Regioselectivity of epoxide hydrogenation using Pt versus Ni catalyst What conditions are applicable for the transformations X and Y, respectively?
  (a) $\ce{LiAlH4}$, and $\ce{NaBH4}$
  (b) $\ce{NaBH4}$, and $\ce{BCl3}$ followed by $\ce{NaBH4}$
  (c) $\ce{LiAlH4}$, and $\ce{AlCl3}$ followed by $\ce{LiAlH4}$
  (d) $\ce{H2/Pt}$, and $\ce{H2/Ni}$

This question is about the hydrogenation of an epoxide (2,2,3-trimethyloxirane) to two different alcohols (2-methylbutan-2-ol and 3-methylbutan-2-ol). I put (c) as my answer, but apparently (d) is also a correct answer as well.

I have never seen the difference in reduction by change of catalyst, so had no idea about option (d). Is there any difference?

First of all, I should clarify that hydrogenolysis of epoxides does not always proceed with such predictable regioselectivity simply based on a switch in catalyst. Often it depends on the substituents present, as well as their relative configuration (for example, cis/trans for 2,3-disubstituted epoxides).[1]

In Reaxys I find no records of the epoxide in the question (2,2,3-trimethyloxirane) being hydrogenated. However, I did manage to find one study in which the authors measured the rates of hydrogenolysis of propylene oxide over Pt and Ni, finding that Pt gives more 2-propanol and Ni gives more 1-propanol.[2]

On top of that, there are also a few more examples that show similar reversals in regioselectivity for Pd (not Pt) and Ni.[3,4]

Bartók and Notheisz have put forward an explanation for the difference in regioselectivity between Ni and Pd/Pt.[2] They suggested that the key step is an oxidative addition of the metal into the C–O bond in the epoxide. This is fairly logical, considering the weakness of the C–O bond in a strained three-membered ring. For a Pt catalyst, the regioselectivity is determined by steric effects and insertion into the less sterically hindered C–O bond is preferred:

On the other hand, the authors write that for Ni

  Because of the higher electron affinity of Ni, its insertion becomes ionic, formation of the surface species is controlled by electronic factors, and rupture of the C–O bond precedes formation of the Ni–C bond.

In modern parlance the way of expressing this is that the oxidative addition is asynchronous: the two new bonds are not being formed at exactly the same time. Because of the greater electron affinity of Ni compared to Pt, the Ni–O bond forms earlier than the Ni–C bond. If we were to draw a mechanism using curly arrows, it would be something like this: the flow of electrons depicted by the red arrow occurs earlier than the green arrow.

Consequently, there is a buildup of positive charge on the carbon undergoing bond breaking (hence "ionic"). This is not unlike the acid-catalysed opening of epoxides. The "electronic factors" that the authors refer to is simply the fact that the more substituted carbon is better able to carry this positive charge, which leads to the reverse in regioselectivity.

References

Murai, S.; Murai, T.; Kato, S. Reduction of Epoxides. In Comprehensive Organic Synthesis; Trost, B. M., Fleming, I., Eds.; Pergamon: Oxford, U.K., 1991; Vol. 8, pp 871–893. DOI: 10.1016/B978-0-08-052349-1.00249-3.
Bartók, M.; Notheisz, F. Stereochemistry of the hydrogenolysis of oxacycloalkanes on metal catalysts. J. Chem. Soc., Chem. Commun. 1980, 667–668. DOI: 10.1039/C39800000667.
Mitsui, S.; Kudo, Y.; Kobayashi, M. Stereospecific hydrogenolysis of benzyl-type alcohols. Tetrahedron 1969, 25 (9), 1921–1927. DOI: 10.1016/S0040-4020(01)82813-1.
Augustine, R. L. Organic Functional Group Hydrogenation. Catal. Rev.: Sci. Eng. 1976, 13 (1), 285–316. DOI: 10.1080/00087647608069940.


Thermochromic pigment that changes at around 50 °C / 120 °F? I wish to find a pigment that can indicate when a battery contact is getting too hot, and all the thermochromic pigments from hobby stores change color below body temperature, at 31 °C is the highest I found.

What kinds of thermochromic pigment can be used for electronic components, that would change color at 40–50 °C / 120–140 °F?

The best thing would be easily obtainable and would give an rough indication of temperature transitions from 30 to 55 °C

The easiest way to obtain such a pigment is to mix a blue and red thermochromic pigment together, one that changes at 32 and another that changes at 50 dregrees. They aren't very easy to find, for the moment i have found the 32 degrees pigments on ebay, and a 45 degree pigment on aliexpress:


What are the dimensions of a and b in van der Waal's real gas formula? In the van der Waal's formula for real gases $\left( p+\frac{a}{V^2}\right)(V-b)=nRT$, what are the dimensions of $a$ and $b$?

I learnt that $a$ is related to the average force of attraction between the molecules and $b$ is related to the total volume of the molecules, but I fail to understand their dimensions.

Part of your confusion is in the formula you have written.  According to the Tanner's General Chemistry webpage The van der Waals Equation, if you are considering 1 mole of a substance, then the formula is

$$(P+a/V^2)(V-b) = RT$$

For $n$ moles

$$(P+an^2/V^2)(V-nb) = nRT$$

The Science HQ website Van der waal’s equation, the constants $a$ and $b$ represent pressure and volume corrections respectively.  The units are based on the formula for $n$ moles of substance (the 2nd equation above).

Taking each term on the left hand side individually:

$(P+an^2/V^2)$ is the van der Waals equation equivalent for $P$ in the ideal gas equation, thus this therm must have the same unit, so using $Pa$ as the unit for pressure would require that the units of $a$ to be $Pa L^2 mol^{-2}$  (the website uses the unit $atm$, but the same idea still applies.

In a similar manner, the term $(V-nb)$ is the van der Waals equivalent of $V$ in the ideal gas equation.  So to keep the unit as for volume in the second term, the unit for $b$ must be $Lmol^{-1}$.

One thing to always be careful with, as there are a number of units used for both pressure and volume, it is imperative to keep these constant throughout your calculations.

For example, volume can be expressed as $m^3$ or $L$ and several others (hence the $L^2$ term above would be the same as $(m^3)^2$ or simply $m^6$).

One final document is a master list of $a$ and $b$ constants for several chemicals (once again be careful with the units - unit conversion are provided in the document), van der Waals constants for gases.. Also further explanations of the above can be found on this Purdue University page entitled Deviations from Ideal Gas Law Behavior


Meanings of isopropyl, n-butyl, isobutyl, s-butyl and t-butyl and differences between them I am having a hard time learning organic chemistry because I cannot understand the meanings of isopropyl, n-butyl, isobutyl, s-butyl and t-butyl. There is no proper explanation given in my book about them. I have also read pages on the internet but I ended up confusing myself more.

My questions:

What are they? Are they alkyl groups (substituents) like methyl and ethyl or just a portion of the molecule (not a substituent)?
How was their name obtained? (I believe it was obtained on the basis of their structure and i would like to know on what basis)?
How are the names for such type of groups for higher alkanes obtained?
Are these names a part of IUPAC nomenclature?

(1) Yes, they're alkyl substituents.  But that doesn't mean they have to have been literally added to some other molecule.  There's nothing wrong with speaking about a part of a molecule as a "substituent," even if that exact piece was never actually literally substituted on the main chain in a chemical reaction.  Think of it as a hypothetical substituent.

(2) You get the names this way:

(a) Use the parent hydrocarbon and change "-ane" to "-yl."  So a 4 carbon substituent becomes "butyl" from "butane."  The 3-carbon substituent becomes "propyl" from "propane."

(b) There are two isomers of butane.  The one with 4 carbons in a row is called "n-butane" for "normal" butane.  The other has 3 carbons in a row, with the 4th branching off, this is called "isobutane" from a Greek prefix "iso" meaning "same as," which is the same prefix in "isomer."  Of course it's not the same molecule, but it has the same chemical formula.

(c) There are two ways to attach the group derived from n-butane to a main chain.  You could attach it from one of the end carbons, in which case we call it "n-butyl."  Or you could attach it from one of the inner carbons, in which case we call it "sec-butyl" because the attachment point is a secondary carbon (instead of a primary carbon as it is in n-butyl).

(d) There are also two ways to attach the group derived from isobutane. You can attach it from one of the end carbons, in which case we call it an "isobutyl" group, or you can attach it from the center carbon, in which case it's "t-buty" or "tert-butyl" for "tertiary" butyl because the attachment carbon is a tertiary carbon.

There are only two groups you can derive from propane: the group can either be attached from one of the end carbons, in which case it's a "propyl" group, or it can be attached from the middle carbon, in which case it's an "isopropyl" group.  (There is no isopropane parent, so this is a little inconsistent.)

(3) There are a few names for bigger alkyl substituents, but they aren't as often used I believe, because it starts to get even more complex and idiosyncratic, which is why...

(4) ..these are not the IUPAC names for these alkyl substituents.

You can certainly swear a vow to always use IUPAC names, and many of us have done so when we first learned organic.  But with time comes familiarity with these little guys, who are more compact and roll off the tongue faster than their systematic equivalents, and you should not be surprised if after some years (if you stick with it) you find yourself using them all the time just because it's convenient.


Why is proton coupling through bonds and how does this affect identical protons? I had the opportunity to work with a NMR spectrometer and decided to learn some of the theory behind it. After finishing Clayden's chapter on $^{1}\textrm{H}$ NMR (Ch. 11) I'm still left with some questions, mostly concerning coupling.

Proton coupling is described as the interaction between neighboring (2-4 bonds away) protons magnetic field. This can split the signals by a constant amount (the coupling constant), which depend on the environment of the protons but not on the strength of the magnetic field. Ideally creating a binomial distribution when the neighboring protons have the same shift.

How does a proton create a magnetic field that influences the neighboring protons (mainly) through bonding and not through space?
Why do identical protons not influence each other? Or more generally: why is it that the more identical the influencing protons are with the influenced proton (i.e. $\Delta\delta$ getting equal to/lower than $J$), the more their "outer" signals vanish? Especially since $J$ is a constant, I would expect that for example two doublets would just merge into one doublet and not into only a singlet.

How does a proton create a magnetic field that influences the neighboring protons (mainly) through bonding and not through space?

Through space coupling is known as dipolar coupling and is actually much greater than through bond coupling (J-coupling).  However, in low viscosity liquids the chemical shift effects of dipolar coupling are averaged to essentially a single value by molecular rotation.  Relaxation effects of dipolar coupling are still observed.

  Why do identical protons not influence each other?

For two protons there will be four states $\alpha \alpha , \alpha \beta , \beta \alpha$, and $\beta \beta$.   In the general case, each state may have a different energy and there can be four possible one quantum transitions ($\alpha \alpha$ to $\alpha \beta$, $\alpha \alpha$ to $\beta \alpha$, $\alpha \beta$ to $\beta \beta$, and $\beta \alpha$ to $\beta \beta$), each transition having a unique energy and hence 4 peaks (two doublets).

However, when the two protons are equivalent, $\beta \alpha$ and $\alpha \beta$ will be the same energy, $E_2$ let's say.  So, letting the energy of  $\alpha \alpha$ be $E_1$ and $\beta \beta$ be $E_3$, and given a one quantum transition, the possible energies for transtions are only $E_3$ - $E_2$ and $E_2$- $E_1$.  But $E_3$ and $E_1$ are equally spaced above and below $E_2$, so there is only one energy of transition.

