attacks C2
It is stabilized by +I effect of a methyl group and a propyl group.

How do I decide which one is more stable, and ultimately, which one is favoured?

Here what you should consider first is hyperconjugation, rather than inductive effect (a brief explanation by me).

You're right that $\ce{Nu-}$ will either attack on $\ce{C-3}$ or $\ce{C-2}$, and hence two carbocations would be formed. Now, we'll analyse the stability of these carbocations on the basis of hyperconjugation which is as follows,

carbocation at $\ce{C-2}$ will have a total of 5 HC structures
carbocation at $\ce{C-3}$ will have a total of 4 HC structures

Based on the above difference, it is easy to conclude that $\ce{Nu-}$ will attack on $\ce{C-2}$ carbon. Whereas based on inductive effect, those carbocations are nearly equally stable.

Note: HC is short for "hyperconjugative structures".


Internal energy change for reactions at constant volume vs constant pressure The reaction of cyanamide, $\ce{NH2CN(s)},$ with dioxygen was carried
  out in a bomb calorimeter, and $∆U$ was found to be $\pu{–742.7 kJ mol-1}$
  at $\pu{298 K}.$ Calculate enthalpy change for the reaction at $\pu{298 K}.$

$$\ce{NH2CN(s) + 3/2 O2(g) → N2(g) + CO2(g) + H2O(l)}$$

$ΔH$ could be found from

$$ΔH = ΔU + ΔnRT$$

I want to know whether $ΔU$ found for constant volume conditions (bomb calorimeter) can be used in constant pressure conditions (while calculating $ΔH$ pressure is constant). In other words, is $ΔU$ same for both constant pressure and constant volume conditions, and why?

I searched the entire net and several standard physical chemistry texts but couldn't find any explanation.

The key point to keep in mind in this type of problem is that the energy of an ideal gas is only a function of temperature, not of volume or pressure$^\ast$. We also assume that the dependence of the energy of the condensed phases on p and V is negligible. Therefore the final pressure or volume is not going to affect $\Delta U$ for the reaction, provided n and T are constant. Since here the reaction refers to a conversion of a stoichiometric amount of reactants into products at the specified T, we therefore do not expect much of a change in $\Delta U$ with change in p or V.

$^\ast$Note that the total differential for U can be written as
$$\begin{align} dU &= \left\{ T \left( \frac{\partial p}{\partial T} \right)_V -p\right\} dV + C_VdT \end{align}$$ Therefore at constant T,
$$\begin{align} dU = &= \left\{ T \left( \frac{\partial p}{\partial T} \right)_V -p\right\} dV  \end{align}$$
The term in parentheses is zero for an ideal gas.


Why are limes alkaline, but lemons are acidic? They taste about the same So, this is kinda a dumb question, but I'm curious. Both are citrus fruits and very related to each other and taste sour which corresponds to both being acidic. Why then are limes said to be alkaline, but lemons are acidic?

The word "lime" has two meanings and they have essentially no relationship between them.

The "lime” that is inorganic oxide quicklime, CaO, reacts with water to give a relatively strong-base hydroxide::
$$ \ce{ CaO + H2O -> Ca(OH)2 }$$
$$ \ce{ Ca(OH)2 <=> Ca^2+ + 2OH-}$$
Due to the presence of these hydroxyl ions, the solution is alkaline.

"Limes" that are the green, hybrid citrus fruits are acidic with citric acid, just like lemons.  Both have juice with a pH in the range of 2-3, for limes as in fruit a typical pH level is about 2.8.  Hence the tart flavor for which both fruits are known.  Like all citrus fruits, lemons and limes are a rich source of ascorbic acid, vitamin C.


Carbocation Rearrangement in SNi In Peter Sykes' A Guidebook to Mechanism in Organic Chemistry  regarding the retention of configuration of reaction of alcohols with $\ce{SOCl2}$:

  The rate at which the alkyl chlorosulphite intermediate breaks down to $\ce{RCl}$ is found to increase with increasing polarity of the solvent and also with the increasing stability of the carbocation $\ce{R+}$: an ion pair $\ce{R+SOClO-}$ is almost certainly involved.

But it is nowhere mentioned whether a carbocation rearrangement is possible or not.

In "NCERT Chemsitry Class 12" in the section of Preparation of Alkyl halides from Alcohols:

  The hydroxyl group of an alcohol is replace by halogen on reaction with concentrated halogen acids, phosphorous halides or thionyl chloride. Thionyl chloride is preferred because the other two products are escapable gases.

