Who devised the term elements? Who devised the term elements? Who was the one who brought the term elements to us? Also please give a bit information on background of the man.

One answer is that the ancient Greeks, specifically Plato, devised the term "element".  Quoting from wikipedia article on Classical elements:

  Plato seems to have been the first to use the term "element (στοιχεῖον, stoicheion)" in reference to air, fire, earth, and water. The ancient Greek word for element, stoicheion (from stoicheo, "to line up") meant "smallest division (of a sun-dial), a syllable", as the composing unit of an alphabet it could denote a letter and the smallest unit from which a word is formed. A similar alphabetic metaphor may be the origin of the equivalent Latin word elementum (from which the English word comes), possibly based on the names of the letters 'l', 'm', and 'n', though the validity of this idea is debated.

If you had a more modern, chemical definition of "element" in mind for your question, then another possible answer is Robert Boyle.

Quoting from wikipedia again, this time from its article on chemical elements:

  In 1661, Robert Boyle proposed his theory of corpuscularism which favoured the analysis of matter as constituted by irreducible units of matter (atoms) and, choosing to side with neither Aristotle's view of the four elements nor Paracelsus' view of three fundamental elements, left open the question of the number of elements. The first modern list of chemical elements was given in Antoine Lavoisier's 1789 Elements of Chemistry, which contained thirty-three elements, including light and caloric.

I can't resist adding a reference to my favorite English-language modern chemistry text that omits the word "element" entirely.

  For most of its being, mankind did not know what things are made
  of, but could only guess. With the growth of worldken, we began
  to learn, and today we have a beholding of stuff and work that
  watching bears out, both in the workstead and in daily life.

  The underlying kinds of stuff are the firststuffs, which link
  together in sundry ways to give rise to the rest. Formerly we
  knew of ninety-two firststuffs, from waterstuff, the lightest and
  barest, to ymirstuff, the heaviest. Now we have made more, such
  as aegirstuff and helstuff. [...]

If the Normans had never invaded England in 1066, we might not have "elements" at all, but "firststuffs" instead.  Either way, it's just a debate about the right translation for "stoicheion", a concept which Plato seems to have given us.


How many valence electrons do elements in the d block have? I just noticed that on the periodic table it doesn't say how many valence electrons there are for each column in the d block.

How do I find out how many valence electrons elements in the d block have?

The simplest answer is: d-block elements have number of valence electrons equal to their group number, which is equal to the number of electrons in the "valence shell". This works if you are using the definition of valence shell to be the outermost shell.

But it is wrong to apply the concept to determine the valency of the element, because that varies with the compound it is present in. For example, iron can show two valencies, $2$ and $3$ in the compunds iron(II) sulphate ($\ce{FeSO4}$) and iron(III) sulphate ($\ce{Fe2(SO4)3}$) respectively. But from the above definition, iron has $8$ valence electrons.

As a matter of fact, this applies even to many elements outside d-block too, for example chlorine typically shows a valency of ($-1$) in compunds like $\ce{NaCl}$, but it can also show a different valency (more appropriately, oxidation state) in ions like perchlorate anion ($\ce{ClO4^-}$), where its valency is $\mathbf 7$ (while the oxidation state is $\mathbf{+7}$, notice the difference that there is no plus or minus sign in valency).


Predicting colors of aqueous solutions While going through my AP chem prep book again, I came across questions asking about the color of certain aqueous solutions. For example:

Which of the following substances will produce a colorless aqueous solution?
A) $\ce{Zn2(NO3)2}$
B) $\ce{Cu(SO4)}$
C) $\ce{K2Cr2O7}$
D) $\ce{Co(NO3)2}$

I know that zinc compounds are generally colorless, so I picked that (and it was correct), but I have no idea how to predict the colors of the rest of the compounds.
So, to answer this question (and questions similar to this), would I need to use pure memorization of the common elements that produce color?
If so, where can I find information about these compounds and the color they produce?

There is no simple way to predict beforehand as far as I know. This is a general knowledge type of chemistry question, you will have to know or memorize the colors of common anions and cations. It will be much easier, if you do so group wise and know the colors of the first row of transition element ions.

Any object which absorbs a certain portion of visible light's spectrum, will appear colored to us. For colored solutions, we will always consider absorption of light in the visible region. So if you see a green solution, it is not reflecting green light, it is absorbing a certain portion of the visible spectrum and the transmitted light [from the solution] appears green to us. Keep this concept in mind because a lot of students have this misconception.

You should know that sulfate ion is colorless but copper (II) ion is blue so most likely copper sulfate solution is blue. In part c, you should know the color of potassium ion and dichromate ion and then guess the color. Do the remaining exercise for others.

It will be very difficult to predict if both ions are colored because it is not like that we are mixing two paints, a yellow paint with blue paint will show a green color. You will have to study the absorption spectrum of that salt. The theory of color of ions and that of solids is quite complex. It was developed by physicists. You can start from crystal field theory if you wish to see how the color of transition metal ions is explained but I will not recommend that you explore it if you are in school.


Why do orthoesters react with Grignard reagents, but acetals and ketals don't? My book mentions the synthesis of aldehydes and ketones using alkyl ortho esters. It's called Bodroux-Chichibabin aldehyde synthesis.

Now, later, the book mentions this:

R- displaces EtO-, giving an acetal or ketal, which are unreactive with Grignard reagents, so a secondary alcohol is not formed.

Why do orthoesters react with grignard reagents, but acetals and ketals don't? They have essentially the same functional groups, just two instead of three.

Grignard reactions are performed in ether or THF.
There you have only one -OR group on a carbon and, since it is the solvent, it is not reacting with it.
This shows that not always a -OR group is substituted in a Grignard reaction.

When you have three -OR's on one carbon, the formal positive charge ($\delta+$) on it is higher than when there are only two -OR's.
Therefore, the reaction proceeds on orthoesters and not on acetals or ketals.


What is the function of tosylchloride in the synthesis of an ether? Considering the following reaction:

Today my teacher explained this reaction as follows:
The electron pair of $\ce{OH}$ (in the first compound) will "attack" the $\ce{S}$, making the $\ce{Cl}$ "jump out". This is something I didn't understand clearly:
Why does the electron pair attack the $\ce{S}$ and not one of the oxygens, or even the ethanol? Is there any particular reason?

I'm not a chemical engineering student (I study engineering physics) so I don't have a great background in organic chemistry.

There are probably only two principal mechanisms in organic chemistry, one being electrocyclic reactions (not applicable here) and one being nucleophile/electrophile reactions. In the latter, which is what is happening in your case, you must always look for the group with either the highest electron density or the ‘best’ (I know, a hand-wavery term) nucleophilic properties — this group is the electrophile — and for the group with the lowest electron density or the ‘best’ (same hand-waveryness) electrophilic properties — termed the nucleophile.[1]

The first step to determining this would be to check electronegativities and oxidation states. Groups with high electronegativities and low oxidation states are typically nucleophiles and vice-versa. In your example, all alcoholic oxygens have a much higher electronegativity than all atoms they are connected to (carbon and hydrogen), are in the $\mathrm{-II}$ oxidation state and would thus act as nucleophiles. The oxygens and the chloride of tosyl chloride also have a larger electron density than their surroundings. For reasons outlined above that I cannot go into details with, they will not react nucleophilically, though.

The carbon atoms which are directly attached to the alcoholic oxygens all lack electron density due to the neighboring electronegative oxygen. They are electrophiles. And also the central sulfur of tosyl chloride — which not only features a $\mathrm{+VI}$ oxidation state but should also be drawn with a double positive formal charge to satisfy the octet rule[2] — is an electrophile.

To synthesise the ether from the two alcohols, we would now somehow need either of the two alcoholic nucleophiles to attack the other carbon electrophile and displace the other alcoholic hydroxy group — this reaction is termed $\mathrm{S_N2}$ reaction or bimolecular nucleophilic substitution. However, it is not likely to occur directly: a direct side-product would be the removal of a hydroxide ion, which is termed a bad leaving group.[3] We need to turn it into a better leaving group.

This is where tosyl chloride comes in. But first another little excursion: Notice the little 1. and 2. in your reaction scheme? They stand for first and second, implying that we are dealing with two separate reactions interceded by a workup and typically purification of an intermediate. So first we react the 1-phenylpropan-2-ol with tosyl chloride, collect the product and then react said product with ethanol. As I outlined before, a pretty good electrophile in tosyl chloride is the sulfur atom, and with the adjacent chloride, we have a good leaving group. So under a nucleophilic substitution,[4] this chloride can be displaced (along with a proton) leading to our intermediate 1-phenylprop-2-yl tosylate.

Technically we would now need to make a fresh analysis of nucleophiles and electrophiles. But it may suffice to note the key differences: there is no longer a good leaving group on sulfur (the alcoholate which would be a leaving group again is the conjugate base of a very weak acid), but if ethanol now attacks the carbon adjacent to what is now a tosylate group we have the tosylate anion as a good leaving group. In a second bimolecular nucleophilic substitution, tosylate and a proton are displaced and the ether is formed.

Notes:

[1]: Actually, which group is nucleophilic and which is electrophilic depends on the frontier orbitals, and that can also be used to unify both reaction types into one general scheme, but that analysis is very hard to do a priori.

[2]: Organic chemistry’s preference to write sulfonic sulfur atoms as doubly bonded to two oxygens, creating a dodectet gained popularity probably because of one of Pauling’s not-so-bright moments. However, evidence points towards d-orbitals (which would be required for the dodectet) not taking part in bonding for non-metals and thus the charge separation depiction being a more accurate description of the truth. This is documented a lot across this site.

[3]: Again, I won’t go into detail here when leaving groups are good or bad, but just leave as one principal pointer that conjugate bases of strong acids are bad leaving groups. The conjugate base of $\ce{OH-}$ is $\ce{H2O}$ which is a very weak acid; thus, hydroxide is a very bad leaving group.

[4]: Technically, this is no longer an $\mathrm{S_N2}$ reaction. Since sulfur is participating as the central, electrophilic atom, the transition state is more long-lived involving a four-electron three-centre bond. The difference is marked enough to call this bimolecular nucleophilic heavy atom substitution or something along the lines. It will usually not be referred to as an $\mathrm{S_N2}$ reaction.