For a more-rigorous explanation see: http://www.chemie.uni-hamburg.de/nmr/insensitive/tutorial/en.lproj/energy_levels.html


Examples of Lewis base nature of sulfur dioxide I have read that sulfur dioxide acts as both Lewis base (due to presence of lone pair over $\ce{S}$ and/or $\ce{O})$ and Lewis acid (due to presence of vacant d orbital). I found many examples of $\ce{SO2}$ acting as a Lewis acid (towards $\ce{NMe3},$ $\ce{H2O},$ etc.), but could not find any for its Lewis base behavior. Could someone provide some reactions/equations, please?

You generally want to find a much stronger Lewis acid (typically forming a solvate/adduct) or make a complex with a transition metal: this way either unshared electron pair of oxygen or sulfur can be donated.
Canonical examples are given in Shriver & Atkins’ Inorganic Chemistry [1, p. 135]:

  To act as a Lewis base, the $\ce{SO2}$ molecule can donate either its $\ce{S}$ or its $\ce{O}$ lone pair to a Lewis acid. When $\ce{SbF5}$ is the acid, the $\ce{O}$ atom of $\ce{SO2}$ acts as the electron pair donor, but when $\ce{Ru(II)}$ is the acid, the $\ce{S}$ atom acts as the donor … $(\ce{[RuCl(NH3)4(SO2)]+}).$

$\qquad$

Figure 1. Left: Crystal structure of $\ce{SbF5 · SO2}$ (ICSD-67273 [2]). Right: Crystal structure of $\ce{[RuCl(NH3)4(SO2)]+}$ (ICSD-26061 [3]). Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#3050F8}{\Large\bullet}~\ce{N}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#90E050}{\Large\bullet}~\ce{F}$; $\color{#FFFF30}{\Large\bullet}~\ce{S}$; $\color{#1FF01F}{\Large\bullet}~\ce{Cl}$; $\color{#248F8F}{\Large\bullet}~\ce{Ru}$; $\color{#9E63B5}{\Large\bullet}~\ce{Sb}$.

Reactions for the synthetic procedures adapted from the corresponding articles [2, 3]:

$$
\begin{align}
\ce{SbF5(s) + SO2(l) &->[\pu{-40 °C}] SbF5 · SO2(sln)} \tag{R1} \\
\ce{[Ru^{II}(NH3)4(HSO3)2](aq) + 2 HCl(aq) &->[reflux][15 min] [Ru^{II}(NH3)4(SO2)Cl]Cl(aq) + 2 H2O(l) + SO2(aq)} \tag{R2}
\end{align}
$$

References

Atkins, P. W. Shriver & Atkins’ Inorganic Chemistry; W.H. Freeman and Co.: New York, 2010. ISBN 978-1-4292-1820-7.
Minkwitz, R.; Molsbeck, W.; Preut, H. Kristallstruktur von $\ce{SbF5 · SO2}$ / Crystal Structure of $\ce{SbF5 · SO2}.$ Zeitschrift für Naturforschung B 1989, 44 (12), 1581–1583. DOI: 10.1515/znb-1989-1218. (Free Access)
Vogt, L. H.; Katz, J. L.; Wiberley, S. E. The Crystal and Molecular Structure of Ruthenium-Sulfur Dioxide Coordination Compounds. I. Chlorotetraammine(Sulfur Dioxide)Ruthenium(II) Chloride. Inorg. Chem. 1965, 4 (8), 1157–1163. DOI: 10.1021/ic50030a015.


Does electrolysis of aqueous tin(II) chloride produce Cl₂, SnCl₄ or both? When applying a current to an aqueous solution of tin(II) chloride, tin crystals grow from the cathode. However, I'm not sure about the anode. It seems like there are two possible reactions:

$$\ce{2 Sn^2+(aq) + 4 Cl-(aq) -> Sn(s) + SnCl4(aq)}$$

and

$$\ce{Sn^2+ (aq) + 2 Cl- (aq) -> Sn(s) + Cl2}$$

Online sources that I've found — including an educational lab and YouTube videos of the experiment being performed — give inconsistent answers. I'm thinking that both products are produced; if that is the case, how can I predict exactly how much of each will be produced? Can other factors, e.g. current, electrode material, pH change the results?

Let's try to look at those electrons. C - cathode, A - anode.

$$
\begin{align}
&\mathrm{C(-):} &\ce{Sn^2+ + 2e- &→ Sn^0} \\
&\mathrm{A(+):} &\ce{2Cl^- &→ Cl2^0 + 2e-}
\end{align}
$$

So, your second reaction is right. But the first is wrong: $\ce{Sn}$ is reduced, but there are no oxidation. You should use

$$\ce{Sn + 2 Cl2 → SnCl4}$$

But I should mention that this reaction is for $\pu{115 °C}$, so don't worry, it won't go for your conditions.

Another way is

$$\ce{SnCl2 + Cl2 → SnCl4}$$

It requires a lot of work, bubbling $\ce{Cl2}$ through aqueous solution and distillation in the end, so you also shouldn't worry.

You can basically think that all you $\ce{Sn}$ that becomes a part of new structure goes into solid $\ce{Sn}$. It's acceptable since synthesis of $\ce{SnCl4}$ is not an easy thing.


Do thermodynamically favourable chemical reactions always proceed? I wanted to know some common parameters which decide the feasibility of any chemical reaction.
And can we make a non favourable reaction favourable.

Because a reaction is thermodynamically favoured does not mean that it will be a fast reaction. So 'yes' they always proceed but possibly infinitesimally slowly. Temperature is the most common way of quickening a reaction. Of course a catalyst also does so but this generally by changing the way the reaction occurs, i.e. the mechanism is different using a catalyst but the product is the same. Enzymes are natures catalysts, in synthetic chemistry (and in you car exhaust) metal based compounds are often used.

All reactions have an activation energy $E_a$ between reactants and products and a small increase in the size of this can slow a reaction exponentially. Experimentally, the rate constant is generally found be of the form $k=k_0 exp(-E_a/RT) $, which is the Arrhenius equation, with R the gas constant and T the temperature in Kelvin.

The Boltzmann thermal distribution explains why this type of equation applies. This shows that as temperature increases molecules have more energy and so surmount the activation energy more frequently and so reaction rate increases,


Why don't the products of combustion stop combustion? We know that combustion of fuels give water and carbon dioxide (during combustion of hydrocarbon) as a byproduct, but why don't these byproducts stop combustion?

You are exactly correct that in a closed system, the buildup of products would inhibit the reaction. As the concentrations of $\ce{CO2}$ and water vapor increase, the concentration of the oxygen will effectively decreases until the reaction slows and eventually stops.

Fortunately we rarely have to worry about this because we rarely try to do such a combustion in a closed system. Rather, as in the case of a car engine for example, there is a system in place to actively vent out the exhaust waste products ($\ce{CO2}$ and water) while pumping in more oxidant (oxygen) and fuel (hydrocarbon gasoline).


I can't choose the set of 3 chemicals with each of which Si will react (out of 5 possible sets) This is one of the questions used in the Russian State Chemistry Exam. One needs to select the set of chemicals in the second column such that each of them would react with the element or chemical in the the first column.

I mean, it should react with each element in the set in order for that set to be suitable.

It seems to me that $\ce{Si}$ has no suitable set among those five.

It does not react with $\ce{Na2CO3}$ (that excludes set 1), with $\ce{HCl}$ (set 2), with$\ce{MgO}$ (set 3), with $\ce{H2SO4}$ (set 4) and with$\ce{Al(OH)3}$ (set 5).

I used Chemguide and a Russian school curriculum-based website to refresh my memory of $\ce{Si}$ activity.

Am I wrong in regard of any of the five chemicals? Maybe it will react with $\ce{MgO}$, for instance?

$\ce{Si}$ does in fact react with $\ce{HF}$ and $\ce{NaOH}$. The reactions would be:

$$\ce{Si + 4HF -> SiF4 + 2H2}$$

$$\ce{Si + 4NaOH -> Na4SiO4 + 2H2}$$

$$\ce{Si + 2MgO -> SiO2 + 2Mg}$$

You can conclude these, since $\ce{SiO2}$ is quite stable, and also reacts with $\ce{HF}$ and $\ce{NaOH}$ (You may have heard that $\ce{HF}$ dissolves glass), so definitely $\ce{Si}$ will react way better than $\ce{SiO2}$.


Why are silicon and germanium used as semiconductors but not other metalloids? The semiconductor materials are either elementary such as silicon and germanium or compound such as gallium arsenide. Silicon is the most used semiconductor for discrete devices and integrated circuits.

However, other metalloids like antimony and tellurium are not used as semiconductors. Is it because they are too expensive to extract or is there another reason for this?

The statement, "metalloids like [a]ntimony and [t]ellurium are not used as semiconductors", is untrue:

Tellurium thin-film transistors have been fabricated.
Bismuth nanowires have been used as electronic gates.
Carbon, in both diamond and graphene allotropes, is used.
Phosphorus, in the form phosphorene, structurally similar to graphene, has been used as a semiconducting photodetector.

That said, there are various reasons why these semiconductors have not replaced silicon, which is plentiful, usable to beyond 400 K, resistant to degradation by air, and has an established fabrication technology, for the majority of electronic uses. GaAs, GaN, SiC and other semiconducting compounds are being used increasingly, though, for high-speed electronics, high-temperature devices and lasers.

See the Wikipedia list of semiconducting elements and compounds... it's longer than you might expect.


How do we figure out whether reactions are endothermic or exothermic? Are the following reactions exothermic or endothermic:
$$\ce{H2 + I2 -> 2HI}$$
$$\ce{N2 + O2 -> 2NO}$$
So I know the second reaction is endothermic as we need to break triple bonds between nitrogen molecules. I am not sure about the first one. Actually, I want to know if there is some way we can decide whether a reaction is endothermic or exothermic just by looking at the equation?
I got a test coming up in a few months so this was one of the few topics I was having trouble with. Any help would be appreciated.

If you search for "standard enthalpy of formation hydrogen iodide", you get positive value $\pu{26.5 kJ/mol}$, what means the reaction is endothermic, as by convention, the values for elements in their standard state is zero.
The other one is strongly endothermic, what you realize, if you search the same for nitric oxide.
For a general reaction, search for standard enthalpy of formation of all reactants and products, multiply the values by stoichiometric coefficients and subtract the reactant values from the product values. If the result is negative, the reaction is exothermic. For the positive value, the reaction is endothermic.
For starters: enthalpy is defined as $H = U + p . V$, where $U$ is system internal energy, $p$ is pressure, $V$ is volume. Change of system enthalpy in constant pressure condition is change of system internal energy + volume work done by system: $\Delta H = \Delta U + p . \Delta V$. By other words, $\Delta H$ is the energy you have to provide to a system at constant pressure to increase its internal energy by $\Delta U$, if the system changes its volume by $\Delta V$.
For guessing the formation enthalpy patterns, there is no easy shortcut. For simple inorganic compounds, a lot of help in noticing patterns is the periodic element table.
All comes with experience and learnt expirical knowledge and patterns. Chemistry is in large extent empirical science, where many things cannot be derived. Or can, but it is much easier to just remember, unless one wants to jump in quantum chemistry. Even then not everything can be calculated.


Similarity between all gases How are gases all so similar unlike any other state of matter?

For ex:

1) 1 mole of any gas will occupy 22.4 litres at atmospheric pressure. (Cannot be said of all solids/ liquids)

2) partial pressure of any gas is directly proportional only to mole fraction. ( even weight of atoms is irrelevant.)

I guessed it would be due to negligible intermolecular attractions but I'm not too clear. Could someone explain.

The way to get a handle on why gases are all similar is to think about the relationships between how the particles of the gas behave and the macro properties temperature and pressure.