If that method is preferred, then I assume rearrangement does not occur.

So, why does not a carbocation rearrangement (like Wagner–Meerwein rearrangement) occur in the $\mathrm{S_Ni}$ pathway eventhough the intermediate is involved? Is it because the intermediate breaks down too quickly before any rearrangement is possible?

The Mechanism of the reaction
The reaction of an alcohol with thionyl chloride under base-free conditions is one of the most significant examples of the SNi mechanism (DN+ANDe to give it it’s correct IUPAC designation, which is perhaps more instructive in this case).
There are two steps to the SNi mechanism, as shown below (figure taken from Bruckner, Organic Mechanisms [1]):

Step 1: The alcohol reacts with thionyl chloride to afford an alkyl chlorosulfite, which can often be isolated.[2] At this stage, the oxymethine stereocentre hasn’t been touched– it’s the second step that will determine the stereochemistry.
Step 2: The chlorine is delivered intra-molecularly, with extrusion of sulfur dioxide gas. Since the sulfur is being delivered from the same face that the hydroxyl was on, the reaction occurs with retention overall.  This can be seen in the dotted lines in the figure above.
Two things are worth noting at this stage:

If pyridine is added to the reaction mixture, we see overall inversion via an SN2 mechanism.[2] This occurs because the pyridine interacts with the intermediate alkyl chlorosulfite formed, liberating a free chloride (nucleophilic) and hence allowing the substitution to take place  inter-molecularly. This can be seen in the solid lines in the figure above.

The second step in which I described as happening intra-molecularly doesn’t quite happen intra-molecularly. In reality, the C-O bond begins to fragment in an SN1 fashion, before the chloride attacks. The reason why we still see retention (rather than racemisation as is common with  SN1) is due to a phenomenon called contact ion pairs.[#]

Carbocation rearrangement
The simple answer to your question is that at no stage of the reaction is it thought that there is any free carbocation concentration, making rearrangement less likely on kinetic grounds (along with elimination, which could equally be argued by your logic).
The ion pair theory is such that the carbocation remains closely associated with the leaving group, allowing the nucleophilic attack to occur rapidly without the need for an additional collision to take place.
[1]: Bruckner, R. Organic Mechanisms- Reactions, Stereochemistry and Synthesis; Springer: Berlin, 2007
[2]: J. Am. Chem. Soc. 1952, 74, 308
[#]: You can read about this in any advanced organic chemistry text such as Carey or March, and is a bit too long of an explanation for me to give satisfactorily here


What are the magnetic quantum numbers for the three real p orbitals? The form of the p orbitals that we are familiar with are the $\mathrm{p}_x$, $\mathrm{p}_y$, and $\mathrm{p}_z$ orbitals:

(source: ChemTube 3D)

I also know that the p subshells have the quantum number $l = 1$, meaning that the magnetic quantum number can take the three values $m_l = -1, 0, +1$. Which orbital has which value of $m_l$?

The answer is... it is not so simple. Some quantum mechanics follow, but the TL;DR version is that while $m_l=0$ corresponds to $p_z$, the orbitals for $m_l=+1$ and $m_l=-1$ lie in the $xy$-plane, but not on the axes. The reason for this outcome is that the wavefunctions are usually formulated in spherical coordinates to make the maths easier, but graphs in the Cartesian coordinates make more intuitive sense for humans. The $p_x$ and $p_y$ orbitals are constructed via a linear combination approach from radial and angular wavefunctions and converted into $xyz$. Thus, it is not possible to directly correlate the values of $m_l=\pm1$ with specific orbitals. The notion that we can do so is sometimes presented in introductory courses to make a complex mathematical model just a little bit simpler and more intuitive.

From Physical Chemistry by Atkins and DePaula, the three wavefunctions for $n=2$ and $l=1$ are as follows.

$$\begin{align}
&\Psi_{2,1,0}&&=r\cos{\theta}f(r)\\
&\Psi_{2,1,+1}&&=-\dfrac{r}{\sqrt{2}}\sin{\theta}\mathrm{e}^{\mathrm{i}\phi}f(r)\\
&\Psi_{2,1,-1}&&=\dfrac{r}{\sqrt{2}}\sin{\theta}\mathrm{e}^{-\mathrm{i}\phi}f(r)\\
&\end{align}$$

The notation is $\Psi_{n,l,m_l}$, $r$ is the radius, $\theta$ is the angle with respect to the $z$-axis and $\phi$ is the angle with respect to the $xz$-plane.
$$f(r)=\sqrt{\dfrac{Z^5}{32\pi a_0^5}}\mathrm{e}^{-Zr/2a_0}$$

in which $Z$ is the atomic number (or probably better nuclear charge) and $a_0$ is the Bohr radius.