Two different methods to express the equilibrium constant give different results $\pu{2.0M}$ of $\ce{I2}$ and $\pu{2.0M}$ of $\ce{Br2}$ are initially present in a reaction mixture. Calculate the equilibrium concentration of $[\ce{I2}]$ , $[\ce{Br2}]$ and $[\ce{IBr}]$ at $\pu{425K}$. $K_c=100$
Equation: $\ce{I2 + Br2 -> 2IBr }$

1st method

Equation:$\ce{I2 \ \ \ + \ \ \ Br2 ->\ \ \ \ 2IBr}$
Initial: $\ \ \pu{2.0M}\ \ \ \ \ \ \pu{2.0M} \ \ \ \ \ \ \ \ \ \ 0$
Change:   $\ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ 2x$
Equilib:     $\ \ 2-x \ \ \ \ \ \ \ 2-x \ \ \ \ \ \ \ \ \ \ 2x$
$K_\mathrm c =\frac{(2x)^2}{(2-x)^2}=100$

2nd method what if I separate $2\ce{IBr}$

Equation: $\ce{I2 \ \ \ + \ \ \ Br2 ->\ \ \ IBr \ \ + \ \ \ IBr }$
Initial:     $\pu{2.0M}\ \ \ \ \ \ \ \pu{2.0M}\ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ 0$
Change:    $\ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ x$
Equilib:       $2-x \ \ \ \ \ \ 2-x \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ x $
$K_\mathrm c =\frac{(x)(x)}{(2-x)^2}=100$
So which one is actually right and which one is wrong? Why?
The idea to use the second method came to my mind because of the the definition:

The equilibrium constant is a ratio of the concentration of the products to the concentration of the reactants

If $\ce{aA +bB \ —> cC + dD}$ (1)
$K_\mathrm c = \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$
If I separate : $a\ce{A} = \ce{A}+\ce{A}+...+\ce{A}$ ($\ce{A} a$ times )
Same thing with $b\ce{B}, c\ce{C}$ and $d\ce{D}$, the equation (1) becomes:
$(\ce{A}+\cdots+\ce{A})+(\ce{B}+\cdots+\ce{B}) \ce{—>}(\ce{C}+\cdots+\ce{C})+(\ce{D}+\cdots+\ce{D})$
$K_\mathrm c = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}$=$ \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$ (Proved)
So, by using the second method, how and why is it not right ?

So which one is actually right and which one is wrong? Why?

The second method is wrong.  In your second method, you've written: $\ce{I2 \+ Br2 ->IBr  + IBr }$ and then assigned the concentration $x$ and $x$ to each of the two (identical!) $\ce{IBr}$'s. Till now, it's fine.

The $K_\mathrm c$ has been used by you as:

$$K_\mathrm c=\frac{[\ce{IBr}][\ce{IBr}]}{[\ce{I2}][\ce{Br2}]}$$

which is okay but a better way to write it would have been

$$K_\mathrm c=\frac{[\ce{IBr}]^2}{[\ce{I2}][\ce{Br2}]}$$

as you've taken the single concentration term for $\ce{IBr}$ (=$2x$) from the correct formula, split it up into two parts ($x$ and $x$), and then used those two parts separately in $K_\mathrm c$, as if there were not one but two distinct $\ce{IBr}$'s being formed! It is just confusing.

Anyways, the first formula will still work. The main problem here is that $[\ce{IBr}]=\color{blue}{\text{total }}\text{concentration of}~\ce{IBr}$. When you put any reactant/product inside the square brackets, you're supposed to write their total concentration, even if they have been split up in parts. So, when in the derivation of $K_\mathrm c$, you write:

$$K_\mathrm c = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}= \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$$

for the reaction: $$\ce{aA + bB —> cC + dD}$$

each of those individual $[\ce{C}]$'s will refer to the total concentration of $\ce{C}$. That means, as $\ce{C}$ was separately written $c$ times by you in the reaction, the final $[\ce{C}]$ that you use will be the sum of all those $c$ $[\ce{C}]$s.

So, in the end of your second method, while applying the correct formula for $K_\mathrm c$, you must add those $x$ and $x$ together, and then write:

$$K_\mathrm c=\frac{(x+x)\times(x+x)}{(2-x)\times(2-x)}$$

into the formula you were using.


Are there natural conditions that could enable the formation of noble gas compounds? Noble gases were considered to be inert until compounds that include them, such as xenon trioxide (as an example) were found.

My question is, what natural conditions allow the formation of noble gas compounds?

As noticed in the comments, there are probably no natural conditions that would allow formation of noble gas compounds; although it should be noted that definition of 'natural' conditions should be specified. Lightning bolts surely carry enough energy to force such a reaction, but it is impossible for them to meet right mixture of gases, due to scarcity of both noble gases and super-strong oxidants and abundance of other substrates; as noted in comments (again), substrate able to create a compound with noble gas will create it with everything else first, and noble gas last.


Cs Point Group's "Horizontal" Reflection Plane Why is it that $C_s$ character tables always list a $\sigma_h$ operation despite the definition of $C_s$ not necessitating the reflection plane be horizontal?

In fact, how can the reflection plane even be characterised without a principal rotation axis to base it on?

There is always a principal rotation axis, even if it's only a $C_1$ axis. As there are an infinite number of these one can always be found perpendicular to the reflection plane, and as horizontal planes take precedence over veritcal ones, the operation is listed as $\sigma_h$.


Reaction of H3BO3 and NaOH I have been searching around and have found varying answers to the question. What exactly would reaction of $\ce{H3BO3}$ and $\ce{NaOH}$ give? Shouldn't it be a neutralisation reaction as $\ce{H3BO3}$ is an acid and $\ce{NaOH}$ is a base?

This isn't something you could know except by experiment, but boric acid doesn't exist as solely $\ce{H3BO3}$ in solution. In addition to being an Arrhenius acid, $\ce{H3BO3}$ is also a Lewis acid. This means it is capable of accepting a lone pair, and will undergo the following reactions in aqueous solution:

$$\ce{B(OH)3 + H2O <=> B(OH)4- +  H+}\tag1\label{Eq1}\\$$

$$\ce{4B(OH)4- + 2H+ <=> B4O7^{2−} + 9H2O}\tag2\label{Eq2}\\$$

By combining \eqref{Eq1} and \eqref{Eq2}, we get the full reaction:

$$\ce{4B(OH)3 <=> B4O7^{2−} + 2H+ + 5H2O }\tag3\label{Eq3}\\$$

Boric acid does not produce tetraborate exclusively, however, and many other species species are also present in solution. The system is actually so complex that it remains difficult to characterize, one paper mentioning that there are some ten different equilibrium reactions in solution$^{[1]}$. In performing an acid-base calculation, however, it is okay to simply treat it as a solution of $\ce{H2B4O7}$, neutralizing according to the equation:

$$\ce{H2B4O7 + 2NaOH <=> Na2B4O7 + 2H2O}\tag4\label{Eq4}\\$$

Titration confirms that $\ce{H3BO3}$ reacts with $\ce{NaOH}$ in a $\text{2:1}$ ratio, and crystallization of the product confirms $\ce{Na2B4O7}$ is produced.

$[1]$ Trejo G., Frausto-Reyes C., Gama S.C., Meas Y., Orozco G. Raman Study of Benzylideneacetone on Silver. Int. J. Electrochem. Sci. 2012, 7, 8436–8443.


The time scale of calculations in computational chemistry Is it possible to determine or get a range of the time required for a specific calculation with knowledge of the type (MM or QM), class (semi-empirical, ab initio or DFT), the degree of theory (MP, CI, etc.) and the basis set (STO-3G, 3-21G, 6-31G(d), etc.)
 used in the calculation and the computer ability (processor, RAM, etc.)?

No, that is not possible. What is possible is to estimate how much more time you need to calculate the electronic energy if you increase the system size by some factor. For any method which uses LCAO-MO expansion, it is the number of basis functions $m$ which primarily determines the computational cost, so that it is usually used as a measure of the system size. For instance, HF (without any tricks like density fitting, or taking the spatial symmetry into account) scales approximately as $m^4$, i.e. if you double the number of basis functions by using a basis which is twice as large, the computational cost will be approximately $2^4=16$ times larger. But that is it: estimating the time for a calculation using the well-known scaling behavior is possible only if you already know some timings on the same setup. And besides, this is all about just the electronic energy, which is quite often not the end of the story.


Reaction rate ambiguity Let’s suppose the following reaction:

$$\ce{ 2 H2O2 —> 2 H2O + O2}\tag{1}$$ with reaction rate $v$

Now let’s suppose the same equation with different balance:

$$\ce{ H2O2 —> H2O + 1/2O2 }\tag{2}$$

I assume that the reaction rate must be the same ($v$), right? It should not change with stoichometry.

However, depending on the balancing, I get different rates for each substance as shown in the equations below, different $k$ constant and so on.

$$\begin{align}v_1 &= -\frac12 \frac{\mathrm d [\ce{H2O2}]}{\mathrm dt} = \frac12 \frac{\mathrm d[\ce{H2O}]}{\mathrm dt} = \frac{\mathrm d[\ce{O2}]}{\mathrm dt}\\[1em]
v_2 &= \frac{\mathrm d [\ce{H2O2}]}{\mathrm dt} = \frac{\mathrm d[\ce{H2O}]}{\mathrm dt} = 2\cdot \frac{\mathrm d[\ce{O2}]}{\mathrm dt}\end{align}$$

Any help?

I assume that the reaction rate must be the same (v), right? It should not change with stoichometry.

No, the reaction rates changes when you double all coefficients. Nothing real changes though (i.e. the rate of disappearing $\ce{H2O2}$ etc.).

You have a similar case with equilibrium constants: when you double or halve all the coefficients, the new equilibrium constant will be the square or the square root of the original one. The molar reaction enthalpy will also change when you change coefficients. It is a choice, and it still correctly describes the reaction.

  Any help?

There is no need for help. If you use either chemical equation format to calculate a measurable quantity (changes in concentration, equilibrium concentrations, enthalpy change for a specific amount that reacts, for the three topics respectively), you will get the same result. It is almost like the choice of describing a volume using liters or milliliters - it does not change the result even though it might look different to someone not used to the milli prefix..


Reagent choice in the formation of Wittig reagent During the formation of Wittig reagent, BuLi is used as it is a VERY strong base (pKa = 50). According to Wikipedia, however, sulfur ylide is 'especially conveniently' made from dimsyl anion (pKa = 35), which is a relatively weaker base although it is also used for preparation of Wittig reagent.
If there is a stronger base is allowed (BuLi), why is dimsyl anion used 'conveniently'? To prevent side reactions or just by what is available at hand?

Typically phosphorus ylids are made from PPh3. What is the purpose of having 3 phenyl groups linked to the phosphorus atom in Wittig reagent? Can't be as small as sulfonium ylide molecule? I have no idea what is the use of adding 3 bulky and sterically-hindered groups in the ylide.