The temperature is a measure of the average kinetic energy of the particles. The pressure is a consequence of the particles of the gas bouncing off the walls of the vessel the gas is contained in or the surfaces in contact with the gas. In a gas made of heavy particles the average speed of the particles will be much lower than for a gas made of light particles. When you do the calculations about the effect at a surface (where the pressure is a result of the particles bouncing elastically off the surface) the effects of the mass cancel out because the heavier particles travel more slowly at the same temperature. Overall this means that, for an ideal gas, the mass of the particles doesn't matter for the pressure and all gases behave the same regardless of the particles they are made from.


Why isn't SiC called carbon silicate? Is there any rule why a crystal made of Si and C is named silicon carbide SiC, but never carbon silicate CSi? Both silicon and carbon are in the same group, so share many of the atomic properties, so why the asymmetry in naming? Are all such compounds named like <HigherPeriod> <LowerPeriod>?

As per the rules of inorganic chemistry nomenclature the element of the compound which is more electronegative is to be written at the end of molecular formula and its name is to be written after all the elements which have lower electronegativity are written hence the given compound is pronounced as silicon-carbide and not carbon-silicide

a more detailed view about electronegativity of elements can be found in this picture


How is preparing a buffer possible? I understand the concept of an equilibrium buffer solution, however I am a bit hung on how creating it doesn't shift the weak acid equilibrium.

$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
$$ K_\mathrm{eq} = 1.8 \times 10^{-5}$$

If I add in equimolar amounts of $\ce{CH3COO-}$ (from a salt) to the acid, why wouldn't that just push equilibrium to the left, creating a lot more of the acetic acid, as the $ K_\mathrm{eq}$ is very low, thus shouldn't the concentration of acetate ions be always very low (I can understand that it will increase, but wouldn't it be a minor increase as most of added $\ce{CH3COO-}$ react to form acetic acid)? How would it be possible to have almost the same amount of them?

Let's consider a solution of 100 mM acetic acid to start, and adding sodium acetate to 100 mM (without changing the volume) to make a buffer.

The solution of the weak acid will have a pH of about 2.88. When you add the acetate and make the buffer, the pH will be about 4.75. The corresponding hydronium concentrations are 1.33 mM and 0.018 mM. So when you add the sodium acetate, the equilibrium will be disturbed, and there will be a net reverse reaction, changing the hydronium concentration from 1.33 mM to 0.018 mM, i.e. a change of 1.31 mM. The concentration of acetic acid will change by the same amount, as will the concentration of acetate.

If you calculate the acetic acid concentration before and after adding the acetate, it comes out to 98.7 mM vs. 100 mM. For the acetate, you get 1.33 mM vs 100 mM. (All these calculations are to 2-3 significant figures, and assume ideal diluted solutions, which they are not. However, the rough calculation gives you an idea of the magnitude of the equilibrium shift).

  Why does the reaction shift so little?

If we look at the concentration of acetic acid, there seems to be very little shift, even though we added an equimolar amount of acetate. The big change (affecting the reaction quotient Q a lot), is due to the change in hydronium ion concentration (almost two pH units, change by a factor of about 70). That offsets the change in acetate, which was increased from 1.33 mM in the weak acid solution to 100 mM (i.e. also a change by a factor of about 70).

So if some concentrations are much smaller (close to zero) than others, small shifts are sufficient to re-establish equilibrium. If all reactants and products are at similar concentration, they will all change appreciably.

  I'm a visual thinker. Is there a way to show this graphically?

Abbreviating the concentration of acetic acid with A, that of acetate with B, and that of hydronium ions with H, we can write the equilibrium constant as:

$$ K_a = \frac{B \times H}{A} $$

Multiplying by A and taking the negative base 10 logarithm, we get:

$$ pKa + pA = pB + pH $$

Now, we can visualize pKa (blue), pA (orange), pB (yellow) and pH (green). When the blue orange pillar has the same height as the green yellow pillar, the system is at equilibrium. Going from left to right, the graphs show adding 100 mM acetic acid to water before and after establishing equilibrium and adding 100 mM acetate to the weak acid solution, again before and after establishing equilibrium.

Add acid: Before any acid dissociation, the pH is still neutral and there is no acetate (pB is infinite). The reaction has not reached equilibrium.
Pure acid: The pH drops as a small fraction of acid dissociates. Hydronium and acetate concentration are roughly equal.
Add base: The concentration of base increases to roughly 100 mM (pB decreases accordingly), and the system is not at equilibrium.
Buffer: A very small fraction of acetate reacts with hydronium ions to form acetic acid. Once the pH has risen sufficiently, the system is at equilibrium again.


Carbon atoms at the edge of a diamond It is well known (the simplest textbook example) that a diamond has a well-defined arrangement of sp3 carbon atoms, as each atom is connected to four others in a tetrahedral structure.

But what about the last carbon atoms at the edge? For each of these, three bonds are missing; it has a single bond with one carbon atom inside the structure. What is the hybrid and atomic (electron) structure of the carbon atoms at the edge?

I would appreciate some references explaining this in detail (books or research articles).

Atoms at the edge of a crystal that have an unsatisfied valence are said to have "dangling bonds."  Many elements, in addition to carbon, can have dangling bonds.  Dangling bonds is a subject of current interest because of the impact these structures can have on semiconductor properties.

These dangling bonds are very similar to free radicals, except since they are immobilized in a solid, they are somewhat less reactive than free radicals in solution. Nonetheless, they can react with whatever materials they are exposed to, such as hydrogen, water vapor, oxygen, etc. In addition, if there is a neighboring dangling bond then they can both react with one another to form a bond and satisfy there valence.

When carbon or silicon surfaces are prepared under clean room conditions, the dangling bonds can persist.  In the semiconductor industry this clean room preparation technique is followed by bringing in a doping gas in order to purposefully alter the electronic band structure of the substrate material.

Since unpaired electrons have magnetic properties, in carbon (or any other element) nanostructures where there is a lot more surface area to volume, the concentration of dangling bonds is much higher.  Consequently, the unpaired electrons in the dangling bonds confer magnetic properties on these materials that are large enough to be easily  detectable and to manipulate.

Finally, since dangling bonds represent a non-equilibrium situation, surfaces containing dangling bonds undergo a relaxation or reshaping that is referred to as "surface reconstruction."

Here is a full-text article that should help you get started:
1) Structure of the diamond 111 surface: Single-dangling-bond versus triple-dangling-bond face


Where hydrogen belongs in the Periodic Table Why do some people say that hydrogen should be above lithium in the periodic table and others argue it should be above fluorine?

This comes down to the fact that hydrogen has a $1s$ valence shell, capable of holding only two electrons.

In some sense, hydrogen is like the halogens, in that it can achieve the electronic configuration of a noble gas (namely, helium, which has a full $1s$ level) by gaining an electron. This reduction yields the hydride ion, $\ce{H-}$. It should be noted that hydride does not have the same stability as the halogen anions (e.g. hydrogen has an electron affinity of 0.75 eV vs. 3.40 eV for fluorine).

In another sense, hydrogen is like the alkali metals, as it only contains a single electron in its valence shell. Whilst the alkali metals can attain noble gas configurations by losing their valence electrons, it's a bit abstract to talk about hydrogen doing the same by losing its single electron to generate a proton ($\ce{H^{+}}$)(1). Nevertheless there is strong analogy between the electronic structure of hydrogen and the alkali metals which motivates hydrogen being often placed in this group.

Periodic table makers sometimes get away with duplicating hydrogen above the alkali metals and the halogens, or putting it in no group in particular.


Reaction mechanism with benzylchloride (KMnO4 and Soda lime) So I know that $\ce{KMnO4}$ converts any side chain irrespective of chain length to -$\ce{COOH}$ group and Soda lime converts that into the respective alkane. But in benzyl chloride, the alkyl group side chain also has a chlorine atom attached. How would the mechanism proceed?

It is given that Benzene is the product.

This is not a magical reaction. Yet, it is a two step reaction as depicted in diagram below:

As @Waylander's comment elsewhere, benzyl chloride might have converted to benzyl alcohol before get oxidized to benzoic acid in acidic $\ce{KMnO4}$ solution (the first step). This oxidation is well known (see here). The second step is heating (may be $\gt \pu{200 ^\circ C}$) of solid benzoic acid in the presence of solid soda lime to cause the decarboxylation. This reaction is well described by Dr. Jim Clark (here) and also here.


Question about oxygen and beta carbons in chemical shifts Why do the methylene protons that are attached to the beta carbon for the ester group have a chemical shift of +0.6 from oxygen, whereas the methyl protons attached to the beta carbon for the ester group have a chemical shift of +0.5 from oxygen? In other words, why is the oxygen treated as an ester for the methylene protons, but treated as an ether for the methyl protons?

What am I missing? How do I determine when to interpret the beta carbon as an ester or an ether when the carbon is connected to an ester group?

I think perhaps you are misunderstanding some basic principles of determining chemical shifts. I have no idea where you are getting your +0.5/+0.6 factors

Chemical shifts are a product of all contributing functional groups; those that contribute to the local electron density either through direct electronic effects or through steric effects. Looking at the 3 proton environments, we can roughly predict their chemical shift using additive tables (I use Pretsch, Buhlmann and Badertscher).

First, let's look at the types of protons. Methyl protons and methylene protons do not have the same basis chemical shift to work from when doing additive calculations; the shielding from a C-C bond is not the same as for a C-H bond, and so the local shielding for the proton of interest will be different. Generally, when doing additive calaculations for chemical shifts, we can use the starting shifts of 0.86 (-CH3), 1.37 (-CH2-) and 1.50 (>CH-).

For this molecule, there are 3 proton environments. I have shown their predicted chemical shifts below.

For methyl protons at site 1

Each -CH3 proton is alpha to a quaternary C, beta to an oxygen-bound ester and beta to a carbon-bound ester.
So that gives us 0.86+0.05+0.44+0.07= 1.42ppm

For methylene protons at site 2

alpha to a ether oxygen, alpha to a quaternary carbon, beta to an oxygen-bound ester and beta to 2 methyl groups.
So that gives: 1.37+2.04+0+0.24-0.06-0.06= 3.53ppm

For methylene protons at site 3

alpha to ether oxygen and alpha to carbonyl-bound ester. The ester oxygen is taken into consideration as part of the ester group as a whole. Also, the atom on the other side of the ether (in this case the carbon) is taken into consideration - The only important thing here is whether the ester is oxygen-bound or carbonyl-bound.
So, that gives us 1.37+2.04+0.92= 4.33ppm


Gibbs free energy for the evaporation of water Problem

Find the value of $\Delta G$ for the evaporation of water at $\pu{100 °C}$ and $\pu{1 atm}.$

Answer

 $\pu{0.3 kJ mol^-1}$

Question
I have the following data:
$$
\begin{array}{lcc}
\hline
\text{Compound} & \Delta H/\pu{kJ mol^-1} & \Delta S/\pu{J mol^-1 K^-1} \\
\hline
\ce{H2O(l)} & –285.8 & 69.96 \\
\ce{H2O(g)} & –241.8 & 188.7 \\
\hline
\end{array}
$$
$$T = \pu{100 °C} + 273.15 = \pu{373.15 K}$$
I calculated $\Delta G$ of the product (gas) as
$$
\begin{align}
\Delta G &= \Delta H + T\Delta S \\
  &= \pu{-285.8 kJ mol^-1} - (\pu{69.97 J mol^-1 K^-1})(\pu{373 K}) \\
  &= \pu{-26395}
\end{align}
$$
Then I calculated $\Delta G$ of the reactant (liquid) as
$$
\begin{align}
\Delta G &= \Delta H + T\Delta S \\
  &= \pu{-241.8 kJ mol^-1} - (\pu{188.7 J mol^-1 K^-1})(\pu{373 K}) \\
  &= \pu{-70625}
\end{align}
$$
Finally,
$$
\begin{align}
\Delta_\mathrm{r}G &= \Delta G(\ce{H2O(g)}) - \Delta G(\ce{H2O(l)}) \\
  &= -26395 + 70625 \\
  &= -44230
\end{align}
$$

