text:
Reduction of alkyl halides to alkanes

For all the three parts, only methane has been given as the product in my textbook.
In 1) and 2), according to the mechanism I wrote, I get both ethane and methane as products. I think both these products should be formed in good yield. Then, why my products do not match?
In 3) , I got methane as the only product as given in the textbook. Can you please check the mechanism?

You won't form a free methyl carbanion under those conditions because it is far too unstable. If it did form it's likely reaction would probably be to deprotonate the methyl chloride rather than attack as a nucleophile because it is such as strong base and the methylene chloride anion is somewhat stabilised by the electron withdrawing chlorine.

@bon what makes the carbanion unstable under those conditions? Are you suggesting a free radical mechanism in the first two reactions?

Do you mean that the carbanion will deprotonate methyl chloride rather than act as a nucleophile because it is a better base than a nucleophile? If not, how else did you decide that?

It's just a really unstable species in the first place because carbon has relatively high energy orbitals. Yes I think it will be a better base than a nucleophile (if it were ever to exist at all). As Klaus' answer suggests the mechanism involves metal insertion into the C-X bond which I expect does proceed via a radical mechanism.

Okay, I get it now. If the reaction can proceed through a free radical mechanism, it is less likely that unstable carbanion should get formed at all in the reaction.  But if we are saying that the reaction will proceed through a free radical mechanism and not an ionic mechanism, we are in a way asserting that free radical here is more stable than carbanion. Why will that be? Also, I can't find the mechanism of reduction of alkyl halides by LiAlH4 anywhere, so I don't know if the mechanism in the third part is right. Can you please take a look?

@bon Which answer of Klaus are you referring to?

@ApoorvPotnis He deleted it after I posted the comment (I can still see it because I have >10k rep).

Your answer for question #3 is correct. However, there is a fundamental misconception you used in #1 and #2 that led to an incorrect answer.
In your notes, it seems that electrons from the metal ($\ce{Zn}$ and $\ce{Mg}$) somehow "attack" the alkyl halide and generate the methine ion ($\ce{CH3-}$). This does not happen for two reasons: Firstly, electrons are not free species in solution (like ions and molecules), they are only transferred from one molecule (reductant) to another (oxidant), so they cannot attack anything by themselves. Secondly, the methine ion is an extremely strong base (the pKa for methane is around 50, so the methine ion, its conjugate base, has to be VERY basic), which could only be produced by using an even stronger base to abstract a proton from it.
What actually happens in both cases is what is called metallic insertion. The metal ($\ce{Zn}$ or $\ce{Mg}$) is inserted between the carbon-halide bond, forcing the carbon atom to hold a partial negative charge (the author of this picture forgot to include the $\ce X$ atom bound to $\ce M$ in the organometal compound, but it is not that relevant here).

https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alhalrx4.htm
When it happens, you end up having a nucleophilic carbon, and it could indeed react with a $\ce{CH3Cl}$ molecule to yield $\ce{CH3-CH3}$ under a unimolecular substitution mechanism ($\ce{S_N1}$). However, such substitution reactions are very slow compared to acid-base reactions, and since you have acidic hydrogens available in both cases ($\ce{HCl}$ in question #1 and $\ce{H2O}$ in #2) the acid-base reaction (abstraction of a H atom from $\ce{HCl}$ or water) would happen fast enough to prevent the formation of $\ce{CH3-CH3}$ adduct.

And sir ,if we does not follow mechanism of op, how does metal insertion occur by itself?.


text:
Are there any reactions where two triatomics form three of the same diatomic?
A chemical equation: $\ce{A}+\ce{B}\to\ce{3C}$
Where $\ce{A}$ & $\ce{B}$ are triatomic molecules and C is a diatomic molecule. Find out $\ce{A}$, $\ce{B}$, $\ce{C}$?
So far the most worthy options for answer I've found are  $\ce{S_3}$ (trisulfur) and ozone ($\ce{O_3}$) as $\ce{A}$ & $\ce{B}$ and $\ce{SO}$ as $\ce{C}$. But I am not sure if this reaction will occur : $\ce{S3 + O3 \to 3SO}$
So, are there any better options? I am even worried about if this even has an answer in chemistry known to this day.

$\ce{N2O + NO2 -> 3NO}$. While still unrealistic, this one is more realistic than yours.

Is there any stipulation that A and B cannot be the same? Because $\ce{2 O3 -> 3 O2}$ feels like the most realistic thing here.

@Ivan does that reaction occur in reality?

@OrthocresolDoes that occur in reality?

The words "realistic" in orthocresol's comment and "unrealistic" in mine were there for a reason.

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text:
Hybridisation of terminal nitrogen in diazomethane
I have a few questions about the terminal nitrogen (highlighted in red) in diazomethane, $\ce{CH2N2}$.

Is that nitrogen $\mathrm{sp}$ or $\mathrm{sp^2}$ hybridised?
What type of orbitals do the lone pairs on that nitrogen reside in?
Does the negative charge on nitrogen undergo resonance?

Another resonance form involves negative charge on carbon.

Terminal atoms are always sp hybridised (at most; if hybridisation is at all applicable). In general, atoms are never hybridised, only orbitals can be hybridised.

I did not get it.. How terminal atoms are always sp hybridised? Like if there is a methyl group it wont be sp.. Pls elaborate

@surabhii The carbon atom in the methyl group isn't the terminal atom. The hydrogen is. And the hydrogen orbitals are not hybridized.

In da wikipedia  you can see that the C=N=N fragment is linear AND another resonance structure involves triple N-N bond. So, the answers

it is sp in one resonance structure and probably sp2  in another
resonance structure.
depends on the resonance structure
considered
yep.

Diazomethane typically reacts with its negatively charged carbon, so this is of little consequence though.

Compounds wouldn’t change hybridisation from one resonance structure to the other.

@Jan Orly ? Why not, if the angles do not change?

Because resonance is only our weak term to better understand intrinsic features of molecular orbital theory. What we see as resonance hybrids is, in fact, a single MO structure. Within a  single MO structure, hybridisation does not change.