I am fairly sure that for both, there are practicality issues at hand.
This is almost certainly so for the choice of phosphine. Even though PPh3 is absolutely terrible from an atom economy point of view, and even though PPh3O can be a real pain to remove after the reaction is done, at least it's a nice white solid which doesn't smell. On the other hand, the simpler alternative PMe3 is a volatile, pyrophoric, and extremely stinky liquid. PPh3 is also substantially cheaper than other alternative phosphines. (There are similar considerations for the Mitsunobu reaction which I have some experience with.)
As for the base, $\ce{BuLi}$ is just one of the most accessible strong bases to use. I don't think you can buy dimsyl sodium from the average chemical supplier. My suspicion is that the main reason why dimsyl anion is somewhat associated with the sulfur ylids is because E. J. Corey and his postdoc M. Chaykovsky popularised them together (e.g., Ref.1). Also note that $\ce{NaH/DMSO}$ mixtures are associated with significant explosion hazards (Ref.2).
Others with more experience will be able to give fuller answers.
References:

E. J. Corey, Michael Chaykovsky, “Methylsulfinyl Carbanion ($\ce{CH3-SO-CH2-}$). Formation and Applications to Organic Synthesis,” J. Am. Chem. Soc. 1965, 87(6), 1345–1353 (https://doi.org/10.1021/ja01084a033).
Qiang Yang, Min Sheng, James J. Henkelis, Siyu Tu, Eric Wiensch, Honglu Zhang, Yiqun Zhang, Craig Tucker, David E. Ejeh, “Explosion Hazards of Sodium Hydride in Dimethyl Sulfoxide, N,N-Dimethylformamide, and N,N-Dimethylacetamide,” Org. Process Res. Dev. 2019, 23(10), 2210–2217 (https://doi.org/10.1021/acs.oprd.9b00276).


Does ethylene glycol sink in water? I’m a little concerned that there is a far stronger mix of antifreeze in my car's coolant system than I am aware. I check the antifreeze level in my car (using a refractometer) when the engine is cold (after being left overnight).

During the night, can the antifreeze settle to the bottom of the cooling system, thus leaving a weak mixture of water and antifreeze within the coolant reservoir up top, where I take a sample with the refractometer?

Ethylene glycol is completely miscible with water in all proportions (as the commenters point out). Therefore, once mixed, the glycol will not separate from the water, ever.

The only exception is if it gets so cold the mixture starts to freeze. Freezing will cause some degree of separation as the water will solidify first. But if it is that cold, separation will be the least of your problems.


Reduction of a ketone in the presence of an aldehyde An aldehyde group can easily be reduced to an alcohol using the mild reducing agent $\ce{NaBH4}$ at low temperature in the presence of a ketone group. This chemoselectivity is due to the fact that an aldehyde is less sterically hindered compared to a ketone, which makes it react faster than the ketone.

Is it possible, however, to chemoselectively reduce a ketone in the presence of an aldehyde using $\ce{NaBH4}$ as reducing agent?

The best strategy is probably to chemoselectively protect the aldehyde before the borohydride reduction.  Protecting groups are derivatives of certain moieties in a molecule that are far less reactive under most conditions, but which can be easily removed under very specific conditions to regenerate the original moiety.

For carbonyl compounds -- this includes both aldehydes and ketones -- the most common protecting groups are all introduced by ketalization/acetalization.   Common examples are dimethyl acetals, 1,3-dioxolanes, and 1,3-dioxanes.

However, in general, aldehydes are far more prone to acetalization than ketones are to ketalization.  That offers a means to chemoselectively protect aldehydes.  Acetals such as 1,3-dioxanes are stable to borohydride reductions, but, as you note in your question, ketones are sensitive.

Poking around reveals several potential strategies for selective chemoprotection of an aldehyde.  Here are two examples.

Tetrabutylammonium Tribromide (TBATB) as An Efficient Generator of HBr for an Efficient Chemoselective Reagent for Acetalization of Carbonyl Compounds:

  This convenient, mild, chemoselective method allows acetalization of an aldehyde in the presence of ketone...

Chemoselective Protection of Aldehydes in the Presence of Ketones Using RuPVP Complex as a Heterogeneous Catalyst

Putting it all together, a recipe for selective borohydride reduction of a ketone in the presence of an aldehyde would be something like:

Protect the aldehyde.  The best choice is probably as a 1,3-dioxane acetal by reacting your compound with 4 equivalents of 1,3-propanediol, 1 equivalent of triethyl orthoformate, and 0.01 equivalent of tetrabutylammonium tribromide.  Under suitable conditions, the ratio of protected aldehyde (desired) to protected ketone (undesired) should be >50.
Borohydride reduction of the aldehyde-protected compound.  Only the unprotected ketone will react.
Deprotection.  Acetals can be removed in acidic conditions.  A number of catalysts that will effect deprotection without requiring acid have also been developed.


Is preference given to long alkyl group or branched cycloalkane? Could you tell me whether it will be 1-(1-ethylcyclopropyl)hexane or 1-ethyl-2-hexylcyclopropane ? I don't know the rule or the preference order.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), two methods are possible for naming systems composed of rings and chains: either the ring or the chain can have seniority. To construct the preferred IUPAC name (PIN), the ring is selected as the parent structure.

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, the preferred IUPAC name (PIN) of the compound that is given in the question is 1-ethyl-2-hexylcyclopropane.

In general nomenclature, however, the context may favour the chain, so that the parent structure has the greater number of atoms in the principal chain. The corresponding systematic name for the compound given in the question is 1-(2-ethylcyclopropyl)hexane.

Note that according to the obsolete 1979 recommendations, a hydrocarbon containing a small ring attached to a long chain was generally named as a derivative of the acyclic hydrocarbon. Many books are based on these nomenclature rules and thus may expect an answer that favours the chain.


Recovering and casting indium I bought some scrap indium online, and I melted it down into an ingot on my kitchen stove, but my losses were very high as molten indium has a surprising ability to stick to things.

I have a lot of indium stuck to a stainless steel pot, and, unfortunately, I dropped a glass dish that had indium in it into the steel pot and it broke. So now I have broken glass that is covered in indium, and it really likes to stick to glass as well.

Is there a good way to recover the indium? Perhaps increase its surface tension so that it beads up and coalesces? I've seen something similar with sodium metal in hot oil; adding a few drops of isopropanol makes it coalesce. Is there a similar method for indium?

Also, is there a good material to cast indium in? I see that teflon melts at a higher temperature than indium, so maybe a teflon coated muffin tray? Any thoughts on that? Recommendations?

Through experimentation, I've come up with satisfactory answers to my problems.

1) What is the best material to cast indium in? Silicone rubber (the kind used for cookware).

2) How can I recovery indium that has "wetted" a stainless steel pot? Put the pot on the stove and heat until the indium is molten. Use a silicone rubber spatula to scrape the indium off, let it harden on the spatula, then pick it off. You can also use a wet sponge as long as it's the kind used for cleaning solder irons (make sure it's a new clean sponge!). I folded it in half and held it with a large clamp. That gave me something to hold onto.

3) How can I recover indium that has "wetted" glass, broken glass, or is contaminated with other odd bits of material?

Number 3 is more involved, and I will break it down into steps:

1) Go through the material with a pair of forceps and discard any bits of glass that have no indium on them.

2) Place the material in Pyrex beaker.

3) Fill the beaker with mineral oil until the material is fully submerged (water may also work, but the oil will probably help keep the indium from re-wetting the glass bits).

4) Place the beaker on a hot plate and slowly heat up to about 160 degrees C (320 F). Periodically stir while heating (a screwdriver worked fine for me; I didn't use a magnetic stir rod because the broken glass might chew it up). Also, make sure your area is well ventilated; the oil will produce vapor.

5) Once you have reached 160 C, poke the indium and make sure it is molten. Increase temperature if it is not molten. Stir for about 30 seconds to break up the indium. Keep on heat; try to maintain ideal temperature.

6) Use a glass dropper to drop in 31.45% Hydrochloric acid (I used 4 drops total when the oil and material together was about 100mL). Add one drop at a time, and let the drops run down the inside of the beaker into the oil. (WARNING: This can result in splashing! Don't over heat the oil, add the acid drop-wise, and use proper safety gear: gloves, long sleeves/pants, face shield, and respirator with acid vapor cartridges). Adding the acid will react with the indium oxide layers and help break the indium free from other materials.

7) Stir vigorously with your implement of choice. The oil will become a nearly opaque brown with all the oxides. The stirring should also free up most of the indium from the glass bits.

8) When the oil seems saturated, thoroughly decant the oil into a waste receptacle (I used a mason jar). Save as much solid material as you can.

9) Fill the beaker with clean oil; do NOT begin stirring yet. Keep on heat.

10) Use your forceps to pick out the larger pieces of glass that seem mostly clean. Use another implement to scrape off little bits of indium from the glass bits as you are pulling them out of the oil.

11) Once most of the glass is removed, use strong stirring once again until the oil is mostly brown.

12) Using your stirring implement, try to push a good amount of the molten indium together into one blob on one side of the beaker. Now decant the oil into a waste receptacle.

13) Place beaker back on heat for a short time just to ensure the indium is molten. Make sure most of the material is in one corner of the beaker.

14) Use a broad-shaped implement (I used the flat head of a long "elevator bolt") to hold back most of the scrap material and pour off the indium from the beaker into a silicone rubber mold. The poured off indium should be MOSTLY clean. Repeat this step until you feel you cannot get anymore. Put the scrap material off to the side.

15) Once the clean blobs have hardened, dab them with a paper towel and cut them into tiny chunks with a pair of wire snips.

16) Take each individual chunk and crush it paper-thin using a flat metal implement and a flat metal surface (I used an elevator bolt and a piece of sheet metal). This step is to find any tiny bits of glass. Glass has quite a different "crunch" than the typical "crackling" of the indium's crystal structure. You will know when you find one. Fine tweezers and some sort of magnifying glass helps.

17) You can now re-melt the clean indium into one chunk. I use a stainless steel shot glass and pour it off into a silicone rubber mold. If it's still oily, wash it with Ronsonol/Zippo lighter fluid, then rinse it with water. I actually chose to keep this piece of indium separate from my other indium pieces that I KNOW are clean. You never know, there might be one last sand-grain sized piece of glass in there.

18) As for the scrap material, there's still a bit of dirty indium in it. You can decide how hard you want to work and continue salvaging it. Use tweezers to pick out as much "clean" glass as possible. Wash it with lighter fluid and rinse it with water. Perhaps re-melt it in a steel shot glass and try to pour a little bit more out of it. Try putting a blob of solder flux in with the molten material and get it to stick to a copper wire. It's all going to be loss anyway, so you may as well try different things.