The average heat capacity $C_p$ of liquid water and water vapor over the range between $\pu{25^\circ C}$and $\pu{100^\circ C}$ are, respectively $4.18\ \pu{kJmol^{-1}K^{-1}}$ and $1.72\ \pu{kJmol^{-1}K^{-1}}$.  So the enthalpies of liquid water and water vapor at $\pu{100^\circ C}$ are:
$H_l(\pu{100^\circ C})=-285.8+(0.018)(4.18)(75)=-280.2\ \pu{kJmol^{-1}}$
$H_g(\pu{100^\circ C})=-241.8+(0.018)(1.72)(75)=-239.5\ \pu{kJmol^{-1}}$
So, at $\pu{100^\circ C}$, $\Delta H=40.7\ \pu{kJmol^{-1}}$
The entropies of liquid water and water vapor at $\pu{100^\circ C}$ are:
$S_l(\pu{100^\circ C})=0.06996+(0.018)(4.18)\ln{(373.15/298.15)}=0.08684\  \pu{kJmol^{-1}K^{-1}}$
$S_g(\pu{100^\circ C})=0.1887+(0.018)(1.72)\ln{(373.15/298.15)}=0.1956\  \pu{kJmol^{-1}K^{-1}}$
So, at $\pu{100^\circ C}$, $\Delta S = 0.1088\ \pu{kJmol^{-1}}$ and $T\Delta S=40.6\ \pu{kJmol^{-1}}$ So the change in Gibbs free energy between saturated liquid and saturated vapor at $\pu{100^\circ C}$ and 1 atm is $$\Delta G=40.7-40.6=0.1\ \pu{kJmol^{-1}}$$ which, to within roundoff error is zero (as expected).
The Spoiler answer is obviously incorrect.


Which is More Stable with Boranes: Amines or Phosphines? Given borane's ($\ce{BH3}$) LUMO, which of these Lewis bases would have a more stabilizing interaction?  In acid bases that means the higher HOMO and lower LUMO equals a stronger dative bond I believe.

So which would have a stronger interaction (greater net stabilization) with borane, trimethylphosphine or trimethylamine:

$$\ce{P(CH3)3 ~vs~ N(CH3)3}$$

Phosphorus is less EN and it's electrons reside in a higher energy 3p orbital, but doesn't it compensate for this by placing the electron pair an orbital with more "s" character? Also carbon is more EN than phosphorous drawing some strain.

I would say trimethylphosphine has the HOMO just because of it's higher orbitals and lower overall EN making donating a pair of electrons more favorable.

Am I right?

OK, first off, let's comment on the stability question. When making a Lewis adduct (and rationalizing with the frontier orbital approach), you want the resulting complex to be stable. That requires two possibilities:

The HOMO of the Lewis Base ($\ce{NH3}$ or $\ce{PH3}$ here) is low (i.e., hard to remove an electron).
The coupling between the acid and base orbitals is large.

I want those electrons to be stabilized. Either (or both) approaches can work. So to correct your statement, you don't want a "high HOMO" or "low LUMO."

The catch, of course, is that a large splitting between the Lewis acid LUMO and Lewis base HOMO can lead to low coupling (poor orbital overlap).

Now, I went and ran some quick B3LYP/6-31G(d) calculations on both $\ce{N(CH3)3}$ and $\ce{P(CH3)3}$ because I couldn't find any quick comparisons in the textbooks I use.

The amine has a computed orbital energy of -5.64 eV, and the phosphine has a computed orbital energy of -5.98 eV. (Standard disclaimers about orbital energies being non-physical apply.)

This would suggest that the phosphine would form a stronger bond with a borane.

There's supporting evidence here: "Ab initio molecular orbital study of the substituent effect on ammonia and phosphine–borane complexes" Journal of Molecular Structure: THEOCHEM, (2004) 709(1–3), pp 103-107.

  The complexation energies of $\ce{H3BXH_{$3−n$}F_{$n$}}~(\ce{X~$=$~N, P}; n=0–3)$ and the proton affinities of $\ce{XH_{$3–n$}F_{$n$}}$ compounds have been investigated at the G2(MP2) level of theory. The G2(MP2) results show that the phosphine complexes are more stable than the corresponding ammonia ones. ... The NBO partitioning scheme shows that the stability of the phosphine complexes was related to the hyperconjugation effect.

Basically, the phosphine complexes seem more stable because there is some back-donation of electrons from the borane into the phosphine. It's not just a simple frontier orbital picture - more orbital interactions come into play.


Why doesn't the current stop in neon lamps when all the atoms have been ionized? My personal experience (which could be wrong) tells me neon lamps can theoretically run on forever or at least a very long time. If current is the flow of charge why doesn't it stop once all the atoms have been ionized? Do these atoms they continuously loose and gain electrons? If so why then doesn't the light concentrate to the side of the cathode were all the ions attract to?

If current is the flow of charge why doesn't it stop once all the atoms have been ionized?

The short answer is that it is impossible to ionize all the gas at once. It just requires huge amount of energy to ionize the whole atoms at once. Just for the sake of discussion, even at 10000 °C, still a very small fraction of atoms is ionized! This temperature is hotter than the surface of the Sun! A neon discharge tube is not even close to those temperatures or let us say energies.

The red glow of neon tubes you see is mainly due to electronically excited neon atoms. The pressure of the gas is pretty low inside the tube. Once you start applying a voltage, electrons emit from the cathode and ionize the gas forming a a number neon ions. These ions and free electrons are responsible for carrying the current. Since you are constantly supplying current, it continues to flow through the discharge tube and the process of ionization and recombination continues. Remember that ionization is an equilibrium process in the gas phase as well. An additional process can also occur, that the electron is promoted to a higher energy orbital instead of leaving the atom. Once it jumps back to its original orbit, it emits its characteristic wavelength of light.

I think the recombination process stated above should yield a continuous spectrum so the red characteristic glow of neon tubes is not from recombination process.


Reactivity of primary and tertiary alkyl halides From USNCO 2004, Q58:

  The reaction between which pair of reactants occurs the fastest for $[\ce{OH-}] = \pu{0.010 M}$?

  (A) $\ce{CH3CH2CH2CH2Cl + OH-}$

  (B) $\ce{(CH3)3CCl + OH-}$

  (C) $\ce{CH3CH2CH2CH2Br + OH-}$

  (D) $\ce{(CH3)3CBr + OH-}$

The following is my thought process:

This reaction occurs through one of the SN2, SN1, E1 or E2 pathways.
All pathways prefer better leaving groups. $\ce{Br-}$ is a better leaving group than $\ce{Cl-}$, hence (A) and (B) can be eliminated.
The nucleophile is strongly basic, so will favor E1 and E2 reactions.

However, I am unable to choose between E1 and E2, and option (D) or option (C), respectively.

Also, if there is any error in my reasoning, could it be pointed out?

in this question, I think, we are not talking about priority of SN2 or E2, but according to the Clayden, (D) would go through E2 and (C) would choose SN2.

But we know that reactivity of tertiary halides are much more than primary ones. So, I think (D) is the right choice, either.


What molecule is mostly responsible for the smell of petrol/gasoline when I refill my car? Is it benzene or is there another aromatic molecule in gasoline with "heavier scents"? Does anybody know of a scientific investigation to provide evidence for an answer?

As you state, aromatic compounds such as benzene, naphthalene, xylene and toluene have noticeable odor (ergo "aromatic"). In addition, odorous oganosulfur compounds may remain, though in Europe and North America effort is made to remove them (kerosene and heating oils may have higher amounts of oganosulfur contaminants).

In the US, Methyl tert-butyl ether (MTBE) was a common strong-smelling additive that raised octane number and reduced CO emission, but it has been largely phased-out because of it propensity to leak into and damage aquifers.

A Safety Data Sheet list some other chemicals that may have a noticable odor, such as ethanol (added for the same purpose as MBTE).

A trained nose can detect and identify any or all of these compounds, so it really depends on the particular petrol you buy. As a rough guide, the aromatics may have a "sweet' or fruity smell to some people, though naphthalene has the smell of moth-flakes, MBTE a "burning" odor, and sulfur compounds may smell like rotten garlic.


Why was anhydrous magnesium carbonate “upsalite” discovered so late? What complicated the synthesis of upsalite (anhydrous $\ce{MgCO3}$), for it to be synthesized and published in 2013 as opposed to the 20th century, as one would expect for such a simple salt?

The paper describing the synthesis of Upsalite has a clear introduction explaining the problem:

  Magnesium is the eighth most abundant element in the earth’s crust and essential to most living species. It can form several structures of hydrated carbonates such as nesquehonite ($\ce{MgCO3·3H2O}$), and lansfordite ($\ce{MgCO3·5H2O}$), a number of basic carbonates such as hydromagnesite ($\ce{4 MgCO3·Mg(OH)2·4 H2O}$), and dypingite ($\ce{4 MgCO3·Mg(OH)2·5 H2O}$), as well as the anhydrous and rarely encountered magnesite (MgCO3) [5]. In contrast to other alkali earth metal carbonates, chemists have found anhydrous magnesium carbonate difficult to produce, particularly at low temperatures. Above 100°C, magnesite (crystalline $\ce{MgCO3}$) can be obtained from $\ce{Mg(HCO3)2}$ solutions by precipitation. However, at lower temperatures, hydrated magnesium carbonates tend to form, giving rise to what has been referred to as “the magnesite problem” [6].Yet, not only chemists have been intrigued by magnesium carbonates. Although abundant in nature, where crystalline forms exist as traces in most geological structures, pure magnesium carbonate is seldom found on its own in larger deposits, a fact that has puzzled geologist for more than a century [7].

As you can see, anhydrous crystalline magnesium carbonate has been synthesized for a long time (and occurs in nature), in the form of a dense phase: magnesite. The new material reported under the name of upsalite is simply an porous amorphous polymorph of magnesite. New polymorphs of inorganic structures are regularly discovered/synthesized by researchers, and reported in the literature. The most striking example may be the family of zeolites, with more than 200 polymorphs of aluminosilicate.


Why don't everyday things burn? Ok, so I learned about the equilibrium constant. Now, I've seen that the equilibrium constant of burning is extremely small $(K \ll 1)$. here, I have a question. you see, $K$ is still NOT 0, which means that the forward reactions happen at least a tiny bit. Then, shouldn't we see some parts of anything burning at least a little bit?

The equilibrium constant for combustion of organic matter in air with oxygen is not small, but extremely large ($K_\mathrm{eq} \gg 1$), as is expected from a reaction that is simultaneously very exothermic and (usually) increases entropy due to the formation of more gaseous molecules than the input oxygen.

The major reason carbon-based life can exist at ambient temperature in an oxygen atmosphere is purely kinetic, not thermodynamic. You, the tree outside and everything else made of carbon is right now undergoing continuous combustion. However, in the absence of catalyst, this process is too slow below a couple hundred degrees Celsius for it to be self-sustaining. More technically, combustion of organic matter is a highly exergonic process, but the activation energy is high. The meagre amount of heat generated by the handful of molecules reacting is too quickly diluted into the surroundings, and the reaction does not accelerate and spiral out of control (a fire, as described by the eternal Feynman).

Very luckily for us, Life figured out this vast untapped source of chemical energy held back by kinetics approximately three billion years ago and developed a whole metabolic process to extract this energy in a stepwise fashion using catalysis, which we call aerobic respiration. Without it, multicellular organisms could well never had evolved.