In switching from spherical to Cartesian coordinates, we make the substitution $z=r\cos{\theta}$, so:
$$\Psi_{2,1,0}=zf(r)$$

This is $\Psi_{2p_z}$ since the value of $\Psi$ is dependent on $z$: when $z=0;\ \Psi=0$, which is expected since $z=0$ describes the $xy$-plane.

The other two wavefunctions are unhelpfully degenerate in the $xy$-plane. An equivalent statement is that these two orbitals do not lie on the $x$- and $y$-axes, but rather bisect them. Thus it is typical to take linear combinations of them to make the equation look prettier. Linear combinations are allowed by the maths of quantum mechanics. If any set of wavefunctions is a solution to the Schrödinger equation, then any set of linear combinations of these wavefunctions must also be a solution. We can do this because orbitals and the wavefunctions that describe them are not real physical objects. They constitute a mathematical model.

In the equations below, we're going to make use of some trigonometry, notably Euler's formula:

$$\mathrm{e}^{\mathrm{i}\phi}=\cos{\phi}+\mathrm{i}\sin{\phi}$$
$$\sin{\phi} = \frac{\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi}}{2\mathrm{i}}$$
$$\cos{\phi} = \frac{\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi}}{2}$$

We're also going to use $x=\sin{\theta}\cos{\phi}$ and $y=\sin{\theta}\sin{\phi}$.

$$\begin{align}
\Psi_{2p_x}=\frac{1}{\sqrt{2}}\left(\Psi_{2,1,+1}-\Psi_{2,1,-1}\right)=\frac{1}{2}\left(\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)=r\sin{\theta}\cos{\phi}f(r)=xf(r) \\
\Psi_{2p_y}=\frac{\mathrm{i}}{\sqrt{2}}\left(\Psi_{2,1,+1}+\Psi_{2,1,-1}\right)=\frac{1}{2\mathrm{i}}\left(\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)=r\sin{\theta}\sin{\phi}f(r)=yf(r)\\
\end{align}$$

  So, while $m_l=0$ corresponds to $\Psi_{p_z}$, $m_l=+1$ and $m_l=-1$ cannot be directly assigned to $\Psi_{p_x}$ and $\Psi_{p_y}$. Rather $m_l=\pm1$ corresponds to $\{\Psi_{p_x},\Psi_{p_y} \}$. Put another way, I suppose we could say that $m_l=+1$ might correspond to $\Psi_{p_{x+y}}$ and $m_l=-1$ might correspond to $\Psi_{p_{x-y}}$.


Gauche effect and bond length Which of the following compounds has the longest bond length?
(A) $\ce{H3C-CF3}$
(B) $\ce{FCH2-CH2F}$
(C) $\ce{F2CH-CHF2}$
(D) $\ce{F3C-CF3}$

My logic was based on the relative electronegativities of the two carbons. The more fluorines there are on one carbon, the more polarised the C-C bond will be, and the longer the C-C bond length will be (since the bond is weakened). Thus, I think that the answer is A.
Is this right?

Actually, the more the polarization of a bond, the stronger it is, since both the atoms are drawn close together due to electrostatic attraction, decreasing the bond length and increasing the bond stability (think about ionic bonds. Your reasoning would dictate them to be very weak, while we know how strong they are!). Actually, I would predict $\ce{CH3-CF3}$ to have a really small bond length (the bond is probably the strongest, would someone confirm this please?), since the fluorinated carbon has quite a positive charge, while the other one is relatively negatively charged (not negative in the absolute, just relative to the other carbon).

All the remaining options have equal number of fluorines on either carbon, differing only in their number. These three are tricky. Consider $\ce{CF3-CF3}$. Both of its carbons are severely deficient in electrons. Each badly needs electrons, and since it can't argue with fluorine (fluorine is an unreasonable bully), so both carbons try to drag the shared pair of electrons towards itself. This ultimately results in them both coming closer to each other. As we reduce the number of fluorines on each carbon, their desparation is reduced, thus increasing the bond length. Thus the answer is (B).

Edit: If @pentavalentcarbon's calculations of bond lengths are correct, then my reasoning was too simplistic, and indeed it seems so now. So if we dig deeper, $\ce{CH2F-CH2F}$ could have the second smallest bond length and $\ce{CF3-CF3}$. Why? Well, it is because of hyperconjugation, similar to the gauche effect (read it up to get better understanding of the gauche effect, although I explain it in short here also).