DISCLAIMER: I am not a professional! This procedure is not advisory. It is an account of my experiment and what worked for me.

If anyone has suggestions or constructive criticism, please feel free to comment. I know there are probably better ways to do this.


Conflicting definitions of Lewis acids I have struggled in attaining a solid grasp of the Lewis acid/base concept largely because so many sources seem to offer conflicting definitions.

For instance, Klein's Organic Chemistry mentions that Brønsted–Lowry acids and bases are strictly a subclass of Lewis acids and bases. Other sources say this is true of bases but that a Brønsted acid need not be a Lewis acid and vice versa.

A common example of a non-Lewis acid provided by Wikipedia is $\ce{HCl(g)}$. The rationale being that it has no available empty atomic and/or hybrid orbitals to accept an electron pair (it does have an empty antibonding orbital though — see below).

Another source says hydronium is only a Lewis acid if it is "informally" treated as a hydrogen ion $(\ce{H+})$.

There appears to be further disagreement on whether the adduct formed in the Lewis acid/base reaction is necessarily a product or whether it can effectively appear as a transition-state before splitting apart to give the products. For example, say something like

$$\ce{NH3 + H2O <=> NH4OH <=> NH4+ + OH-}$$

Other users on this site suggests the approach should be to consider molecular orbitals. I will further note that $\ce{HCl}$ has an empty antibonding orbital so this definition, although arguably the best I can think of, still necessarily conflicts with multiple other sources labeling $\ce{HCl}$ as a non-Lewis acid.

Any help on resolving some of these apparent conflicts would be appreciated.

Moeller's Inorganic Chemistry textbook from 1952 has a fairly long discussion of different theories of acids, citing a lot of primary literature. For Lewis acids, it cites some reviews by W. F. Luder from the 1940's and some of Lewis' original work. Specifically, it says that Lewis in a paper in 1938 (J Franklin Inst. 226:293) described "phenomonological criteria" for Lewis acids and bases, which include the ability to displace weaker Lewis acids and bases.

My impression from Moeller's description of Lewis' work is thus that he defines an acid as a molecule in which one atom has an incomplete valence shell of electrons. He further views a Bronsted acid as a complex between a Lewis acid ($\ce{H+}$) and a Lewis base (the conjugate base). In a reaction like $\ce{NH3 + AcOH -> NH4+ + AcO-}$, he would say that the stronger Lewis base $\ce{NH3}$ is displacing the weaker Lewis base acetate from the Lewis acid $\ce{H+}$.

If I have time, I'll try to dig up some of those primary sources. Here's a quick list if anyone else wants to go after them:

G. N. Lewis; Valence and the Structure of Atoms and Molecules. The chemical catalog co. (1923)
G. N. Lewis; J Franklin Inst 226:293 (1938)
W. F. Luder; Chem Rev 27:547 (1940)
W. F. Luder and S. Zuffanti; The Electronic Theory of Acids and Bases, Wiley and Sons (1946)
W. F. Luder, W. S. McGuire and S. Zuffanti; J. Chem. Ed. 20:344 (1943)
W. F. Luder; J. Chem. Ed. 22:301 (1945)

UPDATE: The original 1923 Lewis text is available for free from Google Books here:https://books.google.com/books?id=2hxDAAAAIAAJ&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false

The most relevant passage is from p.142:


Why does electron velocity increase with the increase of atomic number? I know that for atoms with big values of Z, relativistic effects have to be applied. Why does electron velocity increase with the increase of atomic number?

Why does electron velocity increase with the increase of atomic
  number?

Newton's second law tells us that for an electron moving in a stable circular orbit around a nucleus the centripetal force pushing out must equal the electrostatic attraction pulling in, or

$${\frac{m_e~ v^2}{r}} = {\frac{Z~k_e ~e^2}{r^2}}~~(1)$$

where $m_e$ is the electron's mass, e is the electron charge, $k_e$ is Coulomb's constant and $Z$ is the atom's atomic number.  Rearranging equation (1) yields

$$v = \sqrt{ Z~k_e~ e^2 \over m_\mathrm{e} r}~~(2) $$

From equation (2) we can see that the electron velocity increases as the atomic number, $Z$, increases.

Edit: Response to OP's comment

  I am expecting some quantum mechanical explanation

As the electron velocity given by equation (2) becomes comparable to the speed of light, the mass of the electron increases due to relativistic effects.

$$m_{rel}=\frac{m_{e}}{\sqrt{1-(v_e/c)^2}}$$

As the electron mass increases, the s (and somewhat for p) orbital radius contracts.  So relativistic effects affect the electron mass and the orbital radius.  In terms of electron velocity, these 2 relativistic effects offset one another and the electron velocity is not affected; equation (2) is still valid.


Formation of MnO2 vs [Mn(H2O)6]4+ in aqueous solution I am looking for reasons why $\ce{Mn(IV)}$ wouldn't be stable as a hydrate. I know that from standard reduction potentials it exists as a ppt of $\ce{MnO2}$, but what is stopping it from being a hydrated $\mathrm{4+}$ aqueous cation? Is it that the water ligands would be too polarised and therefore $\ce{H+}$ dissociation would occur very easily? Are there other factors that make it a solid rather than a solvated ion, like lattice enthalpy etc.?

Solid $\ce{MnO2}$ can be hydrated, but it's the solid which precipitates from solutions as a product of redox reactions of other $\ce{Mn}$ compounds. There is also for example crystalline $\ce{[MnF4(py)(H2O)]}$ What about aquo cations you mentioned?

One thing is that $\ce{[Mn(H2O)_n]^4+}$ cations should be indeed quite strongly acidic, but that's no problem as another strong acid in solution would prevent precipitation (similar thing like with $\ce{Fe(OH)3}$). The problem is oxidative power, what may be surprising knowing that Mn has higher oxidation states.

Why do I think so? Well, let's see what happens if we try this trick with acid to make a solution. Concentrated $\ce{H2SO4}$ and $\ce{HCl}$ react with $\ce{MnO2}$, but lets see the reactions:

$$\ce{2 MnO2 + 2 H2SO4 → 2 MnSO4 + O2 + 2 H2O}$$
$$\ce{MnO2 + 4 HCl → MnCl2 + Cl2 + 2 H2O}$$

In first case oxygen gets oxidised and in second chlorine!

Now what if we used $\ce{HF}$? Concentrated solutions are really powerful stuff! Unfortunately the problem persists, as you could see earlier $\ce{Mn^{4+}}$ can oxidise even water if acidity is high enough. With difficulty $\ce{MnF4}$ was synthesised, but guess what, Mn can gradually oxidise fluorine in this compound, making $\ce{F2}$ gas! That was first chemical reaction known to do that.

So, generally not much chance for simple aquo complexes.

Check out these articles on Wikipedia for more info and citations to articles: about $\ce{MnF4}$ and about $\ce{MnO2}$.


How could we figure out the recipe of Coca Cola? I know that the ingredients and the manufacturing process are both important, as discussed in this question. What would be a modern approach to figuring out the ingredients?

The ingredients listed on the website are: "Carbonated Water, High Fructose Corn Syrup, Caramel Color, Phosphoric Acid, Natural Flavors, Caffeine." My ideal answer would provide information about how to get more detail about the specific ingredients (not just 'natural flavors') and their proportions.

I'd say some combination of a GCMS / LCMS. The chromatography separates the components & can often be relied upon to ppm level identification.

Then compare the MS peaks for hits against one of the standard databases e.g. NIST.

I suppose that strategy would identify most of the peaks. The left over peaks, which don't give a good match against a GCMS database, would be more challenging to identify. One would need to screen potential raw materials one suspects may have gone into the blend to see if one generates a hit.

Of course, this identifies individual chemical species. Any practical recipe on the other hand is unlikely to be based on pure species amounts but would use available ingredients which are each complex mixtures of pure species.

The GCMS  / LCMS approach won't tell you what pseudo-components (e.g. lemon oil  ) went in & in what proportions. So you figured out the chemical composition but not really the recipe.

Sometimes, rarely, two compounds can have very similar MS peak patterns. In those cases (or to reinforce a GCMS analysis) one could also compute the Kovats Index (KI) of each peak on the GC based on retention times alone (and the injection of a standard Hydrocarbon mix). There's a good amount of info. and databases on the KI flavor / natural product ingredients.

That would be another strategy to confirm peak identity.

On the molecular biology side of things I'm sure there are techniques using PCR etc. that could amplify and characterize traces of DNA segments that came via ingredients of plant origin. e.g. One might be able to say a certain plant oil extract was used. Although at high dilutions & after extensive processing it would be a challenge to get enough starting matter for these techniques to work.


Is bamboo based plastic less likely to leak chemicals? Some plastics have been known to leak plasticizers (bisphenol and phthalates) into the food they contain, especially when heated.
Because such compounds are suspected of being endocrine disruptors it seems best to avoid eating hot food from a plastic container.

Recently bamboo bowls, cups and plates have become available. As far as I can tell these are made from some kind of composite material which contains bamboo fibers and something that glues these together.
The only information  is that it's: renewable, dishwasher safe, bio-degradable (the last two claims seem contradictory), food safe, but not suited to be microwaved.

Is it possible that these kitchen-wares are safer than other plastic bowls, plates and so on? Is it possible to even assess the risks?

Inspired by the other answer and the comment to my question, i have dug a little deeper. The tableware in the picture is sold by a company called Zuperzozial and made from bamboo and corn reinforced with melamine resin.

Many similar products are sold, almost always without a brand name and always without any specification of the actual material. So I will answer my question for this specific set of cups, there is no way to tell if the answer holds for 'bamboo' tableware in general.

Melamine from tableware may leak into the food, especially when heated, according to the FDA. It is ok to serve cold drinks in a melamine resin cup. The claim that these cups are micro-wave safe is not true.

As for the amount of BS of the rest of the marketing claims: bamboo and corn are renewable sources. But i'm not sure how degradable a resin is.


What makes formaldehyde 'safe' or 'dangerous'? I know that formaldehyde naturally occurs as a metabolite and can be found in our own and other animals' bodies, plants, and unprocessed food.

But formaldehyde is also registered as a carcinogenic VOC and is considered bad when it's in plywood and cosmetics.

Is this judgement based on dosage, form (e.g. gas/liquid), or amount released?

The toxicity of formaldehyde is dose-dependent. The famous saying "the dose makes the poison" is one of the rules of thumb about toxicology. Your body can effectively detoxify small amounts of formaldehyde in a given amount of time, however if you are exposed to concentrations that result in you ingesting/inhaling more than what your body can metabolize, then toxicity will begin to occur.