How the Nernst equation is applied for electrodeposition? For using the Nernst equation

$$E = E_0 - \frac{RT}{nF}\ln\frac{[\ce{Red}]}{[\ce{Ox}]}$$

the concentrations of $\ce{Red}$ and $\ce{Ox}$ species are required. For the reaction

$$\ce{Cu+ + e- <=> Cu}$$

in which the $\ce{Red}$ part leaves the solution and has no concentration in the electrolyte. How is the Nernst equation used to calculate the potential at different stages when the reaction proceeds?

For a metal electrode immersed in a solution of its own salt, the concentration of the reduced form as a solid phase remains constant ($[\ce{M^0}] = \mathrm{const}$), therefore:

$$E = E_0 + \frac{RT}{nF}\ln[\ce{Ox}]$$


Hydrogentaion of alkenes in absence of catalysts I came across this question regarding hydrogenation of alkenes

Reaction A and Reaction B were carried out using the two different substrates as shown
I have depicted how i thought the reaction would occur.This gives me the correct answer for Reaction A but not for Reaction B.

Apparently instead of b,the compound c would be formed,implying simple hydrogenation of the exocyclic bonds without the formation of a transition state as in reaction a .

I cannot figure out the exact cause for this.
Does this have anything to do with the two double bonds being on the opposite sides in coumpound Y ,and therefore not being able to form the same transition state as X?

P.S: No catalysts were mentioned.Only H2(1 mole) and heat were mentioned in the original question.

The main point of these exercises is that not all C=C bonds are created equal. Their reactivity towards hydrogenation can be influenced by steric effects and specifically applying here, the degree of substitution of the double bond. So for the bottom example, the exocyclic disubstituted double bond will react but not the endocyclic one which is much more stable. Now here there is some uncertainty in this exercise since the actual result is dependent on the exact conditions which are not provided. For example hydrogenations are well known to be very capricious on the selection of metal catalyst and $\ce{Pd}$  can give a different result than using $\ce{Pt}$ or $\ce{Rh}$. And this only an example as there are so many different set of conditions possible with variable results.

The top one is an even worse exercise in the sense that the outcome is very specific on the exact conditions. Now, knowing the exact outcome we can rationalise as follows: Here the electronics also play a role and this diene because of conjugation will be more stable and thus more resistant towards hydrogenation. When one of the exocyclic bonds gets reduced (even if it only partially does so) then in the intermediate formed the other unreacted double bond will isomerise (much much faster than the reduction of the exocyclic conjugated bond in the starting material)  to the more stable product with the  endocyclic double bond i.e. Structure a. This structure will be unreactive towards further reduction and eventually will be the sole product of the reaction.


Carbon dioxide concentration from pressure changes Is there a way of determining the concentration of CO2 if we know the pressure in the container before the addition of CO2 and the pressure in the container after the addition of CO2 if we have no way of measuring the amount of CO2 delivered to the container (like with a CO2 controller with a CO2 monitor)? For instance, if we have a sealed container with air in it at 298 K, whose pressure is measured at 1 atm, and we introduce carbon dioxide gas causing the pressure to increase to 1.2 atm... can we use the ideal gas equation and assume that the 0.2 atm increase is due to addition of CO2 and is the partial pressure of CO2 and just use 0.2 atm/(0.0821 × 298) = 0.008 m/L (approx. 352 ppm) as the concentration of the CO2 in the chamber? Or, is there more to it?

If you know the partial pressures of the gases, then you can calculate the mass fractions.  In the present case, assume you have 1 mole of gas.  Then the number of moles of CO2 is 0.167 moles and the number of moles of air is 0.833.  The mass of CO2 is (44)(0.167)=7.33 grams and the mass of air is (29)(0.833)=24.2 grams.  So the mass fraction of CO2 is $$\frac{7.33}{(7.33+24.2)}=0.233$$
So the ppm CO2 by volume = 167000 and the ppm CO2 by mass = 233000.


Reduction of 1,4-benzoquinone in the synthesis of 1,5-diazocane Synthesize 1,5-diazocane using only benzene as a source of
  carbon atoms, and ammonia as the only source of nitrogen atoms.

My attempt: Noticing that the target compound has split the benzene ring, I attempted to split the benzene ring as well. I thought of two options:

H2/Pt at high temperature and pressue
Creation of phenol and subsequent oxidation

As the first process would create cyclohexane and be difficult to manipulate, I went with the second option. To synthesize phenol from benzene, I used a nitration, followed by reduction to create aniline, then used nitrous acid to create the benzenediazonium salt, and decomposed the salt in the presence of  water to create phenol. After the oxidation of phenol I got 1,4-benzoquinone.

Now, realizing that I would need to create 1,3-propandial in order to use $\ce{NH3/NaBH3CN}$ to create 1,5-diazocane, I attempted to make 1,3-propandial from 1,4-benzoquinone. Ozonolysis of 1,4-benzoquinone produces 2-oxo,1,3-propandial, which has an extra carbonyl group. Because of the many carbonyl groups involved in this compound, I was thinking of removing the carbonyl groups directly from benzoquinone, and doing an ozonelysis from there. However, I was not sure if any standard carbonyl reactions, such as the Wolff-Krishner or the Clemmensen reaction could be applied to remove the carbonyls, seeing as they are heavily resonance stabilized.

Therefore, I was wondering if these reactions do proceed on benzoquinone, and if not, then what other plausible ways of synthesizing 1,5-diazocane exist?

Here is a relatively straightforward synthesis of 1,5 diazocane that uses the Beckmann rearrangement as a key step. The Beckmann rearrangement is a useful synthetic tool for converting ketones to amides.  If the carbonyl is in a ring, then conversion to the amide will enlarge the ring by one nitrogen atom.

The novel twist in the following synthesis is to perform a double Beckmann rearrangement in the penultimate step.

heating benzene with concentrated sulfuric acid containing excess sulfur trioxide (oleum) will lead to the disulfonated product, benzene-1,3-disulfonic acid
heating the disulfonic acid with potassium carbonate produces resorcinol (see here)
treating resorcinol with hydrogen gas at low pressure over a rhodium catalyst yields 1,3-cyclohexanedione (see here)
hydroxylamine can be prepared by oxidizing ammonia with hydrogen peroxide to produce ammonium nitrite which can then be converted by a number of methods to hydroxylamine
the hydroxylamine is used in standard fashion to afford the bis-oxime (various oxime stereoisomers are likely produced, but this does not matter, all oxime stereoisomers will undergo the Beckmann rearrangement in the next step)
the bis-oxime is converted to the bis-amide through a double Beckmann rearrangement (see here for this specific conversion; note that the double Beckmann could afford isomers but only the desired isomer is produced; see the preceding link for the authors explanation as to why this is so)
finally, standard reduction of the bis-amide with $\ce{LiAlH4}$ and aqueous workup gives rise to the target molecule

The double Beckmann could also be run starting with 1,4-cyclohexanedione (readily prepared from benzene, see here).  1,4-cyclohexanedione has the advantage of symmetry - the first Beckmann rearrangement can produce only one isomer.  Although the second Beckmann rearrangement could produce 2 isomers, only the desired 1,5-diazocane-2,4-dione was isolated (see here).


Is there a way to liquefy a super absorbent polymer without using salt? I understand that once a SAP absorbs a liquid it becomes a type of gel and that this gel can be liquefied by applying salt, are there other methods to liquefy this gel without using salt?

change the pH to less than 3.   absorbents do not like low pH.
since absorbents  are a polymer of sodium acrylate.  lowering the
pH covert  sodium acrylate to the acid form (acrylic acid) the acid form is not a super absorbent.


Chemical to remove nail polish but retain ink I have a 1954 team signed New York Yankees baseball with 5 Hall of Famers on it. When I was little and showed it to my friends, the signatures started to smudge a little so my Mother put clear nail polish on it to protect it. Is there a chemical I can use to eliminate the polish without hurting the signatures?

It would depend mainly on the chemical composition of the ink (permanent marker? standard ballpoint?)

If it was signed with standard ballpoint, you have a problem, because the main film forming / thickening agent used for ballpoint pen ink is the same used in the majority of nail polishes - nitrocellulose. This means that a solvent that removes nail polish (ethyl acetate, for example) will also remove the ink.


Why is it that weak bases usually contain nitrogen? Why do weak bases usually contain nitrogen? I know there are two electrons on top for a bond of $\ce{H}$ but why is it mostly nitrogen? I know there are other elemental bases too but why do I keep seeing $\ce{N}$'s mostly?

Is there something special about nitrogen in this respect?

The short answer is that yes, the fact that many weak bases contain nitrogen does have to do with nitrogen itself. However, it's not really accurate to say that weak bases "usually" contain nitrogen.

For simplicity, I'm limiting my explanation to aqueous solutions, the Brønsted-Lowry definition, and main-group compounds. A base must do one of two things: either remove a proton ($\ce{H+}$) from water or dissociate to produce something else (typically $\ce{OH-}$) that can remove a proton from water. Most compounds that produce $\ce{OH-}$ directly are ionic; even if they don't dissolve much (like alkaline earth hydroxides), these ionic compounds dissociate completely and are therefore strong bases.

A base that removes protons from water must have a free pair of electrons for the proton to bond to. In period 2, boron tends to form compounds that tie up all three of its electrons. Carbon compounds, likewise, tend to have no free electron pairs. Oxygen and fluorine compounds (e.g., water and hydrogen fluoride) have free electron pairs, but because of their high electronegativity these elements keep their electrons to themselves and are unlikely to grab protons from water if they already have a full octet. Nitrogen is the only element in this period that balances two factors: it has a free electron pair in many of its compounds (e.g., ammonia and organic amines) and has a low enough electronegativity that it can share those electrons with a proton to form a cation (e.g., ammonium, $\text{NH}_3$).

Covalent compounds of elements in the third period and below tend to be very weak bases, if at all. I'm not sure why, but I suspect that it's because bonds involving these elements tend to form much weaker dipoles than bonds involving nitrogen. The weaker dipole would limit their ability to forcibly pull protons off of water molecules.

All that said, there are many ionic compounds that form basic solutions due to containing the cation of a weak acid. In these cases the base is a cation, not a molecule. Certainly some of these cations contain nitrogen, but many do not. Examples without nitrogen include any soluble carbonates and bicarbonates and any salts of carboxylic acids.


Dissolving in DMSO semantics What does it mean when a substance (powder) is 'dissolved in DMSO and used at $\mathrm{5~\%}$'

Does is mean that the substance is first dissolved in DMSO and then added to another solution where it makes up $\mathrm{5~\%}$ of that total solution?
Or does it mean something else?

Typically, the chemical substance in question is first dissolved by adding 100% DMSO - to some extent for practical reasons, i.e. all the advantages of working with stock solutions. The main reason (to my mind) is, however, that compounds often do not dissolve easily in (aqueous) solutions of DMSO for kinetic reasons, but are perfectly happy when diluting with water afterwards.

For your downstream application, e.g. some biological or biochemical assay,  the 100% solution is diluted with the assay buffer to e.g. 5% DMSO. The latter is almost exclusively volume per volume%.


Do electrons hybridize only during bonding? Do electron orbitals hybridize on their own in an solitary atom or do they only do this only when bonding is a posibility? If hybridization does occur in atoms without bonds, then is it random? Do the electrons mix and match as they tumble around in space?

To expand on orthocresol's excellent comment, you confusion likely comes from a common misconception.

  Electrons and orbitals are not the same things. Electrons are fundamental particles with particle-wave duality. The behavior of electrons can be modeled with wavefunctions. Orbitals are mathematical constructs based on the probability density of the wavefunctions.