@Jan to be precise, sp + p lone pairs would give exactly same electron charge distribution as two sp2 lone pairs. Hybridization here is a matter of convenience and simplicity, and speaking about higher matter (non-resonance view) is way above the level of the question. And no, resonance view is not talking about group MO systems anything, you are thinking too good about this particular way of thinking.

As usual with these small molecules, the bonding situation is a lot more complicated than expected. In this very molecule this is to the extend that not only the Lewis formalism, but also the conventional resonance formalism breaks down.
However, Jan's description is sound, and provides a good starting point for further analysis.
Unfortunately formulating a clear cut valence bond description is difficult, because it involves strongly coupled electrons. This is in part due to the high symmetry of the molecule, $C_\mathrm{2v}$, and the even higher local symmetry of the $\ce{CNN}$ moiety, $C_\infty$. Because of the nature of molecular orbital theory, i.e. being better at describing delocalisation, it is actually easier to understand.
Here are the molecular orbitals computed at the DF-M06L/def2-SVP level of theory:

And here is the same thing in numbers

 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-14.323 is N2-s=1.00
 Alpha occ 2 OE=-14.252 is N1-s=1.00
 Alpha occ 3 OE=-10.099 is C3-s=1.00
 Alpha occ 4 OE=-1.055 is N2-s=0.51 N1-s=0.29 N1-p=0.10
 Alpha occ 5 OE=-0.802 is N2-p=0.31 C3-s=0.31 N1-s=0.14 N2-s=0.12 C3-p=0.07
 Alpha occ 6 OE=-0.518 is C3-p=0.26 C3-s=0.24 H5-s=0.16 H4-s=0.16 N1-s=0.07 N2-s=0.06
 Alpha occ 7 OE=-0.487 is N2-p=0.45 C3-p=0.24 N1-p=0.14 H5-s=0.08 H4-s=0.08
 Alpha occ 8 OE=-0.450 is N2-p=0.61 N1-p=0.29 C3-p=0.09
 Alpha occ 9 OE=-0.407 is N1-s=0.44 N1-p=0.41 N2-p=0.08
 Alpha occ 10 OE=-0.366 is N1-p=0.28 C3-p=0.24 N2-p=0.16 H5-s=0.15 H4-s=0.15
 Alpha occ 11 OE=-0.206 is C3-p=0.57 N1-p=0.38
 Alpha vir 12 OE=-0.082 is N1-p=0.55 N2-p=0.32 H5-s=0.06 H4-s=0.06
 Alpha vir 13 OE=-0.005 is C3-p=0.33 N2-p=0.33 N1-p=0.32

From the visual inspection we can see that the terminal nitrogen has only one orbital that could be classified as a lone pair, i.e mo 9; the hybridisation according to this calculation is approximately sp. There are three bonding π orbitals, i.e. mo 7, 8, 10. And there is also one anti-bonding π orbital, i.e. mo 11. The sigma bond is formed by two bonding (mo 4, 5) and one anti-bonding orbital (mo 6).
This is no coincidence, but pretty much always the case for terminal atoms.
Due to symmetry constraints, the hybridisation pattern at terminal atoms is usually:

for the orbital forming the single bond $\mathrm{sp}^{\geq1}$
two orthogonal (π) lone pairs $2\times\mathrm{p}$
one (σ) lone pair $\mathrm{sp}^{\leq1}$

Localisation of the orbitals is not easy for this compound. I performed a natural bond orbital analysis, which transforms the (delocalised) canonical orbitals into hybridorbital like ones, but it results in a low occupied carbon lone pair orbital and a partially occupied anti-bonding nitrogen-nitrogen π orbital.

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   1. - 3. core orbitals (omitted)
   4. (1.97583) LP ( 1) N  1            s( 67.37%)p 0.48( 32.60%)d 0.00(  0.03%)
   5. (1.33940) LP ( 1) C  3            s(  0.00%)p 1.00( 99.92%)d 0.00(  0.08%)
   6. (1.99655) BD ( 1) N  1- N  2
               ( 43.82%)   0.6620* N  1 s( 33.02%)p 2.02( 66.72%)d 0.01(  0.26%)
               ( 56.18%)   0.7495* N  2 s( 46.00%)p 1.17( 53.92%)d 0.00(  0.08%)
   7. (1.99587) BD ( 2) N  1- N  2
               ( 51.93%)   0.7206* N  1 s(  0.00%)p 1.00( 99.63%)d 0.00(  0.37%)
               ( 48.07%)   0.6933* N  2 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
   8. (1.98025) BD ( 3) N  1- N  2
               ( 39.11%)   0.6254* N  1 s(  0.00%)p 1.00( 99.54%)d 0.00(  0.46%)
               ( 60.89%)   0.7803* N  2 s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)
   9. (1.99580) BD ( 1) N  2- C  3
               ( 65.65%)   0.8103* N  2 s( 53.76%)p 0.86( 46.18%)d 0.00(  0.05%)
               ( 34.35%)   0.5861* C  3 s( 30.11%)p 2.32( 69.75%)d 0.00(  0.14%)
  10. (1.95419) BD ( 1) C  3- H  4
               ( 60.92%)   0.7805* C  3 s( 34.95%)p 1.86( 65.03%)d 0.00(  0.02%)
               ( 39.08%)   0.6251* H  4 s( 99.91%)p 0.00(  0.09%)
  11. (1.95419) BD ( 1) C  3- H  5
               ( 60.92%)   0.7805* C  3 s( 34.95%)p 1.86( 65.03%)d 0.00(  0.02%)
               ( 39.08%)   0.6251* H  5 s( 99.91%)p 0.00(  0.09%)
 ---------------- non-Lewis ----------------------------------------------------
  12. (0.01076) BD*( 1) N  1- N  2
               ( 56.18%)   0.7495* N  1 s( 33.02%)p 2.02( 66.72%)d 0.01(  0.26%)
               ( 43.82%)  -0.6620* N  2 s( 46.00%)p 1.17( 53.92%)d 0.00(  0.08%)
  13. (0.64416) BD*( 2) N  1- N  2
               ( 48.07%)   0.6933* N  1 s(  0.00%)p 1.00( 99.63%)d 0.00(  0.37%)
               ( 51.93%)  -0.7206* N  2 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
  14. (0.07754) BD*( 3) N  1- N  2
               ( 60.89%)   0.7803* N  1 s(  0.00%)p 1.00( 99.54%)d 0.00(  0.46%)
               ( 39.11%)  -0.6254* N  2 s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)

This is usually an indication of a strongly delocalised system such as a three-centre bond. Investigating the molecule with natural resonance theory gives the rather odd contribution of long range N1 - C3 bond.