Most consumer products with detectable levels of formaldehyde (I'm thinking of shampoos and body washes particularly) do not directly have formaldehyde added to them. As formaldehyde is a fairly reactive gas that will polymerize & evaporate easily, formulation chemists will use compounds that decompose slowly to formaldehyde plus other mostly inert material. What this does is provide a constant, low level of formaldehyde present to halt microbial growth. (Formaldehyde is a wonderful disinfectant - few things if any will survive treatment with concentrated formaldehyde - and it turns out you need a LOT of formaldehyde to kill a human as compared to a microbe!)

Formaldehyde is also a natural trace compound formed in fermentation/decomposition of other organics, so it's detectable in the biosphere if you want to go looking for it - I'm sure you can find detectable amounts in fruits, wood products, even the air around a garden.


Why do we use pV term independent of U in the equation H = U + pV? I have read the following definition of enthalpy in my textbook:

A substance has to occupy some space in its surroundings depending upon its volume ($V$). It  does against the compressing influence of the atmospheric pressure ($p$). Due to this,the substance possesses an additional energy called $pV$ energy which is given by the product of pressure ($p$) and volume ($V$) of the system.
The sum of internal energy and $pV$ energy of any system, under given set of conditions, is called enthalpy. It is denoted by $H$ and is also called heat content of the system. Mathematically, it may be put as $$H = U + pV$$

Let's suppose we have some water in a test tube at room temperature open to atmosphere. It will have internal energy which consist of kinetic energy of  molecules, chemical energy, rotational energy and some other forms of energy. In order to occupy some space in its surroundings, the molecules of the water should vibrate in such a way that they cancel out the compressing influence of the air molecules present in the atmosphere.
So if we try to measure the internal energy of the system, isn't the energy required to make space for the surroundings already included in the internal energy of the system. Why do we add an extra term $pV$ along with internal energy $U$ to measure the heat content of the system?

The energy U of a system consisting of a single component and phase is

$ U = -PV + TS + \mu N$

As you rightly point out, U includes the energy required to create the volume by expanding against the surrounding (constant) pressure, namely the -PV term. Removing this contribution gives H.

Therefore

$ H = U + PV = TS + \mu N$

Equating TS with the "heat content" of the system leads to the statement in the textbook (this is my interpretation of this textbook's author's meaning). This kind of algebraic manipulation can lead to all sorts of confusion.

EDIT

Some people recoil (and downvote) when they see an expression such as

$ U = -PV + TS + \mu N$

as if this were somehow unphysical. Well, it does suggest that there is such a thing as an absolute value of U rather than only relative values, and it is hammered into our heads in school that there are only relative, not absolute, energies. But the point here is that there is an absolute value, relative to an empty system, and you can use N (the number of particles) as your simple scaling constant. The expression above captures the extensive property of U. You could equally write

$ \Delta U = -P \Delta V + T \Delta S + \mu \Delta N$

and no one will blink. Well, if you define

$\Delta U = \Delta N \times U_m$

$\Delta S = \Delta N \times S_m$

$\Delta V = \Delta N \times V_m$

with

$\Delta N = N-0$

you obtain an expression equal in spirit to the sacriligeous one above.

Usage of terminology such as "heat content" is potentially confusing, but we can use similar arguments to those used in explaining the PV term's meaning as work to create the system from scratch by increasing the particle number, while keeping intensive properties such as T and P constant, to suggest that the term TS is the heat that must be exchanged with the surroundings during creation of the system (from 0 to N particles). This is more of theoretical than experimental interest, in the same way we can hypothetically charge an object from 0 to a finite charge to compute its energy (relative to no charge).

Note that saying that H represents the heat content is confusing, since the application of the equation is not usually in the spirit I have just outlined, and if you look at

$ H = U + PV = TS + \mu N$

there is an additional term $\mu N$ term which cannot be easily associated with the concept of "heat content".

So, being generous, I would think the author of the book meant by the statement was, very loosely, that changes in H, that is $\Delta H$, are usually associated with the heat exchanged by a system at constant P (subject also to other restraints but more on that in other questions and answers referring specifically to the meaning of H).


Lack of elaboration in IR spectra of compounds I've recently started learning about IR spectroscopy, and a common trend I'm seeing in my textbook and other sources is that not all the features in the IR spectrum are explained in detail. The focus is primarily on features whose results are known, and there is quite a bit of backward reasoning in deriving the structures of compounds from their IR spectra.
Consider the IR spectrum for Decane:

Here, what are the features causing absorption in the $\pu{720 cm^{-1}}$ range? Morrison Boyd states that C-C stretching causes absorption in the $\pu{1500-1600 cm^{-1}}$ range, and the absorption in the $\pu{3000 cm^{-1}}$ is explained in the figure, but I am unable to explain the other absorption.
Another example is the IR spectrum for Butanamine:

Here, what is causing the large absorption in the sub-$\pu{1000 cm^{-1}}$ region? Wouldn't such a large absorption at this frequency indicate unsaturation? How are we deducing then that it is an alkane?
The lack of explanation is a common feature for a lot of IR spectra presented in the text. Is this normal for someone learning IR spectra at this level? Also, any suggestions for other references for learning about IR/NMR spectra in more detail are appreciated.

In IR, wavenumbers below 1500/cm are called the "fingerprint" region. It's very tricky to assign anything there, because the transitions don't relate to specific bonds, but come from larger structures.
If you look deeper into the literature, you will surely find an explanation for that big hump at 800something inverse centimeters.
Generally, two substances with matching fingerprint region in the IR can safely be regarded as identical, but you cannot draw conclusions about the molecular structure. Hence the name.


Neutralisation of weak acid by base I'm still working on my concepts related to chemical equilibrium. I came across this question:

10mL of a weak acid HA is 20%
dissociated in water. This solution is completely neutralised by 10 mL
of $10^{-3}$ M of NaOH.

Is [$\ce{HA}$] = $10^{-3}$ M or is [$\ce{H^+}$] =$10^{-3}$?
Why is it wrong to say that since NaOH completely dissociates, and solution is neutral, [$\ce{OH^-}$] =$10^{-3}$ M= [$\ce{H^+}$]?
Here, [.] denotes concentration.

As @Maurice pointed out, the result of a neutralization reaction is not impacted by the dissociation percentage.

Is $\pu{[}\ce{HA}{] = 10^{−3} M}$ or is $\pu{[}\ce{H+}{] = 10^{−3}}$?

$\pu{[}\ce{HA}{] = 10^{-3}}$.
Always remember, neutralization in aqueous medium depends upon the number of equivalents of acid and base:
$$N_1V_1=N_2V_2 $$  Since $V_1=V_2$, therefore, $\pu{[}\ce{HA}{]=[}\ce{NaOH}{]}$.

Why is it wrong to say that since $\ce{NaOH}$ completely dissociates, and solution is neutral, $[\ce{OH−}]$ $\pu{= 10^−3 M = [}\ce{H+}]$?

The secret lies in the name of the concept itself. Chemical equilibrium. To attain an equilibrium state, all favorable chemical reactions are first carried out. The remaining products/ions constitute the equilibrium state. Coming back to the question, the neutralisation reaction between $\ce{HA}$ and $\ce{NaOH}$ is carried out first:
$$\ce{HA + NaOH ->NaA +H2O}$$
The resultant ions constitute the equilibrium, which, are none unless you consider the hydrolysis of the resultant salt of a weak acid/strong base.


Why does greater orbital overlap mean a stronger bond? According to valence bond theory, orbital overlap produces a bond. However, I don’t understand why having greater orbital overlap renders a bond stronger. It’s intuitive, I suppose, but I haven’t been able to find an actual explanation as to why that is. Are there quantum mechanical effects behind this?

An intuitive explanation is given in the comment above. Better orbital overlap generally indicates more constructive, favorable interaction between the two atoms.

Let's consider a homonuclear diatomic bond (e.g. $\ce{H2}$). The qualitative understanding is similar for two different atoms, but the math from quantum mechanics is a bit easier with a homonuclear bond.

So there's one bonding orbital ($\sigma$) and one anti bonding orbital ($\sigma^*$).

The energy of the orbitals will be:

$e_1 = \frac{H_{11} + H_{12}}{1 + S_{12}}$
$e_2 = \frac{H_{11} - H_{12}}{1 - S_{12}}$

You can put in some numbers and see what happens (e.g., as the overlap $S_{12}$ increases from 0 to 1, the bonding interaction becomes more and more energetically favorable).

N.B. There's a slight asymmetry - the anti bonding orbital is higher in energy than the bonding orbital is lower in energy.


Effective core potential and Gaussian basis set Is using the effective core potential (ECP) and Gaussian basis set simultaneously in ab initio calculation a valid approach? For example, is it correct to use ECP for one of the atoms and, instead of a basis set which is optimized for ECP, to take a normal Gaussian basis set (say, 6-31G) when we perform calculations for a molecule with a Hartree–Fock method?
P.S. I know technically I can do it in the quantum chemistry packages (for example, I have checked it in MOLPRO), but I don't know whether it is valid physically.

Sure, it can be valid physically. You can use the full-electron basis set, like 6-31G*, and the ECP with associated basis set, like Lanl2dz, for instance, in calculations of transition metal compounds or compounds with heavier non-metals (Sb, Te, etc.) as an example.
Important here would be to compare results obtained with available experimental data (such as bond distances/angles, reaction energies, UV-vis spectra, etc.) or results of similar calculations in order to see if the simultaneous use of ECP and full-electron basis sets is valid and would not give you some weird results.


Reduce Decomposition Of Glycerin Background: Glycerin, when heated with a metal coil, decomposes below its boiling point to acrolein, which in turn appears to decompose to formaldehyde and acetaldehyde. (see previous post)

Question: Is there a chemical that could be added, or a process that could be applied, that would lessen the decomposition of glycerine when vaporized?

You could try the mechanism behind mist fountains or other "cold fog" machines:

These use an ultrasonic oscillating diaphragm to produce a fine mist of room temp water, no heating performed. The resulting vaping machine would be larger and require more power, but wouldn't have the harmful aldehydes inherent in current vaping liquids. Nicotine itself is hygroscopic but not really water soluble at room temp, but it can form salts that are water-soluble so it could be introduced to the body this way. Various flavors currently used might be more or less water-soluble. Finally, the fog, being simple water vapor, wouldn't be visible when exhaled which is one reason the glycerol compounds are used; they produce a "smoke" that dissipates more quickly than tobacco smoke.


Predict Entropy based on the Balanced Equation You cannot use values but must determine from looking
The equation I have trouble with is as follows:

  $$\ce{2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2}$$

According to my teacher the entropy is positive. I also got the entropy to be positive, but now that I think of it, the moles is decreasing and one is becoming aqueous. So does turning into a gas trump everything?