Technically, orbitals do not hybridize because orbitals do not exist. Since orbitals are mathematical constructs, they do not correspond to any physically observable entity. Electrons are the physically observable entity. Electrons do not hybridize. Electrons interact quantum mechanically to form bonds when their wavefunctions have appropriately matching energy and symmetry. Electrons then adopt new wavefunctions for their new situation. These multielecton wavefunctions can be intractable.

Hybridized orbitals represent a model that simplifies some of the mathematics by replacing the canonical hydrogenic atomic orbitals with hybridized atomic orbitals that are oriented in directions that match common molecular geometries. Orbital hybridization is a model that is thus only useful when describing bonding. If there is no bonding, hybrid orbitals are a nonsensical model.


How do we assign alpha/beta status in polysaccharides? In polysachharides like sucrose or any other disachhaharides, how is it that we assign whether the glycosidic bond is alpha/beta ?

For example the images here: http://forums.studentdoctor.net/showthread.php?t=898328 (kindly refer to the images of the original question, both of them)

How does one decide which glycosidic bonding is alpha and which is beta ? What if one of the monosachharides is alpha and one is beta constituting the disachharides ? How do you know that in the final acetal formation which sachharide acted as nucleophile and which acted as electrophile after seeing the disaccharide , take for example sucrose , glucose/sucrose (?) which acted as nucleophile any why ?

Here is an explanation for alpha- and beta- glucose.
What it doesn't say is that the alpha-glucose can convert to beta-glucose and back again. Once the linkage is formed, it cannot change from alpha to beta.

Alpha linkage has the oxygen (on the aldehyde or ketone) below the ring and the beta has it above the ring.

Down the page, it shows beta-Maltose. The label beta (for the right most oxygen) is unimportant since the link has to be alpha for it to be maltose. A beta link would result in a molecule of cellobiose.


Mixing solvents, or how does paint thinner work? I would like to understand how a mixture of solvents works together.

For example, there are two solvents combined, and each one alone can dissolve some material, but one acts faster.

How do they interact when used in combination on the same material?

An example would be paint thinner, for example a mixture of toluene and xylene.

One reason to use multiple solvents in a paint thinner is to make it dissolve more kinds of paints etc. (In German, this is called "universal thinner".)

But does the combination of multiple solvents somehow make it work better dissolving a single compound or mixture?

I had dissolved some polycaprolactone in that universal thinner above, and it seemed to dissolve quicker than in either toluene or xylene alone.
I do not know exactly what is in the thinner, so it does not mean much – but it made me wonder about this!

So, in which ways interact multiple solvents in the context of the example, and are there other mechanisms in general, not covered by the example?

Addendum:

For discussing a simple example, let's just assume the thinner contains only the two solvents above.

But for context, I looked up the Sicherheitsdatenblatt (MSDS) for my example thinner, which shows some more solvents on the list of ingredients:

Compound                                Mass %
----------------------------------------------
Toluol                                  15- 45
Xylol                                   10- 40
Ethanol                                  2- 15
Aceton                                   5- 20
Ethylacetat                              5- 20
n- Butylacetat                          10- 20
aromat. hydrocarbon mixture (C8-C10)     5- 15
aliph. hydrocarbons (C7-C9)              2- 10

I had dissolved some polycaprolactone in that universal thinner above, and it seemed to dissolve quicker than in either toluene or xylene alone.

Some of these universal thinners do contain a lot more than just toluene and the xylenes.

If you want to find out about the composition of these mixtures, a material safety data sheet (MSDS, in German: Sicherheitsdatenblatt), provided by the manufacturer gives some information:

As an example, take this MSDS by a swiss manufacturer, which lists the following ingredients:

\begin{array}{lr}
\textrm{compound} & \textrm{amount [%]}\\
\hline
\textrm{toluene} & 25-50\\
\textrm{xylene} & 10-25\\
\textrm{acetone} & 10-25\\
\textrm{propan-2-ol} & 2.5-10\\
\textrm{4-methyl-pentan-2-one} & 2,5-10\\
\textrm{2-methoxy-1-methylethylacetate} & 2,5-10\\
\end{array}

Particularly the latter two might be much better solvents for the polycaprolactone than toluene or xylene.


What is the difference between molarity and concentration? I learned these two words in school but still I don't understand what the difference is between them. From my understanding, molarity and concentration means the number of moles in a certain solution. So, what is the difference between the two?

Concentration is any measurement of the quantity of a solute that is present per unit of solution (in general, there are exceptions like moLality, which measures the number of moles of solute per kilogram of solvent).

Molarity is just one way to express concentrations, which, more specifically, is the number of moles of solute per liter of solution.

But you can give values for concentrations in many different ways, like for example, grams of solute per liter of solution, percentages (grams of solute divided by the grams of solution multiplied by 100), etc.

You can find a wide range of ways of measuring concentrations, including molarity here


IUPAC naming of quinol and hydroquinone This is benzene-1,4-diol. Its common name is quinol and hydroquinone. How is hydroquinone an accepted name since the functional group here is an alcohol and not a ketone? Is this the only such case or does it apply elsewhere too?

This is benzene-1,2,4-triol also known as hydroxyquinol. How does the position of alcohol influence the name of the compound here?

Pay attention to the prefix "hydro-", it can definitely change the meaning of the name.

Quinone is the common name given to cyclohexa-2,5-diene-1,4-dione:

However, hydroquinone is not a ketone. Hydroquinone is basically hydrogen added to ketone. As a result, it's not a ketone, but rather it's a phenol.

I must agree, the naming does cause confusion, as this isn't the systematic nomenclature. The common nomenclature doesn't follow any strict rules, and are often named after the compound they were formed from (such as this case). Don't rely on such names too much.


Can the thermodynamic predictions of redox reactions based on E and dG contradict each other? I am referring to the following three reactions. Two of these reactions are imaginary ($\ce{M}$ and $\ce{X}$ are imaginary).

$$\begin{alignat}{3}
\ce{X- + 2e- \;&<=> X^3-} \qquad &&E^\circ_{\ce{X^3-}/\ce{X-}}=0.9\ \mathrm{V}\qquad &&&\text{(a)}\\
\ce{2H+ + MO^2+ + e- \;&<=> M^3+ + H2O} \qquad &&E^\circ_{\ce{MO^2+}/\ce{M^3+}}=0.5\ \mathrm{V} \qquad &&&\text{(b)}\\
\ce{8H+ + MnO4- + 5e- \;&<=> Mn^2+ + 4H2O} \qquad &&E^\circ_{\ce{MnO4-}/\ce{Mn^2+}}=1.5\ \mathrm{V} \qquad &&&\text{(c)}
\end{alignat}$$

Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution. (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)

To find out possible reactions, following steps are taken.

Equations $\text{(a)}$ and $\text{(c)}$;

$$
\begin{align}
&{-}\text{(a)} \times 5 + \text{(c)} \times 2 : \\
\ce{&16H+ +5X^3- + 2MnO4- -> 2Mn^2+ + 5X- + 8H2O} \qquad E^\circ_1\qquad\text{(1)} \\ \\
&E^\circ_1 = 1.5\ \mathrm{V} + (-0.9\ \mathrm{V}) = 0.6\ \mathrm{V} \\
&\Delta G^\circ_1 = -nE^\circ_1F = -10 \times 0.6 \times F = -6F
\end{align}
$$

Equations $\text{(b)}$ and $\text{(c)}$;

$$
\begin{align}
&{-}\text{(b)} \times 5 + \text{(c)}: \\
\ce{&5M^3+ + H2O + MnO4- -> Mn^2+ + 5MO^2+ + 2H+} \qquad E^\circ_2 \qquad(2)\\ \\
&E^\circ_2 = 1.5\ \mathrm{V} + (-0.5\ \mathrm{V}) = 1.0\ \mathrm{V} \\
&\Delta G^\circ_2 = -nE^\circ_2F = -5 \times 1.0 \times F = -5F
\end{align}$$

My questions:

Are there any errors in above calculations? If so, please suggest corrections.
According to $E^\circ$ values, reaction $(2)$ is more feasible than reaction $(1)$. Am I correct?
According to $\Delta G^\circ$ values, reaction $(1)$ is more feasible than reaction $(2)$. Am I correct?
Is it useless to use $E$ values to predict the feasibilities of redox reactions? Should we always resort to $\Delta G$?
Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict?
If not how can these results be explained?

$\Delta G = -n {\mathcal F} \Delta {\mathcal E} $

$\mathcal F$ is Faraday's constant, which physicists currently believe has been constant throughout the universe for all time.  $n$ is the number of electrons transfered in an electrochemical reaction.  For a given reaction, $n$ is also constant in all places and times.

Thus, $\Delta G$ determines $\Delta {\mathcal E}$ and vice versa.

  Are there any errors in above calculations?

Yes, there must be.

  According to E∘ values, reaction (2) is more feasible than reaction (1). Am I correct ?
  According to ΔG∘ values, reaction (1) is more feasible than reaction (2). Am I correct ?

Both of these cannot be correct.

  Is it useless to use E values to predict the feasibilities of redox reactions ?

No, it is just as good and exactly as informative as $\Delta G$.

  Should we always resort to ΔG ?

No.  Feel free to resort to $\Delta G$ if you like units like kJ/mol instead of volts, but other than personal preference there is no reason to prefer one over another.

  Is it possible for predictions based on E values and ΔG values to contradict ?

No.

  If not how can these results be explained ?

UPDATE: Arithmetic error.

The choice of $n$ is arbitrary.  Once you pick $n$ for a given half reaction, it determines the choice of $n$ for the other half reaction, but that's it.  If you then consider a separate cell with separate reactions, you could write that new cell's equation with an arbitrary $n$.  The $\textbf{mol}$ in $\Delta G$'s units of $\frac{\textrm{ kJ}}{\textbf {mol}}$ is $\textbf{mol}$ of reaction, and thus the value of $\Delta G$ depends on how you write the reaction.  For example you could write equation (b) + (c) as

$$\ce{10M^3+ + 2 H2O + 2 MnO4- -> 2 Mn^2+ + 10 MO^2+ +4 H+}$$

Now, $n$ is twice as large, so $\Delta G$ is twice as large.  But $\Delta \mathcal E$ is unchanged.  This is because $\Delta \mathcal E$ has a basis of "$\textbf{mol}$" of electrons, not of moles of a reaction of arbitrary scale.  In reality the "$\textbf{mol}$" of electrons is converted to a different units (Coulombs) to make things more confusing.  (Remember, a volt is a joule per coulomb.)

So contrary to what I said above, the sign of $\Delta G$ but not the magnitude of $\Delta G$ is fixed, meaning that whether a reaction is absolutely spontaneous can determined by either $\Delta G$ or $\Delta \mathcal E$, but to compare the relative spontaneity of two reactions, either the $n$ values of the two reactions must be the same and then $\Delta G$ values can be used, or else without worrying about $n$ you could just look at $\Delta \mathcal E$.


Why do glow-in-the-dark substances dim gradually? Related: How do things glow in the dark?

Growing up, I had glow-in-the-dark stars on my ceiling and after being "charged" by having my lights on, they would phosphoresce for quite some time.

However, throughout the night, a star would gradually dim. I imagine that the cause of the stars becoming dimmer is that some electrons are relaxing down before others.

I thought the thickness of the star could be part of this as the more inner parts likely receive less light, so perhaps this bit would completely dim out before the surface of the star. However, my thought is that the star should either be completely "charged" or cast no light at all, like a switch.

The dimming effect is gradual. Why?