 TOPO matrix for the leading resonance structure:

     Atom  1   2   3   4   5
     ---- --- --- --- --- ---
   1.  N   2   2   0   0   0
   2.  N   2   0   2   0   0
   3.  C   0   2   0   1   1
   4.  H   0   0   1   0   0
   5.  H   0   0   1   0   0

         Resonance
    RS   Weight(%)                  Added(Removed)
 ---------------------------------------------------------------------------
    1*     33.02
    2*(2)  32.31    N  1- N  2, ( N  2- C  3), ( N  1),  C  3
    3*(2)  28.81    N  1- C  3, ( N  2- C  3), ( N  1),  N  2
    minor contributions omitted
 ---------------------------------------------------------------------------
          100.00   * Total *                [* = reference structure]

According to NBO, the terminal nitrogen has an $\mathrm{sp}^{0.5}$ σ-type lone pair orbital, two $\mathrm{p}$ π-type bonding orbitals, and a $\mathrm{sp}^2$ σ-type bonding orbital.

Hybridisation is a mathematical tool to analyse a bonding situation. There is no correct way of applying it. It is also very important to understand an atom is never hybridised. (Even though organic chemists tend to use this jargon - even in textbooks.) The orbitals at an atom can be hybridised.

Until someone steps up and performs calculations, it is best to assume Occam’s razor for determining the hybridisation of the nitrogen atoms. The simplest solution for the outer nitrogen is to assume no hybridisation.
There are two common resonance structures for diazomethane:
$$\begin{gather}\ce{H2C=\overset{+}{N}=\overset{-}{N} <-> H2\overset{-}{C}-\overset{+}{N}#N}\\\ce{(A)\quad<->\quad(B)}\end{gather}\tag{1}$$
Resonance structure $\mathrm{(B)}$ has a single lone pair on the terminal nitrogen, so applying Occam’s razor tells us to put it into an s-type orbital. Such an orbital would also be present in resonance structure $\mathrm{(A)}$ and it would again house a lone pair.
Resonance structure $\mathrm{(B)}$ has no further lone pair but 2 orthogonal p orbitals which each form a π bond to the neighbouring nitrogen. One of these π bonds does not extend any further as it reaches the $\ce{C-H}$ bonds on carbon’s side; the other extends across to interact with the lone pair on carbon. There are three orbitals of this type; π1, π2 and π3; the first fully bonding, the second partially bonding, the third fully antibonding. The lower two are populated giving putting four electrons (one double bond and one lone pair) into that π system. This is important for resonance structure $\mathrm{(A)}$ that we’re about to come back to.
$\mathrm{(A)}$ must feature the same π system as there is no rehybridisation between resonance structures. Thus, there must also be two orthogonal π orbitals on the terminal nitrogen; one again forming a π bond with it’s neighbour (the one in plane with the hydrogens), the other this time housing a lone pair.
Instead of allowing for an unhybridised terminal nitrogen, the same analysis also permits the assumption that the terminal nitrogen is $\mathrm{sp}$ hybridised. In this case, it would be an $\mathrm{sp}$ orbital housing the lone pair present on nitrogen in both $\mathrm{(A)}$ and $\mathrm{(B)}$.
It is not a good idea to say ‘the negative charge participates in resonance’ in the way you seem to be intending it to be understood. The entire molecule is neutral, so any macroscopic partial negative charges must be balanced by a corresponding partial positive charge. Nitrogen being more electronegative than carbon allows us to assume the the partial negative charge is primarily located on the nitrogens while the partial positive charge is on carbon. However, this related question on the dipole moment of carbon monoxide shows that we should never assume partial charges or dipole moments by looking at the Lewis structures alone.

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text:
Is the reaction between thiosulfate and hydronium an acid-base reaction?
Is the following reaction an acid-base reaction:
$$\ce{2 H3O+(aq) + S2O3^2-(aq) -> 3H2O(l) + SO2(g) + 1/8 S8(s)}$$
And if so, which is the acid and which is the base?

$\ce{H3O+}$ is the acid, $\ce{S2O3^2-}$ is the base, and they react to produce an unstable product that decomposes further, which is not an acid/base reaction.

could you write the reaction of decomposition

Why, it's $\ce{H2S2O3 -> H2O + SO2 + S}$

but the ph levels changes

Ivan,so you say the reaction is 2H3O+(aq)+S2O3-2(aq)=>3H2S2O3=>3H2O(l)+3SO2(g)+3/8S8(s)?

Wikipedia (https://en.wikipedia.org/wiki/Acid%E2%80%93base_reaction) gives at least 7 different definitions of acid-base reactions, and they do not all require the formation of water + a salt.  A good example is in the OP; there is an acid and a base and they react.  Maybe you could call it a neutralization reaction?  And presumably the counter ions not shown in the equation form ionic bonds with each other, as one is positively charged and the other negatively charged.  That still doesn't have to be a salt though.