So, what happens is that in the case of $\ce{CH2F-CH2F}$, the bonding C-H molecular orbital (MO) overlaps with the antibonding C-F MO, transferring some of its electron density to the C-F MO (see diagram). This interaction is stabilizing (since delocalization of electrons stabilizes the molecule), and also introduces slight double bond character in the C-C sigma bond, thus serving to shorten it.

Now in $\ce{CH2F-CH2F}$, there will be two such interactions, but then there are also two such interactions in $\ce{CHF2-CHF2}$! There is a difference though. In the latter case, there are two fluorines on each carbon, which heavily draw electron density away from the carbon through the sigma bonds. This decreases the sigma bond density in the C-H bond, which being poor in electrons, gives relatively low stabilization when overlapping with the antibonding C-F MO. So $\ce{CHF2-CHF2}$ has a slightly higher bond length than $\ce{CH2F-CH2F}$.

Using this same reasoning, we can see why $\ce{CH3-CF3}$ has the smallest bond length (three favourable hyperconjugational interactions!) and $\ce{CF3-CF3}$ has the longest bond length (no hyperconjugation).


How does the entropic force arise within the hydrophobic effect? I know that the hydrophobic effect is usually explained by the entropic effect originating from the disruption of hydrogen bonds between molecules of water and the nonpolar substance. The hydrogen bonds are reoriented to minimize the disruption of the hydrogen bond network of water molecules.
What I don't understand here is how this explains the tendency of non-polar substances to aggregate in water? Supposedly they disrupt the hydrogen bond network, but what forces them to clump together?
Then how does the above explain the tendency of water to decrease its surface area toward the nonpolar substance? Yes, the molecules of water around the nonpolar surface have lower entropy, but how does a decrease in entropy (in sense of a higher structural order) force them to minimize their exposure towards the nonpolar substance?

Whenever a non-polar substance is put into water, water molecules will organise themselves around it in a cage-like formation. The molecules exposed to the non-polar substance at any given time will orient themselves to form as many hydrogen bonds with the rest of the solution as possible.
(Image from here)

The way I understand it intuitively is that these water molecules are restricted in the orientations they can assume, becoming more ordered to maximise hydrogen bonding, so have an overall lower entropy than if they were free to tumble like in the bulk solution. Similarly, the aggregation of non-polar molecules also allows more free movement within the non-polar 'blob'. If they were distributed evenly, there would be a higher surface area and therefore amount of 'cage' water molecules and non-polar solutes would be more restricted in their movement (trapped in a polar environment). This is the entropic factor that drives the hydrophobic effect.
It is also favourable enthalpically; hydrogen-bonding interactions and non-polar interactions are maximised respectively.
These two approaches demonstrate why it is thermodynamically favourable for non-polar substances to aggregate, not disperse. To clarify, it is still thermodynamically unfavourable to mix the two in the first place, that's why they separate; this is just the better option if they were to mix.
It is also worth noting that oil 'blobs' in water tend to be very round, due to the fact that a sphere is the shape with the smallest surface area to volume ratio. This supports the above reasoning, minimising the number of 'cage' water molecules whilst maximising movement of both water and the hydrophobe.


Fourier transform for spectroscopy spectra? I was thinking back to my chemistry major days and remembering the bunch of cool spectra we produced by various means (HNMR, IR, UV/Vis) and I remember all the cool 'self-interference' like couplets, triplets doublets, etc.

My question is basically motivated by the complexity of spectra, and the reductionist instinct to 'break it down' into simpler components. The FFT or DFT is known to separate out some periodic components of signals, as are SVD or tensor decompositions known to isolate independent components.

I am completely not an expert in any of this but I am interested to hear from any coal-face chemists if they regularly use some 'spectra decomposition' method to aid in interpretation or analysis. I remember being taught how to read various peaks in spectra and analyze them, but I can't help thinking, for some very complex molecules, surely it gets pretty hard. And what if, we don't actually know the structure beforehand?

So given that Fourier transform is used in NMR, my question is : is there ever a need to take a spectrum and do a computer analysis of it that isn't just about matching peaks to a database of known signals? What other nifty things can be done with a spectrum once it is obtained?

As Mad Scientist said, many spectroscopy methods are already based on Fourier transform of a time-dependent observable. (This is not true of the basic spectrophotometry measurements, where you directly measure the transmittance as a function of frequency.) NMR is a method based on Fourier transform, as is FTIR.