There are two potential questions here: one is why is dimming a slow drawn-out process (incoherent, unlike a "switch"). The second question is why different stars may appear to lose their intensity at a different rate.

The light-emission process behind your stars involves phosphorescence. The process involves relaxation of an electronically excited system during which it undergoes a "forbidden" triplet$\rightarrow$ singlet transition. For an individual system, light emission is controlled by the probability of the quantum-mechanical interconversion (or equivalently by the lifetime of the triplet state). It is a stochastic event described by Poisson statistics or "shot" noise.

However, the luminosity of one star represents the collective behavior of a large number of emitters (systems). One star, therefore, appears uniform in luminosity and its time-dependent luminosity decays exponentially.

Variability might be expected in the luminosity of different stars. This is not likely to be QM statistics since, as already noted, you are observing a very large ensemble of relaxing molecules, and one fairly uniform decay constant is expected for all stars, assuming the phosphorescent material is uniform. It seems more likely that luminosity differences among or within stars may occur due to size effects or differences in the concentration of phosphorescent material, or both$^\daleth$. Stickers that are thicker or containing a greater concentration of excited phosphorescent material, due for instance to variability in the manufacturing process, should appear to glow longer simply because they contain more emitters$^\dagger$.

$\dagger$ As helpfully pointed out in the commentaries, more emitters lead simply to the detection limit being reached later.

$\daleth$ As helpfully emphasized in yet another comment, and alluded to in the OP, it is the concentration of excited phosphorescent material that determines the luminosity, and excitation may also depend on the dimensions of the star.


Chemically removing rust without leaving any unwanted residues I have this iron pan that got rusty from not being properly dried. Scrubbing it I was able to get rid of most of the rust, but there's still some I just can't remove. I thought I could chemically remove it, but I myself am a bit rusty on the inorganic chemistry subject, so I couldn't think of any way of doing it without leaving any unwanted residue on the pan.

The rust is the brownish-red type, which according to Wikipedia consists of $\ce {Fe2O3·nH2O}$ and $\ce{FeO(OH)·Fe(OH)3}$. Most rust-removal "recipes" I found involve all sorts of acids, but I can't tell what the reaction(s) would be with $\ce {Fe2O3·nH2O}$ so I can't tell if there would be any residues left behind that I could not remove. Anyone with any experience on the matter?

I contacted the manufacturer and the pan is hand-made, just Iron and natural coating (fat).

The common DIY rust remover is cola of some sort -- most colas contain a significant quantity of phosphoric acid, which reacts with the iron oxide to form iron (III) phosphate which you can then scrub off. Any remaining iron phosphate should come off with time and/or more water, and is harmless in domestic quantities.

(Wikipedia:Phosphoric acid - Rust Removal)


Determining sulfate content in Mohr's salt I am practicing problems for my chemistry test and I am stuck on the following problem:

The value of $x$ in $\ce{Fe(NH4)2(SO4)2 ⋅ xH2O}$ can be found by determining the amount, in moles, of sulfate in the compound.
A $\pu{0.982 g}$ sample was dissolved in water and excess $\ce{BaCl2(aq)}$ was added. The precipitate of $\ce{BaSO4}$ was separated and dried and found to weigh $\pu{1.17 g}.$
Calculate the amount (in moles) of $\ce{BaSO4}$ in the $\pu{1.17 g}$ precipitate. And calculate the amount (in moles) of sulfate in the $\pu{0.982 g}$ sample of $\ce{Fe(NH4)2(SO4)2 ⋅ xH2O}.$

I was able to do the first part of the question:
$$n (\ce{BaSO4}) = \frac{\pu{1.17 g}}{\pu{233.38 g mol-1}} = \pu{0.005 mol}$$
But, I am unsure of the second part. Any help would be appreciated.

If there is $\ce{0.00501 mol}$ of $\ce{BaSO4}$ in the precipitate, this means that there was $\pu{0.00501 mol}$ of sulfate in the Mohr's salt solution and $\pu{0.00501 mol}$ of sulfate in $\pu{0.982 g}$ of Mohr's salt. And there was one half, that is $\pu{0.0025 mol}$ of $\ce{Fe(NH4)2(SO4)2.xH2O}$ in $\pu{0.982 g}$ of Mohr's salt. This allows you to calculate the molar mass of the salt, and to deduce $x$.


Does a complex that has no overall charge still have color? The iron complex $\ce{[Fe(H2O)6]^2+}$ is colored but I'm wondering, if a pair of chloride ion is inserted on its secondary valency to form [FeCl2(H2O)6], does the complex loose its color or not?

Tetrakis(triphenylphosphano)palladium(0) or $\ce{[Pd(PPh3)4]}$ is a common catalyst in organic chemistry that features a palladium(0) core with four triphenylphosphane ligands. It is a bright yellow, sometimes described as canary yellow. See the image below (taken from Wikipedia, where a full list of authors is available); however, I feel the picture doesn’t quite do the actual compound justice.

Colour is a feature of electronic transitions. Different orbitals have different energies. If the energy difference between an occupied and an unoccupied orbital happens to correspond to the energy of a photon ($E=h\nu=\frac{hc}{\lambda}$), the photon can be absorbed and the electron excited (subject to a couple of forbidden exchanges as usually defined in the Laporte rule and the spin rule). This absorption of photons is what we perceive as colour if $\pu{400 nm}<\lambda<\pu{700nm}$.

While energy levels of orbitals are affected by the overall charge of a complex, a correct energy difference can happen independent of partial or entire charge.


Single displacement reaction Will sodium metal, if mixed well with iron oxide, displace the iron in a chemical reaction? I know thermite involves the mixture of aluminium powder and iron oxide, and the single displacement reaction results in molten iron and aluminium trichloride. The other day I melted some sodium metal to the point of melting but reacting with the air. I wonder if either mashing up sodium with iron oxide and then igniting, or adding melted sodium to iron oxide will act like a thermite reaction.

Sodium metal will react with iron (III) oxide to produce sodium oxide and iron.

$\ce{6Na +Fe2O3 -> 3Na2O + 2Fe}$

This is also like a thermite reaction, so try to use the soam\e method as the reaction.


Lithium metal not reacting with water To start off with I want everyone to know that I'm not doing this for nay kind if drugs or anything illegal, just having some fun. So I know that lithium( Li) is the least alkaline metal in group one and that it will react with water like Na ,k, Fr, Rb and Cesium (Cs). So I decided to take apart some rechargeable D cell Li batteries to gather some lithium, the first time I tried it didn't react with the water at all. The second time I did all my work covering everything in mineral oil to make sure non if it reacted with the moister in the air ,then dried it. But when I placed that in water it still did nothing after 10 minutes. I even tried hot water but still nothing. Can someone give me an explanation if what's wrong with my lithium?
The batteries came out of a cordless drill that no longer took charge in case that has anything to do with it. Thank you ( I'm sorry for any spelling mistakes or chemical symbol errors, been quite a few years since I have been out of chemistry)

The lithium-ion batteries do not contain pure metallic lithium, nor do they contain any metallic alloy of lithium. They contain various intercalation compounds which are relatively harmless by themselves.

When a charged battery is short-circuited (by throwing it into the water or otherwise), it may react quite violently, but that's another story. Any battery would do the same, more or less. It's about physics, not chemistry (I mean, chemical reactions would occur as well, but they don't explain the explosive part; they would occur during the normal operation all the same).


Molecular structure of iodine nonoxide A question in an exam was as follows:

Iodine reacts with ozone gas to form a dark yellow solid $\ce{X}.$ Let the number of lone pair of electrons in un-ionised form of $\ce{X}$ be $m,$ number of lone pair of electrons in the anionic moiety of $\ce{X}$ be $n$ and the positive charge on the cationic moiety of $\ce{X}$ be $p$ units. Then what is the value of $\displaystyle\frac{m - p}{n}?$

This reaction of iodine with ozone is:
$$\ce{I2 + O3 -> I4O9 <=>I^3+(IO3^-)3}$$
Therefore the anionic moiety is $\ce{IO3-}$ and cationic moiety is $\ce{I^3+}$. However the first part of the question states un-ionised form. My assumption is that this is $\ce{I4O9}$ and they ask for the number of lone pairs on the molecular form.
The answer to this question takes into consideration that $\ce{I4O9}$ is an equimolar mixture of $\ce{I2O4}$ and $\ce{I2O5}$ which gives the answer to be $2.5$
However, from the abstract of J. Raman Spectrosc. 1985, 16 (6), 424–426:

The Raman spectrum of $\ce{I4O9},$ formed by the gas‐phase reaction of $\ce{I2}$ with $\ce{O3}$, has been measured. Freshly prepared samples of $\ce{I4O9}$ gave broad band spectra characteristic of an amorphous solid. Vibration bands at $780,$ $740,$ $620$ and $\pu{450 cm−1}$ were observed. It was established conclusively that $\ce{I4O9}$ is a distinct molecular species and not a mixture of $\ce{I2O5}$ and $\ce{I2O4}.$

If $\ce{I4O9}$ is a distinct molecular species, what is the molecular structure of $\ce{I4O9}?$

Regarding its ionic form, I don' think it is as simple as $\ce{I^3+(IO3^-)3}$. It is much more complex than that. From Ref.1:

This compound is likewise to be considered as an $\ce{I(III,V)}$ oxide and reacts with alkali hydroxide to give $\ce{I-}$ and $\ce{IO3-}$.
$$\ce{3I4O9 + 12OH- -> I- + 11IO3- + 6H2O}$$
Structurally, $\ce{I4O9}$ is possibly an iodate $\ce{I3O6+IO3-}$ (more
precisely $\ce{(I3O6+)_n.nIO3-}$) in which the isopolycation
$\ce{I3O6+}$ has a polymeric structure and is formulated as
$\ce{I_^{III}(I^{V}O3)2^+}$ consisting of twice as many pyramidal
$\ce{I^{V}O3}$ groups as square-planar $\ce{I^{III}O4}$ groups thus
$\ce{I4O9}$ would correspond to $\ce{I(IO3)2^+IO3-}$ which would
become $\ce{I(IO3)3}$.

Reference

Inorganic Chemistry By Egon Wiberg, A. F. Holleman, Nils Wiberg, Academic Press, 2001


Does this reduction mechanism of an diazonium via stannic chloride sense? Is this mechanism possible.
I didnt find any sources of a mechanism of this specific reaction.

Agreeing with Jan's comment, there are a few problems, and they all have to do with your reducing agent. That's not tin!

Tin is $\ce{Sn}$ and not $\ce{Zn}$ (zinc). Your mechanism is using $\ce{ZnCl2}$, zinc chloride. Let's say you made a typo and replaced $\ce{Sn}$ with $\ce{Zn}$. Your electron configurations match tin and not zinc. Then your mechanism is good. $\ce{SnCl2}$ is a reducing agent and could do this reaction. $\ce{SnCl2}$ is not stannic chloride, but stannous chloride. Stannic chloride is $\ce{SnCl4}$. Regardless $\ce{SnCl2}$ would work. $\ce{ZnCl2}$ would not work. The zinc in $\ce{ZnCl2}$ is already in zinc's highest common oxidation state. $\ce{ZnCl4}$ does not exist.