What if NaOH was solid? Would it mean still that the entropy would be positive?

Also finally according to the textbook the entropy values is as follows:

  $$\ce{s<l<<g}$$

But where would the (aq) be?

Aqueous salts will always have less order than solid salts. There is much less order when ions are floating around in solution than when they are in their rigid lattice structures. This is what I think you are failing to consider. The reaction is not so much:

$$\ce{2Na_{(s)} + 2H2O_{(l)} -> 2NaOH_{(aq)} + H2_{(g)}}$$

But really:

$$\ce{2Na_{(s)} + 2H2O_{(l)} -> 2Na^{+}_{(aq)} +2OH^{-}_{(aq)} + H2_{(g)}}$$

What results from the reaction of 2 atoms of a very structured metal and 2 molecules of liquid water, is 4 aqueous ions that are free to move within the solution, and one molecule of gas. Not only are there now more species present, but they are also in a relatively more free states than before.


Clarification about the difference of internal energy formula I am trying to understand the concept of internal energy and it's difference. My textbook said that the difference in internal energy of a system is

$$
\Delta U=Q+W
$$
where $Q$ is the heat absorbed or released by system and $W$ is the work done by system to the surrounding or by the surrounding to the system. After that, it mentioned that the signs of $W$ and $Q$ are determined like this:

  The work done by surrounding to system: $(+)$. The work done by system to surrounding: $(-)$. Same on heat.

After that general definitions, it started to explain the internal energy differences in various systems like isochoric system, isobaric system. In the section of isobaric system it says that:

  If a system were free that can expand against external pressure, when heat is given to that system, it will give some energy by doing some work. So the internal energy difference is:$$
\Delta U=Q_p-W
$$ where $Q_p$ is the heat given to a isobaric system.

It is completly OK till the new formula, but isn't the new formula bizarre? Because if we give heat to a system and it will do work. As a result the difference would be like:

$$
\Delta U_{example}=Q_{p_{example}}-(-W_{example})
$$
So, does the work done by system increase its internal energy?

Your book has driven you crazy with its stupid sign convention.  If the gas is expanding against constant pressure, then the equation is still $$\Delta U=Q+W$$ but, in this case, the work done on the system by the surroundings is negative (i.e., W is a negative number, because the system is doing work on the surroundings).  So not all the heat added to the system is being used to increase its internal energy.  The increase in internal energy is less than if the gas was not pushing back its surroundings.


Electrolysis of AgNo3 I have a solution of $\ce{AgNO_3}$ (is diluted with water). I electrolyse it. What products from at anode and cathode.

You would get silver at the cathode, and oxygen at the anode.

Since silver is below hydrogen in the spectrochemical series,it tends to get reduced over hydrogen, similarly, since $\ce{NO3-}$ is above $\ce{OH-}$ , it tends not to get oxidised, therefore giving oxygen.

$\ce{Ag+ + e- -> Ag}$ (at cathode)

$\ce{ 4OH- -> O2 + 2H2O + 4e-}$ (at anode)


Why are organolithium and organomagnesium compounds nucleophilic in nature? My textbook doesn't say anything about this but does say that on addition of organolithium compounds to carbon dioxide, they give ketones and not carboxylic acids like grignard reagents. This has been attributed to the greater nucleophilic character of Organolithium compounds compared to grignard reagents. I see that greater polarity of the C-Metal bond must relate to this according to many websites but I cannot understand why! Also, during formation of the RLi compounds using alkyl halides, what does the alkyl halide dissociate into? Please help!

When Li reacts with an alkyl halide, a salt (lithium halide) is formed, whose stability is the primary thermodynamic driving force of the reaction: $\mathrm{2Li}+\mathrm{RCl} \rightarrow \mathrm{RLi}+\mathrm{LiCl}$.

Due to the larger difference between the electonegativities of C and Li, the C-Li bond in organolithium compounds has more ionic character and C is more negatively polarized. The higher polarization makes R in RLi a harder nucleophile compared to RMgX, which explains why lithium organyls and Grignard reagents react differently when adding to certain carbonyl compounds.


Does a weak acid undergo it's weak ionization upon the addition of any kind of base? For an example let's assume we have a weak acid  HA . And we add either a strong or a weak base. Then what I want to know is whether  the weak ionization of the acid HA taken place?.

Weak ionization of acid HA. ==>

A weak acid does not (contrary to what you show in your equation) predominantly ionize before the base is added. But keep in mind that proton transfer reactions (at least those between oxygen and/or nitrogen atoms) occur almost instantaneously in both directions, so the equilibrium you illustrated is established immediately (for a weak acid, with the big arrow pointing left). So when (strong) base is added, H3O+ is consumed immediately, then more acid is ionized to re establish the equilibrium--and this goes on as you add more base until you have consumed almost all of the weak acid. These reactions are controlled by the relative energies of the reactants and products, and not by reaction rates.


Is the electronegativity value ever used in calculations, or is it just a number assigned to each element? Does the electronegativity only qualitatively describe bond polarity, or does it have any quantitative use? Is it possible to describe electronegativity in terms of other periodic properties, like valence electrons, $Z_{eff}$, etc. (thus eliminating the need of a separately defined value)?

True, with values of electronegativity in hand (and beside the by Pauling, there are others by Rochow, Mulliken, etc. to name some scales) you describe bond polarity. Beyond this number, infrared vibrational frequencies, chemical shifts in NMR spectroscopy and isomer shifts in Mössbauer spectroscopy are mentioned by the corresponding wikipedia article.


What makes for a good 'swimmer' in order see the filling level of Xenon gas in a vessel? Most foams will probably get heavier due to diffusion. Is there glass balls this thin?

You can easily blow large, incredibly thin bubbles of soft glass with a propane torch or alcohol lamp -- so thin that interference patterns color them. Though fragile, it might be fun to make one as part of the demonstration. You'll need to experiment with surface-to-volume ratio to get it to float in Xe (you could also use halocarbon computer duster or $\ce{CO2}$, rather than expensive Xe). N.B. Use goggles, not just safety glasses, because the thin glass, when it breaks, lofts into the air... and possibly, your eye.

Rehearse, though, because removing the bubble from the tubing is likely to break it. After you get a large, thin bubble, aim the flame at the point where it connects to the tubing, and heat it until the tubing droops to a hair-thin tube. Do not melt this thin tube shut, because outside air-pressure will crush the bubble as it cools! Break that thin tube, leaving a short stem on the bubble. Yes, there will be some slow diffusion through the neck, but the bubble should last for weeks. If the demo is in a temperature-stable location, you might be able to seal the neck without the bubble busting.

There are videos showing lampwork here and here, but I suggest using soft glass (not borosilicate "Kimax" or "Pyrex") for your first bubbles, as it takes much less heat to melt.

BTW, a similar demonstration of density is the Cartesian diver, which also makes a good first student project in flamework:

With ~6 - 8 mm diameter soft glass, blow a bubble which can fit through the mouth of a plastic soda bottle.
Cut the bubble off with a triangular file, leaving a ~10 - 20 mm stem.
Optional: flame-polish the cut stem.
Dribble water into the (cooled) bubble until it just floats in a beaker of water.
Put the bubble in a filled plastic soda bottle, cap the bottle, and squeeze.

When I've had student 11-16 years old do this, it went exceedingly well (I actually had to remind them when lunch-break came that they needed to stop).


Test for unsaturation Ethene and ethyne can be distinguished by employing the test
a. $\ce{Br2}$ in organic solvent
b. Baeyer's test
c. Phenyl hydrazine
d. Tollen's reagant

According to me it must be both a and b, but there's no such option.

I see two possible answers. Not (A) and (B) but rather (A) and (D).

Chakravarthy Kalyan is responsible for reminding us that (D) can be an answer too (refer to comments).

For option (D):

$\ce{HC ≡ CH + 2AgNO3 + 2NH4OH→ AgC ≡ CAg↓ + 2NH4NO3 + 2H2O}$

The greyish precipitate of silver acetylide will readily form when ethyne is passed through freshly prepared tollen's reagent.

You may read more from this answer by @ron: Does acetylene give a positive Tollens' test? If not, why is this so?.

For option (A):

While $\ce{Br2}$ looks like a seemingly unsuitable candidate, it can actually be used as a crude way to differentiate them.

Take the same volumes of equal strength of bromine water in two vessels and allow equal amounts of ethyne in one and ethane in the other to pass through them (this shouldn't be difficult, just passing equal volumes at STP will do the trick).

Carefully observing which vessel has the more intense color will help you identify which one is ethene, which would consume half the amount of bromine consumed by ethyne.

If you're not convinced, put the final solutions in a cuvette each and measure the absorption or transmittance in a spectrophotometer at 530 nm.

P.S. Make sure that you have a sufficient amount of bromine water, or you'll end up comparing two colorless solutions.

P.P.S. You could also use titration to find which one reaches the end point first, which would be ethene.


Solubility equilibrium This is a worked example about solubility equilbrium I am new to this subject and I am really confused first of all I know that Q is the product of concentration of Ag and Cl in the solution and ksp is the product of molar solubility of Ag and Cl then why the difference of these two is not amount of precipitation
second: if we are not given ksp value of AgCl and we are given only the last part ( part b) how can we find the ksp value?
P.s: after change the amount of cl become zero and amount of Ag become 99*10^-6 so it means that after mixing two solution precipitate is formed and amount of dissolved Ag ions become 99*10^-6 and all of Cl ion is precipitated am I right?
(Sorry if my questions doesn't make sense for you it is because I am new to subject and I don't have any teacher I really tried hard on this example and I saw lots of videos about this subject but I didn't understand it)

Q is known as the ionic product which is defined as product of concentration of ions of the electrolyte, each raised to the power of their coefficients in the balanced chemical equation in solution at any concentration. $\pu{K_{sp}}$ is the same but it is applicable only for a saturated solution.

To find the solubility for the salt with the given $\pu{K_{sp}}$ value, For $\ce{AgCl}$, we have the equilbrium,
$$ \ce{ AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$
If initally we have $‘a ‘$ moles of AgCl, at equilbrium we will have $’a-s ‘$ and we have $\ce{[Ag+]=[Cl-]=s}$. Hence we get the relation as,
$$ \pu{K_{sp}} = s^2 $$
This is the maximum concentration of the ions present in solution at equilbrium.(because $\pu{K_{sp}}$ = Q).
When Q just crosses $\pu{K_{sp}}$, salt is precipitated out.

Since they are products of ionic concentration, we can’t directly subract Q and $\pu{K_{sp}}$ to find the amount of salt precipitated.

I’m just working backwards to get $\ce{K_{sp}}$.