Regarding NMR signal processing, it's a very wide field, which includes interesting research ranging from academic interest to software development. So, it's hard to answer your question other than saying “yes, of course, further processing of NMR signals is sometime possible”. One thing you have to realize is: while you have probably learnt proton NMR during your studies, this is only one NMR technique, and the simplest of them all. There are plenty of different NMR techniques, depending on the type of excitation and sequence of RF pulses, many of them being multidimensional. Each method can use a variety of signal processing techniques.


Why do higher orbitals have more energy? I have seen in textbooks and videos that an electron must absorb energy (become excited) to enter a farther-away orbital.

The amount of energy that must be gained is equal to the difference in energy between the orbitals. When the electron relaxes and returns to its ground state, it emits that energy.

However, I am not sure I understand why "higher" orbitals are of greater energy, for a couple of reasons.

When I think of the potential energy of the electron, as given by $U = \frac{kQq}{r}$, it seems that $U \propto \frac{1}{r}$, which would indicate less energy as distance increases.
I also know that the electron is attracted to the nucleus due to opposing charge. By this logic, an electron closer to the nucleus would be more stable (have a lower energy state) than an electron farther from the nucleus (which would thus have a higher energy state). This line of thinking is somewhat similar to how a ball that's still high in the air is in a higher energy state than a ball that's closer to the ground.

So #1 is inconsistent with what we are taught about electrons and orbitals, while #2 would support it.

I am attempting to clarify the misunderstanding I must have somewhere (possibly not making a proper analogy) - and better understand why higher orbitals are of greater energy.

potential energy is negative between infinity to nucleus

It varies with .Thus, U∝(−1)1/r so it means the father you go from nucleus the value become less negative that is higher in energy than equilibrium energy.


What is the mechanism by which sugars glycate proteins? I was looking at why it's ok for glucose to circulate through the blood while fructose is prevented from doing so by the liver.  Fructose glycates proteins more readily than glucose.  This is the opposite of what I expected as glucose is an aldose while fructose is a ketose.  It seems reasonable to expect that the aldose would be more reactive than the ketose in many reactions.  What is the mechanism of glycation of proteins?

One mechanism of glycation is the Maillard Reaction.

The mechanism starts with nucleophilic attack of the sugar carbonyl group by a side chain amino group from the protein.

The Milliard Reaction occurs in the human body and has been implicated in various adverse health conditions.

The cornea, lens and retina are known to be damaged by this reaction.

See The Maillard reaction in the human body. The main discoveries and factors that affect glycation Pathologie Biologie, Volume 58, Issue 3, Pages 214-219 for a recent review.


What is the difference between a desiccant and a deliquescent? I understand that both a desiccating solid and a deliquescent solid are "hygroscopic" in the sense that they bind to water.

But what is the difference? And what are the applications of a desiccant vs a deliquescent?

A deliquescent solid will eventually dissolve in water it absorbed, like, for example, calcium chloride does. Molecular sieves are very good dessicants but they do not dissolve in water, so they are not deliquescent. As you can probably guess, operating with dessicants that are deliquescent may be more troublesome than with non-deliquscent ones, but it doesn't mean that they are inferior.

The choice of a proper drying agent depends on the material you want to dry. For example, non-deliquescent dessicants, like aluminium oxide, silicagel or molecular sieves are best to fill columns for solvent drying, because they won't clog when their drying capacity is exceeded. Deliquescent drying agents are perfect for the use in dessicators, because it is easy to remove them when they are exhausted.


Taiwan propylene explosions one week ago Is it wrong way to use WATER to relieve the big disaster(propylene leak).

What is a better way to relieve propylene leak in a street about large area?

Any possible choice to prevent explosion?

Propene gas is flammable and has a higher density ($1.81 \ \mathrm{kg/m^3}$) than air. It therefore easily forms explosive mixtures with air when it accumulates near ground level or, like in this case, inside an underground channel without much ventilation. While propene itself is color- and odourless, it is described in this article that "eyewitnesses reported a smell of gas (...) coming out of manholes". This smell comes from volatile compounds which are added to the gas in small amounts to make the detection of leaking gas easier. However, by the time that the smell was noticed, enough propene had already accumulated underground to exceed explosion limits.

Detection of underground gas leaks by odour alone is therefore not a reliable means to prevent possible explosions, namely when the gas is heavier than air. It is also not possible to exactly localize the leak when the gas escapes from multiple manholes which are spread over a large area. Gas sensors in the underground pipeline network could provide earlier warnings about leaks, and gas accumulation could be prevented in this case by shutting down the affected pipeline. However, prevention of leaks in the first place (by regular inspections) should still have higher priority.