Two of the three acid-base theories (Brønsted and Lewis) also define an acid-base reaction while the third theory (Arrhenius) does not — it merely defines what acids and bases are.
In the Lewis theory, an acid-base reaction occurs when a lone pair donor interacts with a lone pair acceptor to form a new bond without a simultaneous redox reaction happening. The less complicated Brønsted-Lowry theory describes an acid-base reaction as a reaction in which a proton $\ce{H+}$ is transferred from an acid to a base, giving us a conjugate base and a conjugate acid, respectively.
As has been pointed out, your reaction is not a one-step reaction. Rather, we have to break it down into its constituent steps. The first step, a reaction of hydronium $\ce{H3O+}$ with thiosulfate $\ce{S2O3^2-}$ has to be an acid-base reaction because that is practically the only reaction hydronium undergoes.
$$\ce{2H3O+ + S2O3^2- -> H2S2O3 + 2 H2O}\tag{1}$$
The acid is the species that donates the transferred proton, the base is the species that accepts it. Therefore, we can assign the acid and base roles as shown in equation $(1')$.
$$\ce{2 \underset{acid}{H3O+} + \underset{base}{S2O3^2-} -> 2 \underset{conjugate base}{H2O} + \underset{conjugate acid}{H2S2O3}}\tag{$1'$}$$
The second reaction is a redox reaction. Thiosulfate (and therefore also its conjugate acid) has two sulfur atoms in the oxidation states of $\mathrm{+V}$ and $\mathrm{-I}$, respectively. These undergo a redox reaction leading to compounds of the oxidation state $\mathrm{+IV}$ (sulfurous acid $\ce{H2SO3}$) and $\pm0$ (elemental sulfur). We can use the full redox reaction determination algorithm, but I’m skipping that since it is a simple one-electron transfer.
$$\ce{H2S2O3 -> H2SO3 + 1/8 S8}\tag{2}$$
And finally, we have another decomposition namely that of sulfurous acid to sulfur dioxide $\ce{SO2}$ and water. This can be seen as the back-reaction of a Lewis acid-base reaction wherein water is the liberated Lewis base and sulfur dioxide the liberated Lewis acid. Indeed, the forward Lewis acid-base reaction (the reverse direction of this reaction) occurs, too. This reaction is in equilibrium, driven to the product side because of the volatility of $\ce{SO2}$.
$$\ce{H2SO3 (aq) <=>> H2O (l)  + SO2 (g) ^}\tag{3}$$
Therefore, the overall reaction is the sum of one Brønsted acid-base reaction, one redox reaction, and one Lewis acid-base reverse reaction.
$$\ce{2 H3O+ (aq) + S2O3^2- (aq) -> 3 H2O (l) + 1/8 S8 (s) + SO2 (g) ^}\tag{4}$$

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text:
How much more salt can I dissolve in hot water compared to cold water?
Hot water increases not only the speed of solubility but also the amount.
I don't know whether this is for all materials in water, apart from salt.
But, if this is the case, what causes hot water to dissolve more salt (or other solutes)?

Searching for 'salt solubility in different temperatures' in Google yields many results, among them this simple explanation and this table. I believe most basic chemistry books also cover this topic.

Your statement is not generally true. For NaCl, there is only a slight increase in solubility, can be much stronger for others, or even reversed. The reason is thermodynamics, of course, but would be nice if someone worked out when exactly what happens.

There seems to be two parts to your question. First, not all substances have increased solubility as water temperature rises. Gasses for example have decreased solubility in this case.
As for the second part of the question, increased heat adds energy to the water molecules which makes it easier to overcome the bonds between $\ce{Na}$ and $\ce{Cl}$. According to Wikipedia, the solubility of $\ce{NaCl}$ in nearly freezing water is $35.65 \, \mathrm{g} / 100 \, \mathrm{mL} \, \ce{H2O}$ versus boiling water which is $38.99 \,\mathrm{g} / 100 \, \mathrm{mL}\, \ce{H2O}$. Hope this helped.

Note that it isn't just gases that exhibit decreased solubility with increased temperature, but many solids also. Calcium hydroxide for example has a solubility in water of 1.89 g/L at 0C, 1.73 g/L at 20C and 0.66 g/L at 100C.


text:
Hofmann elimination

An unknown amine is treated with an excess of methyl iodide. Two equivalents of methyl iodide react with the amine. The amine is treated with silver oxide and water, and then heated to $120\ ^\circ \mathrm{C}$. The resulting products are trimethylamine and ethylene. The unknown amine is

$\ce{CH3CH2NHCH3}$

$\ce{CH3CH2NH2}$

$\ce{CH2=CHNH2}$

$\ce{CH2=CHNHCH3}$

I tried to get the answer using concept of Hofmann elimination.
The unknown amine has to be a secondary amine because after adding 2 equivalents of $\ce{CH3I}$, a quaternary ammonium iodide will be formed which on treatment with most silver hydroxide and heating thereafter well give tertiary amine, alkene and water. This is what happens when one of the alkyl group is other than methyl.
I have never seen this reaction anywhere working with amine containing a alkyl group having double bond, so I eliminated the fourth option and went for the first option and the answer was correct.

My question is that why the fourth option is not correct? What is the corresponding mechanism?

If the reaction worked, it would produce acetylene (ethyne/$\ce{HC#CH}$) instead of ethylene.

@DHMO can you please provide the mechanism

The mechanism is here but it doesn't matter. The point is that the carbon region would have one additional bond. So if you start with a double bond, you would end up having a triple bond.

The reaction has three steps.
I shall use N-methylethenamine, $\ce{CH2=CH-NH-CH3}$ (the fourth option), as an example.

Step $1$:
$$\ce{CH2=CH-NH-CH3 + 2CH3I -> [CH2=CH-N(CH3)3]+ + I-}$$
This step is an $\mathrm{S_N2}$ step.
The carbon in the methyl iodide is the electrophile while the nitrogen in the N-methylethenamine is the nucleophile.
As the comments have pointed out, this step is not viable. The following tautomerism happens:
$$\ce{CH2=CH-NH-CH3 <=> CH3-CH=N-CH3}$$
Also, the electrophile methyl iodide would attack on the carbon-carbon double bond instead of the nitrogen.

Step $2$:
$$\ce{Ag2O + I- -> AgI + AgO-}$$
$$\ce{AgO- + H2O -> AgOH + OH-}$$

Step $3$:
$$\ce{[CH2=CH-N(CH3)3]+ + OH- ->[\Delta] CH#CH + CH3-N(CH3)-CH3 + H2O}$$
This resembles an $\mathrm{E1}$ step.

According to this forum post, there should be no problem with the first step. The methyl iodide would not destroy the double bond.
There is also no hydrogen available to perform an addition reaction with the iodide ion to the N-methylethenamine to destroy the double bond.
I believe that the second step is also viable, also because of the lack of a good electrophile (silver doesn't count because silver can't react with carbon-carbon double bonds).
The third step should be slower because eliminiation reactions that result in alkynes are slower than elimination reactions that result in alkenes in general.