Let mass of $\ce{AgCl}$ precipitated be $’m ‘$ g.
No. of moles precipitated $n = \frac{m}{M}$.  Let the volume be $’V’$ litres,
$\ce{[AgCl]}$ precipitated $x = \frac{n}{V}$

$$ \ce{Ag+ + Cl- -> AgCl} $$

We know initial concentrations of the ions. Let $\ce{[Ag+]}$ and $\ce{[Cl-]}$ be $’p‘$ and $’q‘$ respectively.
Then solubility product is,
$$ \ce{K_{sp}} = (p-x)(q-x) $$

Hope that makes sense.


Mechanism of homologation of aldehyde to alkyne: Ohira–Bestmann reaction This reaction was part of an organic scheme I was doing, and I was wondering what the mechanism of the reaction would be.

I tried first forming a carbene by letting $\ce{N2}$ leave from the diazocarbonyl. Then, I did a Wolff rearrangement, attacked the ketene with water, decarboxylated the COOH, then did a Wittig reaction with the phosphonate ester. However, I ended up with an alkene and not the alkyne.
I also tried first doing the Wittig rxn, but I didn't know where to go after that (since you can't form the carbene anymore).
Any help with the mechanism of this reaction would be appreciated.

This reaction is the Ohira–Bestmann modification1,2 of the Seyferth–Gilbert homologation.3,4
In the original Seyferth–Gilbert homologation, a diazomethane phosphonate is used to convert carbonyl compounds to alkynes with one extra carbon in the presence of potassium t-butoxide. I used acetophenone as the reactant here just to simplify things, but the same mechanism applies to your substrate.

In the last step, the diazoalkene may be equivalently thought of as a carbene. You could lose nitrogen to form a carbene, which undergoes a 1,2-alkyl shift to generate the alkyne. I have drawn it as being concerted but I am not entirely sure whether it is really concerted (as far as I know, there is still some degree of contention over the Wolff rearrangement — but feel free to correct me if I am wrong).
This procedure has some issues due to the strongly basic conditions. For example, in your substrate, the presence of t-butoxide can easily lead to racemisation α to the carbonyl group. The Ohira–Bestmann conditions are much milder and generally do not lead to racemisation.
As Waylander mentioned in a comment, you need to specify the solvent, but generally the Ohira–Bestmann version is carried out with potassium carbonate in methanol. The combination of carbonate and alcohol generates small amounts of alkoxide, which then attacks the carbonyl group in the diazophosphonate to produce the same reactive species:

However, I looked up your compound on Reaxys and it seems that the authors used caesium carbonate in t-butanol.5 I am very hesitant to suggest t-butoxide attacking a carbonyl group as a nucleophile, so I would go out on a limb and propose an alternative mechanism. The ketene generated can probably be intercepted by the t-butanol to generate t-butyl acetate. I don't have any evidence for this, however.

In general you had a lot of good ideas (Wittig-type chemistry and a Wolff rearrangement) and were mostly on the right track. The missing ingredient was probably the solvent, which is crucial for the generation of the reactive carbanion.
References

Ohira, S. Methanolysis of Dimethyl (1-Diazo-2-oxopropyl) Phosphonate: Generation of Dimethyl (Diazomethyl) Phosphonate and Reaction with Carbonyl Compounds. Synth. Commun. 1989, 19 (3–4), 561–564. DOI: 10.1080/00397918908050700.

Müller, S.; Liepold, B.; Roth, G. J.; Bestmann, H. J. An Improved One-pot Procedure for the Synthesis of Alkynes from Aldehydes. Synlett 1996, No. 6, 521–522. DOI: 10.1055/s-1996-5474.

Gilbert, J. C.; Weerasooriya, U. Elaboration of Aldehydes and Ketones to Alkynes: Improved Methodology. J. Org. Chem. 1979, 44 (26), 4997–4998. DOI: 10.1021/jo00394a063.

Kürti, L.; Czakó, B. Strategic Applications of Named Reactions in Organic Synthesis; Elsevier: Amsterdam, 2005; pp 402–403.

Brenneman, J. B.; Martin, S. F. Application of Intramolecular Enyne Metathesis to the Synthesis of Aza[4.2.1]bicyclics:  Enantiospecific Total Synthesis of (+)-Anatoxin-a. Org. Lett. 2004, 6 (8), 1329–1331. DOI: 10.1021/ol049631e.


Is hot air or hot water more effective at melting a frozen pipe? This question is related to specific heat, and I thought I understood it but I've managed to confuse myself. My initial reasoning was that since in both cases, the pipe is the same, the amount of energy needed to raise the temperature will depend on the specific heat only, and since air has a lower specific heat, it will be more effective. After I started thinking about it more, I thought that the specific heat of the pipe is actually what is important in determining how much energy will be needed to raise its temperature. In this case, the amount of energy needed to unfreeze the pipe should remain constant, and since the specific heat of water is higher, it is more effective at passing on higher amounts of energy than air. Are either of these reasons correct or am I really misunderstanding something? I appreciate any help or illumination you can offer! Thanks!

Specific heat relates to the heat capacity; how much temperature change you get for a specific mass and energy once the heat has finished ‘moving’.

But you also have to consider how fast the heat is moving.

For melting a frozen pipe with a fluid you also need to consider the heat resistivity of the fluid which is the inverse of heat conductivity. Some fluids like air are good heat insulators - much more so than water. The resistivity will limit the rate at which energy can be transferred to the mass.

Since water has a much lower resistivity than air ($\sim 1.7\ \mathrm{m\cdot K/W}$ vs $\sim 50\ \mathrm{m\cdot K/W}$) you can expect a heat transfer rate into the cold pipe about 25 times faster. This all assumes the air and water are at the same temperature.


What happens to concrete if you add lye to the mix? What happens if you’re mixing concrete and you add some lye to it? Does it make it weaker, stronger or does nothing at all? I’m guessing concrete is basic so adding lye to it might expedite the reaction, causing it harden faster?  Alternatively, it may react to with the water and starve the mixture?

The article referenced in the comment above (Ref 1) is not relevant to concrete. It analyzes solutions with a water to solids (cement plus $NaOH$) ratio of 1, whereas ordinary concrete seldom goes above 0.5. This is a paste, not a structural material. The $NaOH$ speeded up early hydration at 1 M, and retarded it at 2 M. It was apparently affecting the aluminates more than the silicates.
The composition of typical portland cements varies, but is usually considered to be mostly $C_3S$, a faster reactor and $C_2S$, a slower, but eventually stronger component. Ref 2

Many admixtures are used in portland cement concrete (Ref 3), but sodium hydroxide is not one of them. $NaOH$ is used as a promoter with silica-rich additives like flyash in highly modified cements, but portland cement is already silica-poor. When portland cement hydrates, $Ca(OH)_2$ is formed, which is not a really good strengthener, except that over long periods of time, it will become incorporated into the silicate gel, strengthening the concrete somewhat.
However, there are adverse reactions noted with silica-rich aggregates and the high alkalinity of sodium hydroxide (Ref 4), so it is always advisable to keep the sodium level low, whether from an alkali, an admixture, or external applications, like salt.
Ref 1: https://www.sciencedirect.com/science/article/abs/pii/000888468990046X
Ref 2: https://en.wikipedia.org/wiki/Portland_cement
Ref 3: https://www.cement.org/cement-concrete-applications/concrete-materials/chemical-admixtures
Ref 4: https://www.concreteconstruction.net/how-to/repair/chemical-attack-on-hardened-concrete_o


Diffusion/Streaky bands on TLC plate These days I am working on a research project where I am supposed to run TLC for samples which are wax extracts from coconut butten nuts using chloroform as the solvent. All the samples have been concentrated to 10% from its original volume. I recently run a TLC for a sample using hexane(90%):diethyl ether(7.5%):acetic acid(1%) as the solvent system (this solvent system has been used to run similar wax extracts). But my TLC run resulted me very diffused streaks  having some curves towards both the sides  of the tlc plate. Can anyone tell me why this happens??

Several things that may help your situation, listed roughly in the order in which I'd try them:

1. Application of the sample to the TLC plate

TLC is most usually ran by applying a small spot of the sample to the plate, and allowing it to run up. Analytically, this gives the greatest resolution, and as such for things such as monitoring reactions or identifying components in  column chromatography fractions, spots are the most widely used.

In some cases, the TLC plate is used to purify components of the sample, this is called preparative layer chromatography, or PLC for short. In PLC, bands are applied to the plate, to allow greater quantities of the sample to be applied and hence separated.

When applying bands to the PLC plate, it's crucial to ensure that the band doesn't get too close to the edges (as it has in the images you've uploaded), as a rule of thumb, at least 1/2 cm from each side should be left clean.

The reason for this is that the solvent doesn't run perfectly level up the plate, this is most evident at the edges of the plate. If you re-run your TLC plates with a shorter band, you should see a decrease in the curving effect that you're observing.

2. Quantity of sample loaded onto the TLC plate

Looking at the images you've provided, you may also have loaded too much sample onto the TLC plate, saturating the silica and reducing the separation.

You could try either increasing the size of the plate, increasing the thickness of the silica layer (special PLC plates are available with thicker layers of silica, allowing more material to be loaded), or simply just diluting your sample and loading less on.

3. Solvent system

Although you say that "this solvent system has been used to run similar wax extracts", its always worth trying different eluents with your chromatography, unless your sample is identical to the prep you're following.  A different modifier (formic acid or trifluoroacetic acid often work well), or different solvent mixtures (methanol in dichloromethane is often a lifesaver) might help tighten up the bands.

Finally, and possibly most of all, make sure you completely dry the plate out before visualising it (leave it sat on the side for a few minutes or use a hot air gun).


What type of isomers are cis-but-2-ene and 2-methylpropene? One of my homework problems is asking whether cis-but-2-ene and methylpropene are cis-trans isomers or constitutional isomers. But I'm not sure about whether cis-trans isomers include those which the two functional groups exist on the same carbon in the double bond. So does it?

I've searched on Google and looked in my textbook, but they doesn't seem to have anything.

(2Z)-But-2-ene and 2-methylprop-1-ene are constitutional isomers as they have different connectivities, i.e. the sequence in which the atoms are bonded to each other is not the same. For example, the carbons highlighted here are different: in (2Z)-but-2-ene it is only bonded to two other carbons (through a single bond and a double bond), but in 2-methylprop-1-ene it is bonded to three other carbons:

Cis/trans isomers are a subset of stereoisomers, which need to have the same constitution, or molecular connectivity. The two isomers of but-2-ene satisfy this:

Note how both highlighted carbons are now bonded to the same partners, but with a different spatial arrangement.