The MSDS data sheet for propene states that shutting down gas supply and letting the fire burn out while cooling the surroundings with water is the recommended way of extinguishing burning propene. So using water to cool down the roads with the aim to reduce ground temperature was not completely off, but trying to fight the fire directly with water is unadvisable.


Consumer choice; harmful environmental aspects of polyethylene vs polypropylene What is the difference between polyethylene and polypropylene in terms of their aspects which are harmful to the environment during manufacture and disposal.

Every material is terrible if not properly collected and processed after use.

Wether you re-use it (like glass bottles), or recycle the material, or burn it to make electricity and district heating, is small potatoes.

What material you use is also not very important, only the actual energy and environmental costs of "bio"materials like paper or cotton are very high. You have to re-use them very often before they make sense!


Can butane-2,2,3,3-tetraol be oxidized by potassium permanganate? Can the compound butane-2,2,3,3-tetraol, $\ce{C4H10O4}$, be oxidized by $\ce{KMnO4}$? If so, in what conditions? As I have learnt in class, normally the hydroxy group would oxidize to form a ketone, but what about this particular case?

You are talking about the dihydrate of biacetyl.  gem-diols are generally not stable, the equilibrium is on the side of the aldehyde/ketone, except when the keto group is next to an electronegative atom.  Glyoxal in aqueous solution is all hydrate and oligomers, as is chloral.

Biacetyl is more hindered than glyoxal, but in aqueous solution one would think they there is still some hydrate in the equilibrium.  What you'd think would happen is that on reaction with permanganate you'd get the cyclic diol and under sufficiently forcing conditions (read: acidic medium) you'd get cleavage to acetic acid.

Biacetyl is awful.  Run away!


Sodium thiosulfate and Hydrochloric acid If I was using this method (http://www.rsc.org/learn-chemistry/wiki/Expt:The_effect_of_concentration_on_reaction_rate) however used pre-prepared concentrations of sodium thiosulfate solution, which intervals would be more suitable, 0.1M, 0.2M, 0.3M, 0.4M or 0.5M, 1M, 1.5M and 2M? Also, what concentration of Hydrochloric acid should be used (I've seen a variety of suggestions elsewhere). This is for a secondary school practical. Thank you.

There's a nice demonstration of the experiment performed by Annis Hapkiewicz.

She uses a 0.15 M $\ce{Na2S2O3}$ stock solution.  She prepares 5 beakers with 50 40, 30, 20 and 10 mL of the stock solution filled up to a total volume of 50 mL with destilled water.

This corresponds to thiosulfate concentrations of 0.15, 0.12, 0.09, 0.06, and 0.03 M.

I couldn't figure out the concentration of the hydrochloric acid used, but the times for the appearance of the sulfur cloud (disapperance of the cross) were given.

The range is between 20 s and 5 min.

I'd probably go with the 2 M HCl for safety reasons and perform the experiment with the thiosulfate concentrations given above. Check the times and adjust the concentrations if necessary.

You don't want it to to fast to be measured by the kids, but you don't want the highest concentration to take half an hour.


Dissolution of Lithium in Methanol Easy question to kick off my chemexchange membership!

I'm learning chemistry in my spare time, and fittingly for a video art project I'm doing, decided to try a simple inorganic(?) procedure to make some crimson flames.

My research lead me to believe that lithium methoxide works wonders for this, especially in solution with methanol which only has a very faint blue flame on its own. I basically already know it works, based on some demonstrations seen on YouTube, but I'm unable to find much anything in literature about the lithium salt of methanol. So I'd like to verify my understanding of the equation:

To my understanding, dissolving lithium metal in methanol should yield:

$$\ce{Li + CH4O -> CH3LiO + H}$$

...Lithium methoxide in solution with methanol, assuming an excess of methanol, and some released hydrogen gas.

Am I correct? Can someone shortly explain the actual reaction that takes place?

Additionally: I think the solution should be relatively stable and easy to store and handle, like methanol, unless there's a reason to believe otherwise?
Also, if one wanted to produce dry lithium methoxide powder from this, what might be a good precipitant to crash it out of methanol?

I have not done it with lithium but many times with sodium. Lithium is the less reactive, so it should not give more problems. We use sodium to dry methanol, so it is a common operation in the lab. The reaction is:

$\ce{2Li + 2CH3OH -> 2LiOCH3 + H2}$

Of course, hydrogen is dangerous, especially in the presence of oxygen and still more in this exothermic reaction. So, be careful.