Therefore, the product would be ethyne $\ce{HC#CH}$ instead of ethylene $\ce{HC=CH}$.

It would not work at all as enamines tautomerise to imines.

Yeah the supposed enamine product doesn't exist. Furthermore even if you could isolate the enamine it is more likely to react with MeI on carbon not nitrogen.

@orthocresol do your remarks contradict with the link to the forum post?

No. That post assumes that the amine and the alkene are isolated. When they are conjugated like in an enamine the reactivity is different.


text:
Why does the addition of oxygen render materials colourless?
In our book, it is mentioned that $\ce{H2O2}$ (hydrogen peroxide) is used as a bleaching agent because of its oxidizing nature:
\begin{align}
  \ce{H2O2 &-> H2O + [O]}\\
 \ce{\text{coloured material} + [O] &-> \text{colourless material}}
\end{align}
Why does the second equation happen? In other words, what role does oxygen play in the absorption of light here?

The "colourless material" is essentially a different compound from the "coloured material". It isn't like mixing yellow and blue gives green. It is like mixing hydrogen and oxygen gives water which has radically different properties.

From Wikipedia, "An oxidizing bleach works by breaking the chemical bonds that make up the chromophore. This changes the molecule into a different substance that either does not contain a chromophore, or contains a chromophore that does not absorb visible light." So the oxygen basically just disrupts the region that can absorb visible like and make it colourless.

Many organic dyes (arguably, all of them) produce color due to electron transitions in conjugated $\pi$ systems. Strong oxidants (not just the peroxide, mind you, but also hypochlorite bleach and others) snap the double bonds and thus destroy these conjugated systems. So the color vanishes.
This is not applicable to inorganic dyes, which usually would not get bleached by peroxide. On the contrary, some transition metals ($\ce{Ti^4+,\;Cr^6+}$, and some others) which have little or no color by themselves would produce intensely colored complexes when hydrogen peroxide is added. These reactions are sometimes used in analytic chemistry to discover their presence.

Sir, Why it is not applicable to Inorganic Dyes? And what happens in the case of Ti and Cr that addition of Hydrogen Peroxide make it colourful?

Inorganic dyes don't have conjugated $\pi$ systems. As for the transition metals, why, they form some complexes with hydrogen peroxide, and it just so happens that these complexes have strong absorption bands in visible region.

"Organic dyes produce color due to electron transitions in conjugated $\pi$ systems." Could you go into more detail on how this happens, or point to a helpful source?

Also, I did read the Wikipedia page on chromophores. It does not explain how resonance helps capture photons in the visible light range.

Well, your second question contains the answer to your first one. Conjugated π systems do not care about our visible range. It just so happens that many of them have their transitions in that range.

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text:
Order of Basicity of Nitrogen Trihalides
I was asked to arrange nitrogen tri-halides in order of increasing basicity. On looking up the answer online, it turned out to be the following: $\ce{NF3}$ < $\ce{NCl3}$ < $\ce{NBr3}$ < $\ce{NI3}$
The reason given was that more electronegative halides attract negative charge & hence reduce its availability.

I thought that the order should be $\ce{NF3}$ > $\ce{NI3}$ > $\ce{NBr3}$ > $\ce{NCl3}$ because:

all the halides except fluorine have vacant d-orbitals. So, back-bonding should reduce the basicity of all but $\ce{NF3}$.

$\ce{NCl3}$ shows more effective back-bonding as compared to $\ce{NBr3}$ which shows more effective back-bonding than $\ce{NI3}$. Better the back-bonding, lesser is the availability of the lone pair on the nitrogen atom.

It seems to imply that electronegativity is a more dominant factor as compared to back-bonding. But when we compare the acidity of boron tri-halides, we consider back-bonding to be the dominant factor. Same goes for Ipso effect.
So, what is wrong with my assumptions?
PS: I got the order from Yahoo! answers(which is pretty unreliable BTW)
PSS: I often come across questions of this sort which ask me to compare acidity & basicity of two or more compounds(both organic & inorganic). I think of looking at the pKb/pKa values of the chemicals can help me to reliably get the order in some cases. But I couldn't find any source/compilation of those values. Do you know any source which lists the pKb/pKa values of chemicals? If so, please add it to your answer. It may prove to be really useful for me & other people.

Which theory of acid/base are we referring to? pKa refers specifically to Brønsted-Lowry while I feel that you are talking about Lewis base...

@DHMO Yes, I am talking about Lewis bases. I see your point. But a source of pKa/pKb values would be useful nonetheless. I'll edit the question. Thanks.

The pKa/pKb values would be quite useless as they are not soluble in water.

@DHMO I see. But anyway just for possible future reference, I want to know a source of the data. Thanks again:)

BTW, how would one actually measure the pH+ of NI3? It would be a bang up experiment!

@DrMoishePippik I agree. But we can measure the values for many other substances. I solve questions asking me to compare acidity & basicity of various compounds on a pretty regular basis. A table would be pretty useful to check my answers.

@DrMoishePippik I see that the last part regarding the compilation of pKa values is confusing a lot a people. I'll edit my question:)

The comment was just meant in humor... NI3 is fun to create in minuscule amounts, but any manipulation ends in a bang.

@DrMoishePippik I saw where you went there.

Electronegativity must be taken into account here. Hence nitrogen trifluoride will be least basic. There are lot of things that one should just accept in chemistry rather than reasoning and thinking a lot. I am not saying thinking is bad, I'm just saying that you can save this thinking for more logical subjects like Physics or Maths :)

See, as far as I know, backbonding will be surely negligible for NBr3 and NI3 as pπ-dπ bonds between 2p-4d and 2p-5d orbitals will be quite ineffective. Also, I have got a source which says that the N-Cl bond in NCl3 is mostly non-polar, so that means there is similar electron density between these atoms, so backbonding is ruled out here as well. Hence electronegativity maybe the only dominating factor here

https://chemistry.stackexchange.com/q/26272/102629

I know I am 4.5 years late but .... I remember studying in Organic Chemistry that I effect of halides is dominant over R effect.