Hydrolysis of Phosphorus Trichloride During hydrolysis of a halide of carbon via $\mathrm{S_N2}$ mechanism, the incoming neutral water molecule attacks the partially positive carbon centre from the side opposite to the halide side and forms a transition state where the hybridisation of centre is $\mathrm{sp^2}$. The nucleophile and the leaving group (halide) thus are attached to opposite sides of the same p orbital (now elongated). Ultimately the nucleophile kicks off the halide from the back.

But when we are carrying hydrolysis on $\ce{PCl3}$ (phosphorus trichloride) why does the water attack an empty d-orbital of the central phosphorus atom and not do the same mechanism as it does in the previous case mentioned. Why can't the attack take place like above leading to formation of an elongated p orbital?

I don’t know where you heard about the water molecule interacting with d orbitals before displacement takes place, but the idea is wrong. Phosphorus’ d orbitals are energetically too far removed to take part in bonding. Remember that according to the aufbau principle, 3d orbitals are higher in energy than 4s! They exist and you can excite electrons into them but that’s it (and you need high energies for that excitation).

Rather, the oxygen of water interacts with the $\sigma^*_{\ce{P-Cl}}$ orbital. Since phosphorus uses an orbital with a very high p-content, it has a notable lobe on the far side allowing for interaction. An intermediate tetracoordinated phosphorus species is formed which displaces the better leaving group (chloride). Water can lose a proton at any point in this mechanism.

The mechanism is basically identical to that of a carbon-centred $\mathrm{S_N2}$ reaction except that the intermediate breaks down much faster in carbon-based reactions.


Why does the hydrogen atom in HCl (when dissociated in water) bond to the oxygen rather than staying with the chlorine atom? Why does the hydrogen atom in $\ce{HCl}$ (when dissociated in water) bond to the oxygen rather than staying with the chlorine atom. They ($\ce{Cl}$ and $\ce{O}$) have roughly the same electronegativity (3.5).

I saw this question:
Dissociation of HCl in aqueous solution
but it doesn't answer what I'm asking.

Look at it from the point of view of the chloride ion. In $\ce{HCl}$, the chloride is attracted to only one hydrogen. But when dissolved in water, the chloride is attracted to the partial positive charges on the hydrogens of several water molecules. Each $\ce{H2O-Cl}$ attraction is less than the $\ce{H-Cl}$ attraction but many of them act together to attract the $\ce{Cl}$ ion away from $\ce{HCl}$. Once the $\ce{Cl}$ is attracted away, that leaves a hydrogen ion to be attracted to the partially negative oxygen of a water molecule.

These forces between multiple waters and an $\ce{HCl}$ are stronger than forces between adjacent $\ce{HCl}$ molecules. Otherwise, $\ce{HCl}$ would be a liquid at room temperature instead of a gas.


What is the effect of dilution on the degree of dissociation of a weak acid or base? Please ponder on an equilibrium of a very weak base, say $\ce{MOH}$. This base undergoes a very low ionization, even in its pure form (pure form implies that there isn't the presence of a solvent). The concentration of  $\ce{OH-}$ ions, that our base generates, can be assumed to be $ \le 10^{-6} $. Now, let us add some water to $\ce{MOH}$ so as to dilute it. Ostwald's dilution law states that if the degree of dissociation ($ \alpha $) is very less as compared to unity, which in this case, is, it can be calculated by the equation $ \alpha = \sqrt{\frac{K_a}{C}} $. This leads to the conclusion that as I make the solution less concentrated (by dilution), $ \alpha $ increases.

But if I think of it in a different way, I get confused. For instance, please consider the simultaneous equilibria:
$$ \ce{MOH <=> M+ + OH-} \tag{1}$$
$$ \ce{H_{2}O <=> H+ + OH-} \tag{2}$$
Since the concentration of $ \ce{OH-} $ ions our base generated in its pure form was comparable to the concentration of $ \ce{OH-} $ ions generated by water, they may cause the first equilibrium to be shifted in the backward direction, thus leading to a decrease in $ \alpha $ upon dilution. Is there any point I'm missing which makes me conclude the opposite of what one concludes from Ostwald's dilution law?

Your argument is valid, though I believe the scenario you describe is not within the range of applicability of Ostwald's law. Allow me to recast your example in a more obvious form.

You have a solution of weak base. You add a stronger base. The increase in hydroxide concentration leads, by Le Chatelier's principle, to a lower extent of dissociation.*

This is beyond the scope of Ostwald's law, which tacitly assumes the lack of what amounts to a common-ion effect. Ostwald's law holds if the solvent which you're using to dilute your solution does not interfere (or interferes only weakly) with your electrolyte's solubility; the added solvent must serve to decrease the concentration of your electrolyte. Evidently, then, adding a stronger base will interfere with the solubility of your weak base, and is not governed by Ostwald's law.

*There is perhaps some subtlety here. As an example, consider the salt $\ce{MX}$, which dissociates weakly into $\ce{M+}$ and $\ce{X-}$. The intuition behind the greater degree of dissociation is that (1) upon dilution, the equilibrium constant for dissociation doesn't change, whereas (2) the concentration of each species decreases. Because dissociation produces more ions than was originally present, this always leads to $Q < K$, and more dissociation will occur.

If we  add a salt $\ce{MY}$, then, we are introducing two effects: (1) decreased concentration of $\ce{X-}$ and (2) increased concentration of $\ce{M+}$. Because these effects pull the equilibrium in opposite directions, we can't conclusively determine what exactly will happen. Going back to your example, if we add a slightly stronger base, then we get exactly these two contrasting effects, and the change in the degree of dissociation is indeterminate. In contrast, if we add a much stronger base, then certainly (2) dominates and the degree of dissociation is reduced.


Why does butter get soft with heat, but not ice? Sometimes a recipe asks me to soften the butter before I use it. This can be accomplished with a few seconds in the microwave, or just by leaving it out at room temperature for a while.

Ice, on the other hand, does not get soft with heat. It just immediately starts melting.

What chemical differences between the two would account for this?

The main difference between warming 'solid' butter and warming solid ice is that one is a mixture and one is a pure substance.

Ice has a crystal structure set in place by the shape and polarity of the water molecule. It requires quite a bit of energy to be removed from 0 degree water to cause the formation of the solid phase. The structure has two options - remain water, or, once cooled enough, become solid.

Butter is a mixture of fat and water and other components. As this mixture warms, the fat component changes viscosity and becomes more liquid and less sludgy. The water does not change much, but is trapped in the layers of the fats. At 40 F, just above refrigerator temperature, butter is 'hard' because the fats are not flowing. As the temperature warms, the fats start to move, thus softening the butter.

The butter softening behavior can be seen in some metals as they get heated close to the melting point. Iron becomes 'soft' and malleable, and can be forged. Ice has a particular, rigid structure, so never has a 'soft solid' stage.


Clarification about Frost diagram y-axis conventions? I was looking at the Frost diagram of Manganese, and on the UC Davis ChemWiki, I see this one:

However, the one straight from my textbook, "Descriptive Inorganic Chemistry" (Overton), looks like this:

In the first image, the y-axis is $\mathbf{nE^o}$ and the x-axis goes from 0 to +7 from left to right, but in the second image, the y-axis is $\mathbf{-nE^o}$ and the x-axis goes from +7 to 0 from left to right.

I'm wondering if reversing the oxidation state ordering changes the y-axis definition between $nE^o$ and $-nE^o$.

It seems like a silly question to me, but I still want to be sure.

The UC Davis version is incorrect. The free energy of the system ($\Delta G$) is equal to -nFE, so y-axis is reported as $\Delta G / F$ or -nE, according to the original report by Arthur Frost (J. Am. Chem. Soc., 1951, 73 (6), 2680-2682).

Note: In the original report free energy was denoted by $\Delta F$, rather than $\Delta G$. The x-axis designation is insignificant.


When naming hydrocarbons, when do you number the branch or double bond? I have realized that not all chains and double bonds are numbered. I have noticed that up until 3 carbons in a chain, the double bond is not numbered, but I am not sure about the chain.

When there are 2 carbons in a chain, the double bond will be at the number 'one' position. Therefore it's not always mentioned in the naming.

For example:  Ethene
It can be written as 1-ethene or eth-1-ene. Since the position of the double bond can be at no other places than the first, the position in the naming you saw, wasn't mentioned.

When 3 or more carbons are present, the double bond can be between any 2 of the carbons, ie. either the first position or the second. (but-1-ene/butene or but-2-ene)

Position of double bonds and that of branches, must always be mentioned while naming organic compounds.


Why is tetramethylammonium hydroxide a stronger base than tri- or dimethyl amine? Of the following, which compound is the most basic??

  $\ce{Me3N}$
  $\ce{Me2NH}$
  $\ce{Me4NOH}$
  $\ce{NaOH}$

Between 1 and 2, 1 is more basic as the 3 methyl groups would produce a +I effect resulting in better availability of the lone pair than in option 1.

But the book says it is 3. I don't understand why. I do make out that it is ionic, but how does this fact make it more basic than option 1?

Between 1 and 2, 2 is more basic as the 3 methyl groups would produce a +I effect resulting in better availability of the lone pair than in option 2.

I agree that more methyl groups would result in a greater +I effect.

  But the book says it is 3. I don't understand why. I do make out that it is ionic, but how does this fact make it more basic than option 1?

The fully methylated ammonium part is not basic. It has no easily labile proton. As a result, tetramethylammonium hydroxide acts as a plain hydroxide salt, like sodium hydroxide. Hydroxide ion is much more basic than ammonia (at least in water).
So we've eliminated the amines with lone pairs as contenders for "most basic" because we've established hydroxide to be much more basic than an amine, regardless of the extent of the +I effect in an amine.

We're left with $\ce{NaOH}$ and tetramethylammonium hydroxide. The only difference between the two is the cation.

$\ce{Na+}$ is electron-withdrawing (electronegative); it has a positive charge. This lessens (stabilizes) the negative charge density on the hydroxide portion, reducing the basicity of the hydroxide ion slightly.

$\ce{(H3C)4N+}$ has a similar effect on the hydroxide ion part. However, tetramethylammonium ion has four inductively donating methyl groups to lessen its electron-withdrawing tendency. I suspect this is what your book is getting at.

Some speculation follows. In addition, $\ce{Na+}$ is likely smaller than $\ce{(H3C)4N+}$, which has a lot of bulk associated with its four methyl groups. This likely allows $\ce{Na+}$ ions to more effectively solvate and thereby stabilize hydroxide ions in solution. This again decreases basicity, but I'm unsure of how much of an effect this has, since the solution molecules will be present in a much greater quantity than the solute ions.