The solution is stable, but attracts water and carbon dioxide, so it must be stored well closed.

To get dry lithium methoxide, the easiest is just to evaporate the excess of methanol.
You put it in a round bottomed flask, attach it to the rotary evaporator, open the vacuum and distill it of. With sodium it works perfectly well. A white solid remains. Again, this solid will attract water and carbon dioxide, so it must be stored properly.

Finally, if the only purpose is to have a methanolic solution of lithium, you can better use lithium chloride which is very soluble in methanol, and much safer (no reaction, no hydrogen, no heating).


Density and solubility Do all saturated solutions have the same density?  Based on your data, which salt is more soluble?

  \begin{align}
&\ce{NaCl} & \pu{11.812 g/10 mL} \quad &\text{or} \quad \pu{1.181 g/mL} \\
&\ce{CaSO4} & \pu{9.783 g/10 mL} \quad &\text{or} \quad \pu{0.978 g/mL}
\end{align}

I know the first part: all saturated solutions do not have the same density. For the second part I believe that solutions with higher density will be more soluble. So in this example $\ce{NaCl}$ will be more soluble that $\ce{CaSO4}$.

Would I be correct and my answer has to be based on density?

Let's think in terms of units. Density is mass per unit volume. Not only will different solutes have different masses dissolved into a unit volume, but the volume of the resulting solution will be different for different solutes.

When people say they can dissolve a unit of mass into a unit of volume (a common measure of solubility) they are not keeping track of the final volume of the resulting solution.

Here are the expressions for the density of the two saturated solutions (we should assume that the temperature and pressure are constant so that the density of the solvent itself is constant).

$\frac{m_{A,sat}}{V_{A,sat}} = \rho_{A,sat}$

$\frac{m_{B,sat}}{V_{B,sat}} = \rho_{B,sat}$

where $m$ and $V$ are masses and volumes, and $\rho$ is the density. $A$ and $B$ are the two solutes.

So if:

$\rho_{A,sat}{}\gt{} \rho_{B,sat}$

that could mean, in one limit where

$V_{A,sat} = V_{B,sat}$,

that

$m_{A,sat}{} \gt {}m_{B,sat}$.

Which is how I'm guessing you're thinking. If the volume change of solvation was similar for two solutes, one could trust that the density of the saturated solution with the (mass) solubility.

But there's another limit where there is a large difference in the volume change of solvation, between the two solutes. Then if:

$\rho_{A,sat}{}\gt{} \rho_{B,sat}$,

as before, and

$V_{A,sat}{} \lt{} V_{B,sat}$,

then it's possible that the unit mass dissolved in the two cases was equal,

$m_{A,sat} = m_{B,sat}$,

so that the two solubilities would be equal.

So no, with the above definitions of solubility and density, one cannot assume that the density of a saturated solution corresponds to the solubility.


Which of the following have the same molecular formula as the molecule shown in the model? This model has 3 hydrogens connected the the first and last carbon and 2 hydrogens connected to the other carbons

Finding the same molecular formula as the alkane model I have to check all that apply

I know the last two options are not correct, because they don't have the same amount of carbons, but when it comes to the first two I'm confused because there is only two carbons attached to 3 hydrogens yet in these options there are 3 different carbons attached to 3 hydrogens.

Would they both be correct or does it matter if it's connected to the second bond instead of the third?

Im assuming since they are isomers that they are equal to eachother.

Remember they are asking for the molecular formulas which are the same, not the structural formulas. The structural formulas are shown in the question, but for example the first option, if you count up the number of carbons and hydrogens you get the molecular formula $\ce{C7H16}$, so you'll need to check if the molecule shown has the same formula.

So for molecular formulas, it doesn't matter how the atoms are arranged; all isomers have the same molecular formula. Only in structural formulas do you get information on the arrangement of atoms in the molecule.

Note there is also condensed structural formulas which look similar to the molecular formula but is still a structural formula. For the first option the condensed structural formula would be $\ce{CH3CH(CH3)CH2CH2CH2CH3}$.


Why convert benzoate in soda to benzoic acid for spectroscopy experiment? I did an experiment where I analyzed the amount of caffeine and benzoate in soda via spectroscopy. I've got the absorbances and did all the math for everything, but I'm having a hard time figuring out why we had to add an acid to protonate the benzoate to form benzoic acid before running the spec.