This can be attributed to the electronegativity of nitrogen and the lone pairs in halogens. Most bonds are all about energy if it releases more energy, it would be the "correct" structure/interaction. In boron, there are no lone pairs and backbonding occurs even if donor atom is more electronegative. But in case of halogens we have lone pairs and they will lessen the magnitude of bond enthalpy. Obviously we want a theoretical approach to check if the backbonding is favourable or not. So it is best to compare electronegativity difference in such cases. The halogens below Cl are less electronegative than nitrogen and would not normally show backbonding with it. In case of chlorine lone pair repulsions outweigh the electronegativity difference. In fluorine there is no vacant orbital so no backbonding.
Thus backbonding isn't a deciding factor in this case.

I would really appreciate if the point that I missed is pointed out or is written as a separate answer. The point of a community as this is to share correct knowledge. The validity of answers can be ensured only when faults are pointed out.


text:
How to draw Fischer projections for carbon chain having more than two carbons?
How can we draw Fischer projections for one-five carbon compounds?
For example, say we have pentane-2,3,4-triol ($\ce{CH3-CHOH-CHOH-CHOH-CH3}$; image below). With wedge and dash structure, the alcohol group is below the plane for the 2nd and 4th carbons but above if for the 3rd carbon. How can we draw the Fischer projection of this compound?

I read an extract from the chapter about carbohydrates in Solomon's book (10th edition) but only understood partially how to draw Fischer projections as it is given very briefly. Also, a brief introduction is given in chapter 5 but it is for two-carbon compounds only.
I have tried a lot but cannot understand how to construct it or what are the conventions on which they are made. Apparently, the web doesn't describe Fischer projections in a very detailed way probably because they are generally only used in biochemistry.

This video is quite informative in my opinion.

Thanks for the link it is good for explaining one chirality centre conversion. But it doesn't explain more than three chirality centres. On the web similarly no one has explained more that three chirality centres in a very elaborative way.

@Resorcinol Please do not use \ce{...} for something that is not a chemical. It can have undesired side-effects as you should have seen in the edit preview. Also, please use the edit preview to check how your edits will end up. You can see what the post looked like here. Finally, quotation markup should be reserved for quotations.

The first things to note when drawing Fischer projections are the general rules for drawing them. The first general rule concerns the choices of ‘top’ and ‘bottom’ while the second concerns how each cross is interpreted.

In a Fischer projection, the carbon atom with the highest oxidation state should be closer to the top than the bottom.
In cases where the highest oxidation state is in the centre or where multiple carbons have the same highest oxidation state, the second-highest applies analogously and so forth.

In a Fischer projection, the vertical bonds are always considered to point behind the paper (or screen) plane (hash) while horizontal bonds are considered to point in front (wedge).
This is exemplified in figure 1.

Figure 1: How stereochemical information is encoded in Fischer projections.
You compound presents a problem according to these rules as there is no most oxidised carbon. Rather, all three carbons 2, 3 and 4 are in the oxidation state $\pm 0$ while the two terminal ones are both $\mathrm{-III}$. To ease this limitation, I am going to discuss the Fischer projection with the related molecule shown in figure 2, which has one terminal carbon oxidised to a carboxylic acid breaking the tie.

Figure 2: L-5-deoxyribonic acid; a replacement molecule with the same stereochemistry on the central hydroxy groups.
For this molecule, it is clear that the acid should be put on top. Going down each bond, we need to rotate them until we have the longest chain in the vertical positions and pointing backwards (cf figure 1). I’ll let you use a modelling kit, pen and paper or your 3D viewing to perform the actual transformations; the final product you should end up with is the following in figure 3.

Figure 3: Fischer projection of the molecule in figure 2.
There should be little debate on the carboxylic acid; and whether the carbonyl group or the hydroxy group of it is pointing to the left does not matter; we can rotate around the $\ce{C{1}-C{2}}$ bond. To see the correct orientation of $\ce{C{2}}$ and $\ce{C{4}}$, you can take the skeletal formula from figure 2 and look from above; for $\ce{C{3}}$ you should be looking from below. We find that all hydroxy groups are pointing in the same direction; here: to the left. This means my molecule can be seen as a derivative of ribose; and since the lowest asymmetric carbon’s hydroxy group is pointing to the left, it derives from L-ribose.
Back to your molecule. Not only is the designation ambiguous (see above on the highest oxidation state) but it is also $C_\mathrm{s}$ symmetric, i.e. achiral. I have drawn your molecule along with its plane of symmetry (dashed) in figure 4.

Figure 4: The original molecule in question; 1,5-dideoxyribitol or (2R,3s,4S)-pentane-2,3,4-triol.
We need to arbitrarily choose a ‘top’ carbon here. For reasons of consistency, let’s choose the one on the left that was an acid in the former structure (figure 2). This means, we have a very easy job of drawing the Fischer projection as we already did it. It is shown in figure 5.

Figure 5: Fischer projection of the triol shown in figure 4.
We may be tempted to assign an L-configuration immediately but remember we realised that the molecule should have a plane of symmetry. Well, it does in the Fischer projection, too; the plane is horizontal and encloses the $\ce{HO-C-H}$ bonds of the central carbon atom. We can also turn around our piece of paper or PC screen and realise that we can thereby match it perfectly with its mirror image, shown in figure 5 to the right of the dashed mirror plane. Thereby we have twice shown that your molecule is achiral, that both a hydroxy group pointing to the left and to the right on the bottommost asymmetric carbon is possible and that therefore assigning a D/L stereodescriptor is meaningless.

But according to other sources the carbon atom which is given number one according to IUPAC nomenclature is placed at top not the one with highest oxidation state . Also you said carboxylic acid should be at top it is also numbered one but it has the same oxidation state of III same as that of terminal methyl ?

Thank you so much jan for your answer. Can you please explain how can we rotate the bonds until the l9ngest chain is in the vertical position ? Also what is the meaning of all the bonds are pointing backwards ?

Also why for c3 we are looking from below and for others from above?