Determine rate equations for entire kinetic scheme I am working with the following kinetic scheme that represents pyrolysis of a biomass particle:

The paper that discusses this scheme provides the following rate constants:

K1 for the pathway of $\text{Biomass} \rightarrow (\text{Volatiles + Gases})_1$
K2 for the pathway of $\text{Biomass} \rightarrow (\text{Char})_1$
K3 for the pathway of $(\text{Volatiles + Gases})_1 \rightarrow (\text{Volatiles + Gases})_2 + (\text{Char})_2$

The rate equation provided in the paper for the conversion of biomass is
$$\frac{\mathrm d\,B}{\mathrm dt} = -(K_1 + K_2)\,B$$

The rate equation for the char 1 component is given as
$$\frac{\mathrm d\,C_1}{\mathrm dt} = K_2\,B - K_3\,C_1$$

And finally the rate equation provided for the char 2 component is
$$\frac{\mathrm d\,C_2}{\mathrm dt} = \delta\,K_3\,C_1$$

where $\delta$ is a deposition coefficient reported as 1.45 in the article. Note that all components of the system are on a mass basis such as $\mathrm{kg/m^3}$.

It is fairly straightforward to calculate the concentrations of $B$, $C_1$, and $C_2$ at each time step given an initial concentration of $B$. However, I would like to determine the concentrations of the volatiles and gases in the system too. Since it doesn't seem possible to calculate the individual volatile and gas components from the given scheme, I would like to determine the concentration of the (Volatile + Gases) group which could be labeled as G. So how would I develop the overall rate equations for B, C, and G?

$$
\frac{\mathrm d\,B}{\mathrm dt} = -(K_1 + K_2)\,B \\
\frac{\mathrm d\,C}{\mathrm dt} = K_2\,B + ? \\
\frac{\mathrm d\,G}{\mathrm dt} = K_1\,B + ?
$$

The Biomass-equation is:
$$\frac{\mathrm d B}{\mathrm d t} = -(k_1 + k_2) B$$
The (volatile + gases)$_1$ equation is:
$$\frac{\mathrm d VG_1}{\mathrm d t} = k_1 B$$
The (volatile + gases)$_2$ equation is equal to equation for (Char)$_2$:
$$\frac{\mathrm d VG_2}{\mathrm d t} = \frac{\mathrm d C_2}{\mathrm d t} = k_3 C_1$$
(It might be that it needs to be multiplied by 1/2.)

And the equation for (char)$_1$ is this:
$$\frac{\mathrm d C_1}{\mathrm d t} = k_2 B-k_3 C_1$$

The equations for the complete set of (char)s and (volatile+gases)s are then the sums of the solutions for $C_1$ and $C_2$ resp. $VG_1$ and $VG_2$.

For initial concentration of $B(0)=1$ and everything else $f(0)=0$, as well as $k_1=2$, $k_2=10$ and $k_3=3$, you should get something similar to this:


Are secondary alcohols more easily oxidised as compared to phenol? Which among the following compounds is oxidised most easily?

  2-propanol ($\ce{CH3CHOHCH3}$)
  phenol ($\ce{PhOH}$)
  ethoxyethane ($\ce{Et-O-Et}$)
  1-methyl cyclohexanol

The answer given in my workbook is option 1. It's obvious to me that options 3 and 4 can't be the right answer (ether, hindrance). But I am confused between options 1 and 2.

I have arguments that can support either of the 2 options. There's less hindrance on phenol so formation of an intermediate ester like chromate ester would be easier their as compared to secondary alcohol. However, phenol is losing aromaticity as well, therefore activation energy of oxidation should be higher as compared to secondary alcohol.

Since $E_\mathrm{a}$ matters more, we would say that secondary alcohol is oxidised more easily, is it?

Hindrance has little to do with this. Loss of aromaticity is a big deal, so the oxidation of phenol to quinone takes a high energy reagent (Ceric Ammonium Nitrate is a good one). Oxidation of a secondary alcohol by contrast is fairly easy.


Name for analog of salt hydrate with non-water solvents? Salts can bind with some water to form hydrates. For example, we can call $\ce{CoCl_2\cdot 6H_2O}$ cobalt chloride hexahydrate. There is no reason that water should be the only solvent that can bind with salts though. Dimethyl sufloxide is pretty polar and can dissolve some salts, so I might hypothesize that we could make salt crystal that has some bound into it, say $\ce{CoCl_2\cdot 6DMSO}$. I might call that cobalt chloride hexadimethyl sulfoxate, but that is totally made up. Is there a name for that sort of thing?

The corresponding rule in the Red Book (‘Nomenclature of Inorganic Chemistry – IUPAC recommendations 2005’) reads as follows:

  The names of the individual components of such a generalized addition compound are each constructed by using an appropriate nomenclature system, whether compositional, substitutive or additive. The overall name of the compound is then formed by connecting the names of the components by ‘em’ dashes; the proportions of the components are indicated after the name by a stoichiometric descriptor consisting of arabic numerals separated by a solidus or solidi. The descriptor, in parentheses, is separated from the compound name by a space.

$\ce{CaCl2.8NH3}$ calcium chloride—ammonia (1/8)

$\ce{AlCl3.4EtOH}$ aluminium chloride—ethanol (1/4)

$\ce{BiCl3.3PCl5}$ bismuth(III) chloride—phosphorus(V) chloride (1/3)

$\ce{2Na2CO3.3H2O2}$ sodium carbonate—hydrogen peroxide (2/3)


Creating 100mg/L aspirin solution I plan on creating 100 mL of 100mg/L aspirin solution by dissolving 10 mg of pure aspirin powder in 100 mL of buffer solution (of different pHs). Problematically, aspirin doesn’t tend towards dissolving in polar solution easily, so I thought about dissolving the 10 mg of aspirin in 1 mL of ethanol first, and then taking that ethanol-aspirin solution and adding it to 99 mL of the buffer solution.
I was wondering if this method seems reasonable/feasible. If not, are there any better suggestions?

According to Wikipedia, solubility of aspirin in water is $\pu{3 g/L}$ or $\pu{3 mg/mL}$ at $\pu{25 ^\circ C}$. Thus, it is not hard to make a $0.3\% (w/v)$ aquious aspirin solution (maximum concentration, which equals to $\pu{300 mg}/\pu{100 mL}$ at $\pu{25 ^\circ C}$). Therefore, your target, $\pu{10 mg}/\pu{100 mL}$ at $\pu{25 ^\circ C}$ can be easily achieved without any modifications. You have not mentioned the what different $\mathrm{pH}$ values of buffers you are planning to use. Since $\mathrm{p}K_\mathrm{a}$ of aspirin is about $3.5$ and the amount of it above your target value would dissolve in water (so final $\mathrm{pH}$ would be around $2.12$ for $\pu{300 mg}/\pu{100 mL}$ soution), it is guaranteed to dissolve your desired amount in any buffer with  $\mathrm{pH}$ above $2.12$.
None the less, dissolve aspirin first in small amount of ethanol and diluted it with desired buffer is not bad idea. As a matter of fact, in aspirin preparation lab, we use ethanol to recrystallize the final product, therefore, would not arise any problem on dissolution. Thus, this would be an ideal alternative procedure.


Are all strong bases alkalis? In my text book, it says that strong bases are alkalis because they fully dissociate in aqueous solution. But I'm not too sure of this. Can you not have a weak alkali? And can you not have a strong base which is not an alkali?

Thank you in advance :)

Are all strong bases alkalis?

No, quaternary amine hydroxides such as tetramethylammonium hydroxide can also be strong bases.

  Can you not have a weak alkali?

According to "Acids and Bases" by Kristi Lew, beryllium hydroxide is the one and only exception to the "alkali bases are strong" rule.

I would add that many of the pKb values of the alkali hydroxides are very close to zero.  So you should only think of the them as strong in the sense of strong means pKb near or less than zero.  If you think of strong as full dissociated even in a concentrated solution then they (NaOH for example) would not be strong by that definition.

Also, for the alkaline earth metal hydroxides (which have two hydroxides per metal ion), it is only the first pKb that is strong.


Carbylamine Reaction In my lab manual, it is given that alcoholic KOH should be the base used in carbylamine test. However as it was not available, I used 10℅ NaOH. The test results came out positive. (Yes, it was actually a 1° amine) Is there any reason to prefer alcoholic KOH over NaOH?

My memory tells me that KOH is more soluble in ethanol than NaOH. The chemistry is the same, just more efficient at the higher concentration.


Are we capable of deliberatly creating DNA from inorganic compounds? Good day
The  Miller–Urey experiment showed that organic compound can naturally form from inorganic substances. What i wish to know is this: Could we as humans deliberately create a DNA molecule starting with only inorganic substances?

Yes we could.
It wouldn't be worth the effort since all the building blocks are available from biological sources, but if you really wanted to you could have a chemist prepare nucleotides from inorganic substances.
Commercial labs routinely build long strands of DNA from individual nucleotides. You can tell them what sequence you need and they will make it for you. There's a Wikipedia article on how it works.


What do you call the number of atoms in a molecule? I just bought a molecular modeling kit for my toddler ("Snatoms").

To use the kit to play with her, I want to start out by memorizing a list of all molecules (isomers) fitting these criteria:

only contains C, H, and/or O
contains a maximum of 6 atoms

I started out by noodling around with Lewis diagrams and wikipedia... but then I ran into the various isomers of C2H2O2:

I wanted to do this search on a chemical database (like chemspider.com) but I couldn't find a criterion that would filter the search by number of atoms in the molecule (isomer).

Is there a name for this "number"?

The "number of atoms"'s name is... the number of atoms. I am not aware of any special word for that. Databases may not let you search/filter that kind of criterion, i.e. the total number of atoms, but usually let you search by "chemical formula" which is, in your case, C2H2O2.

I have just tried with Chemspider by typing "C2H2O2" in the search box and it disappointingly gave me just 3 results, i.e. glyoxal (OHC-CHO, not in your list), 42879-41-4 and 16005-17-7.

DID YOU KNOW? Just to expand the scope, did you know that the last digit in CAS# is actually a check digit? It helps any software checking that you did not mistyped the number.


The decarboxylation of phenylacetic acid via addition of acid I came across a question where the following compound except with two methyls, ortho and meta to the nitro group, is dissolved in methylene chloridde, and then treated with trifluoroacetic acid.

It is given that $\ce{CO2}$ gas is given off in a decarboxylation reaction. I cannot find any sources about this. The solution explains that:

  Treatment of a carboxylic acid with acid results in decarboxylation, and the evolution of $\ce{CO2}$, especially if the resulting compound contains a benzylic or allylic carbon, as is the case here.

First of all, what mechanism would allow for acid catalyzed decarboxylation? I know how decarboxylation of carboxylic acids with beta-carbonyls with heat works, and that acid can help that but that doesn't seem to have relevance here.

Secondly, why does having a benzylic or allylic carbon confer any special stability to promote the reaction?

The decarboxylation of carboxylic acids having allylic or benzylic carbons can also be explained by what may be called the vinylog concept. One of the most common decarboxylations is that of β-keto acids as illustrated by the decarboxylation of β-keto acid 1 to afford cyclohexanone 2. Hagemann's ester (3) is a γ-keto ester, which upon aqueous acid hydrolysis gives acid 4. This acid suffers decarboxylation to afford 3-methylcyclohex-2-en-1-one (5). The red double bond in 4 simply extends the properties of a β-keto ester to the γ-position in a vinylogous (allylic?) fashion.
Nitrobenzene 6, which may be the compound in question (if not, ignore the methyl groups), is doubly vinylic (benzylic?) as indicated by the two red double bonds in structure 6. While this compound does not have a keto group in conjugation with the aromatic ring, the nitro group functions as a viable substitute for a keto group. Protonation of the N=O bond on oxygen leads to decarboxylation and rearomatization to nitrobenzene 7.
The green σ-bonds in structures 1, 4 and 6 must have overlap with the π-framework of the carbonyl (in 1), the γ-position of enone 4 and the aromatic ring of nitrobenzene 6 for decarboxylation to occur.
Finally, nitroacetic acid, as its magnesium salt, decarboxylates at 80 oC to provide nitromethane and CO2.