I've been able to find many similar procedures and many with various pre-lab questions, but none seem to address the reason for the conversion from benzoate to benzoic acid. Here is one such procedure: UV Spectroscopic Analysis of Caffeine & Benzoic Acid in Soft Drinks

The only thing I can think of is that it was in an aqueous solution (soda), so perhaps some benzoate will be in the form of benzoic acid and some will still be in the form of benzoate. By adding $\ce{HCl}$, we can force all the benzoate to be in the form of benzoic acid for better quantification.

Is this the correct line of thinking or am I missing something?

Your assumption is correct, the addition of HCl, as already in the earlier publication by McDevitt (J. Chem. Educ., 1998, 75, 625-629 DOI: 10.1021/ed075p625) ensures benzoates are converted into benzoic acids.

The publication by Guo et al., is instructive as it lists several benzoic acids side by side, for example:

source: J. Phys. Chem. A, 2012, 116, 11870–11879 DOI: 10.1021/jp3084293

The shift of the centre of absorption may be more important than a potential change in $\varepsilon$ if you want to use the UV-Vis spectrum recorded in a principal component analysis (PCA), where overlap of absorption bands of your consitiuents in the analyte should be avoided.


Probabilistic molecular graph models Let a collection of molecular graphs have at most $N$ nodes, $d$ node types (atom type), and $t$ edge types (bond types). A graph from this collection is normally represented by the tuple $(F, E)$ with

\begin{equation}
\text { node-label matrix } F \in \mathbb{R}^{N \times(1+d)}\quad \text { and }\quad \text{edge-label tensor } E \in \mathbb{R}^{N \times N \times(1+t)}
\end{equation}

In a statistical model, $i$-th row in matrix $F$ represents the probability of different node types for a node $v_i$. Similarly, filament $(i,j,:)$ of tensor $E$ represents the probabilities of edge $\epsilon_{i,j}$ in the graph.

In these models, usually, probabilities of nodes and edges are assumed to be independent. This, however, is not a realistic assumption. Type of atoms and bonds are obviously not independent from each other. For a node of certain type, the incident edges collectively must have total type values not exceeding a threshold, where the “type value” is some numeric property of an edge type.

Question:

I was wondering if the type of the node is only dependent on its immediate neighboring edges (incident edges) or we can argue that it also is affected by the edges farther from the node.

Assuming that "atom type" already accounts for the actual nuclei and only refers to the different chemical character, I can think of examples of influence from a distance.

The one with the smallest distance is the acidity of hydrogen attached to carbon. When we compare the hydrogens compared to the central carbon in Propane and Propionaldehyde, we observe a large difference.
Comparing any carbon of Benzene with one of the inner carbons of Butadiene, we observe a large difference in reactivity towards electrophiles.
The same comparison for a carbon of Benzene and the carbon in the 4 position of Pyridine yields a drop in nucleophilicity for the latter.


Substitute for alkali-iodide-azide reagent for BOD test of water samples To measure the biological oxygen demand (BOD) of water samples, I am following a procedure that uses manganese sulphate, $\ce{H2SO4}$, and titration with standard sodium thiosulfate solution. Alkali-iodide-azide reagent is an important part of the procedure, but my school lab does not have this reagent. Is there a substitute for it?

There are two types of oxygen demand of water: easily oxidated substances and  oxidated with difficulties. Thus potassium permanganate, $\ce{KMnO4}$ (permanganate index) are used for first group or potassium bichromate $\ce{K2Cr2O7}$ (Chemical oxygen demand)  for both groups. Both $\ce{KMnO4}$ and $\ce{K2Cr2O7}$ are used in acidic environment.

Alternatives are iodates and cerium(IV) compounds.

I recommend using $\ce{KMnO4}$, but it is prohibited as precursor in some countries. also it is not primary standard, thus standardization by titration of solution of primary standard like oxalic acid needed.

$\ce{K2Cr2O7}$ is primary standard, but it is highly toxic, carcinogen. Thus be very careful with that chemical.

You can try iodates for determination but they are not very common at school. Some bleaching agents contain iodates, but I can not guarantee their usability for analysis.


Donation by inductive effect and withdrawal by conjugation Is there a group that would attached to a benzene ring, donate electrons by inductive effect and withdraw them by conjugation?

I guess a $\ce{-BH2}$ group attached to a benzene ring would have this effect. It has an empty $\ce{p}$ orbital that would withdraw electrons via conjugation, but $\ce{B}$ is less electronegative than $\ce{C}$ and so it would have a positive inductive effect.