@RaghavSingal Not sure about whether it is the oxidation state or the one numbered as number one. I’ld have to find that out. Note that oxidation states are signed, so $\mathrm{-III < +I < +III}$

@RaghavSingal Drawing the rotations on paper is rather hard to do and it should benefit you more if you attempt it yourself.

@RaghavSingal We are looking from below for $\ce{C{3}}$. We want to have the bonds in the chain pointing backwards; if we look from below, they are.

@Jan The carbon at the top should be $\ce{C}1$, not the highest oxidation state. Source:  p. 375 of
Organic Chemistry: Methane to Macromolecules by Roberts, Stewart and Caserio (link).


text:
Balance reaction equation for stabilised PAN cylisation reaction
I am dealing with polymer chemistry and trying to balance one of these reaction steps. I have read a lot about the cyclisation reaction step but it is not clear to me how to balance these equations or what happens between the first step A) and B1) cyclised form of polyacrylonitril?

By looking at the image and counting the H- and C-atoms, I can tell that polyacrylonitril in the first step has 13 H-atoms (the same amount of C-atoms like the cyclised structure) while the cyclised structure gives 15 H-atoms. Can someone help me to balance or comment on this or what is missing in this part?
P.s.: I know that it is an increasing polymer chain but I just would like to truncate it to this step.

Author of this picture didn't feel the need to keep it to the same fragment of the polymer.

So which part is missing in order to get from A) to B1)?I know that B1) can have more C- and H-atoms but I need to know which fragment is missing. I was thinking of a Diels-Alder similiar reaction (even if this is another mechanism) but I am not sure which fragment should be added and increases the polymer chain.

sigh Minimally left or right. It's simple polymerisation of CN groups.


text:
How would you measure the rate of this SN1 reaction?
I am performing the following $\mathrm{S_N1}$ reaction:
$$\ce{t-BuOH + HCl -> t-BuCl + H2O}$$
I would also like to measure the rate of this reaction, but I cannot seem to find a way. I initially envisaged measuring the rate by using an indicator to tell when the conc. $\ce{HCl}$ had been used up.
However, I do not believe the reaction has gone to completion, thus why, when using my method of using an indicator, the indicator never changed colour, due to leftover acid - also Clayden's Organic Chemistry [1, p. 348] cites a 90% yield, presumably meaning not all of the acid reacts.
The reaction should take 20 minutes [1] at room temperature and so I don't think I can measure the rate by looking measuring a change in temperature. I do not know how else I could measure the rate.

Clayden, J.; Greeves, N.; Warren, S. G. Organic Chemistry, 2nd ed.; Oxford University Press: Oxford; New York, 2012. ISBN 978-0-19-927029-3.

Measure the pH, and just measure the initial rate.

As the reaction is very slow you could measure IR spectrum vs time, preferably using total reflection ( ATR) method.

Side-question: terminology-wise, should this reaction be considered an SN1 reaction or an acid-base reaction? Can a reaction be both?

@Don_S The mechanism by which it occurs is SN1, therefore, I would consider it to be an SN1 reaction

You can sample aliquots and use NMR or GC to compare peaks of your starting materials to products.

https://www.chemguide.co.uk/organicprops/haloalkanes/agno3.html

If you want to measure it with an electronic device I would recommend some spectroscopic technique such as IR, Raman, 1H-NMR etc. or using a digital pH meter. For the spectroscopic techniques you should probably cool down the solution to minimize error during the measurement or alkalify it to avoid further reaction.
However if you want to avoid these methods and you have enough soultion-volume you could try titrating it with NaOH solution and a proper indicator.


text:
About the equilibrium of a flowing reaction
Imagine to have a tube (to make a reactor simple) inside which a chemical reactions take place and reactants are constantly fed so overall can be considered in steady-state conditions.
Just to pick one, I assume the reaction to be exothermic.
Now in my mind there are the following two options, one of them is not clear.

Assuming that the tube is cooled and kept to a constant temperature and that the reaction is not controlled by kinetics, I think that at the end of the tube (if long enough) it can be considered in thermodynamic equilibrium.
If the tube is adiabatic then a distribution of temperature inside should arise. For each temperature there exists an equilibrium constant, and thus a different composition. In this case, would there be thermodynamic equilibrium at each single step (infinitesimal) along the tube? Again, this is considering an equilibrium-controlled reaction.

What is the actual question here? Is it are these descriptions correct?

Sure.  If the tube is long enough, the system will reach a final temperature and composition.  You can even precisely calculate it by considering the same reaction mixture in an adiabatic batch reactor, and waiting until the system has equilibrated.

@bon The actual question is related to the second point, if it is correct or not!

@ChesterMiller yes, but how about the second point, each step of the reaction can be considered as in equilibrium? If so it is correct, there exist an optimal temperature to maximize the calories conversion

What you describe is a 'flow reactor' commonly  used to measure rate constants in gas phase reactions by placing the detector at different places along the tube. Question (1) yes; eventually equilibrium is reached at some point at a point distant from the source inlet. (2) If the temperature is not controlled the rate constant will change as the gases travel along the tube and equilibrium reached sooner if the temperature is raised. No reliable rate constant can now be obtained.

@porphyrin Thanks, I have a question more on this type of reactor. Back to your answer, what do you mean by no reliable rate constant can be obtained? The other doubt is related to Le Chatelier's Principle application, in a closed volume at equilibrium, removing the products increases the reactants conversion (to counteract the change, reaction shifted to the right); in a flow system how this principle apply, since reactants and products continuously move? Is the reaction always shifted to products?

You can't get more than 100% conversion, and that would correspond to some specific final temperature.  You can then evaluate the equilibrium constant at that hypothetical limiting temperature, and use it to determine what the conversion actually would be if that temperature prevailed.  It will certainly be less than 100%.  So the actual final temperature will be less than for 100% conversion.  With regard to @porphyrin 's comment on a flow reactor, this type of reactor is widely used in industry to carry out chemical reactions in continuous processes.

What I meant wrt rate constants was that you start at zero conversion at the input point and distance of detector along tube corresponds to time at constant flow. Thus, if the temp changes along the tube, the rate constant changes (Arrhenius ) also so a different amount of  reaction has occurred at a given distance than expected and so rate constant computed will be in error.


text:
Is Henry's Law sufficient for calculating CO2 solubility?
$\ce{CO2}$ reacts with water which is why it enters and exits solution slower than other gasses (if I understand correctly this is why carbonated drinks bubble gradually instead of all at once).
Henry's law is an approximation that assumes the gas solute does not react chemically with the solvent.
Is the Carbonic acid reaction significant enough that Henry's Law does not work for $\ce{CO2}$?
What about in the case of blood, where the reaction is catalyzed by the enzyme Carbonic Anhydrase?

There are multiple forms of Henry's law (see this Wikipedia entry). The net or effective Henry's Law takes into account the other pH-dependent equilibria of $\ce{CO2}$ and carbonate species in water (chemical reactions with water in your question).
In the case of blood, as you said the picture is even more complex. You would have to be careful using, for example, published values of the effective Henry's Law solubility of $\ce{CO2}$ in that case.


text:
Why does the Co³⁺/Co²⁺ couple have such a high reduction potential?
Cobalt is changing from $\mathrm{(3d)^6}$ to $\mathrm{(3d)^7}$ electronic configuration. What's so stabilizing about that? Just by reduction potentials this couple is more oxidising than hydrogen peroxide and that just sounds amazing and crazy at the same time.

The electronic configuration has nothing to do with it. The reduction potentials of $\ce{Ni^3+}/\ce{Ni^2+}$, $\ce{Cu^3+}/\ce{Cu^2+}$ and $\ce{Zn^3+}/\ce{Zn^2+}$, if they have been/could be measured, would be even greater.
The reduction potential for $\ce{M^3+}/\ce{M^2+}$ is most dependent upon the third ionisation energy. If $I_3$ is large then it will be difficult to oxidise $\ce{M^2+}$ to $\ce{M^3+}$ (hence a high reduction potential). Likewise if $I_3$ is small then the reduction potential will be smaller.
$$\begin{array}{c|cc} \hline
\text{Metal} & I_3\text{ / eV} & E^\circ(\ce{M^3+}/\ce{M^2+})\text{ / V} \\ \hline
\ce{Sc} & 24.76 & (-2.6) \\
\ce{Ti} & 27.48 & (-1.1) \\
\ce{V}  & 29.31 & -0.26 \\
\ce{Cr} & 30.96 & -0.41 \\
\ce{Mn} & 33.67 & +1.60 \\
\ce{Fe} & 30.65 & +0.77 \\
\ce{Co} & 33.50 & +1.93 \\
\ce{Ni} & 35.16 & (+4.2) \\
\ce{Cu} & 36.94 & (+4.6) \\
\ce{Zn} & 39.72 & (+7.0) \\ \hline
\end{array}$$

Data from Weller et al. Inorganic Chemistry 6ed; Johnson Some Thermodynamic Aspects of Inorganic Chemistry. Numbers in parentheses indicate predicted values. Graph by me in Microsoft Office 365.
So your question is essentially: why is $I_3$ of $\ce{Co}$ so "large"?
Going across the 3d block, effective nuclear charge increases which leads to the general trend of IE3 and $E^\circ(\ce{M^3+}/\ce{M^2+})$ increasing.
Any stabilisation derived from "special" electronic configurations e.g. the half-filled subshell in $\ce{Fe^3+}$, or the ligand-field stabilisation energy of $\ce{Cr^2+}$, merely lead to relatively small deviations from the general trend.
Moral of the story is: look at the big picture first. Those $\mathrm{d^5}, \mathrm{d^{10}}, \cdots$ configurations can be relevant, but they aren’t responsible for the overall trend, and certainly shouldn’t be the first thing you consider when trying to rationalise this set of data.

Complexes of Ni3+ and Cu3+ are known.

I’m sure they are. But that’s not relevant to this question, because the reduction potentials are those of the aquo ions.

Well, even +4 are known, but yeah, not aquocomplexes :D

As a further clarification, can we explain this on the basis of the crystal field stabilization energy too?

@AbhigyanC no, that cannot explain anything. If anything, the large CFSE of Co(3+) should lower the reduction potential (as it stabilises the oxidised form). Anyway, as described in the main text, these details are minor.

@orthocresol Why is CFSE a minor detail? I mean... We explained in our classes that Cr3+ is a good reducing agent as Cr2+ attains a t2g configuration.. Which lowers it's CFSE for the aquo ion

@YUSUFHASAN, if you’re talking about Cr(II) being a good reducing agent, then fair enough; this is partly explained by CFSE/LFSE, and that is why in the graphs above there is a dip in E° although there is no dip in IE3. But this dip is relatively small, and as already mentioned, the big picture is that the trend in E° is determined primarily by the trend in IE3. If CFSE was really such a major factor, then why isn’t Co(II) as good a reducing agent as Cr(II), since Co(III) is typically low-spin (t2g)6? That’s precisely what this Q+A is about, so please re-read it carefully.


text:
How to name molecules with a hydroxyl group in a side chain?
TL;DR: Are there clear nomenclature rules for naming the below molecule?
Today in school we learned about the nomenclature of alcoholes. Unfortunately, it was not mentioned what to do with hydroxyl groups in side chains.
With the rules we learned I would guess the name of this molecule is something like 2-(methylol)-propane-1,3-diol. But I could not find any specific rule telling me how to name hydroxy groups in side chains. 1
A free molecule builder with integrated naming I found names the molecule as follows:

2-(hydroxymethyl)propane-1,3-diol

But I'm still not sure if this is the right way to do it.
Are there any precise IUPAC nomenclature rules that say that you use "-ol" in the main chain and "hydroxy" in side chains? 2
1 I did some research with no solutions: our schoolbook, some websites, Chemistry Stack Exchange...
2 I mean, isn't this IUPAC's aim? To provide (relatively) clear nomenclature rules?

You might wanna consider narrowing down, or removing your last questions. Why there's a shortcoming in text you find is rather opinionated, and at the very least off-topic. And ''alcohols'' isn't just one or two compound and the nomenclature is really vast, i.e. too broad to fit in an answer.

@M.A.R. Thanks, I edited the question to be more specific and added a TLDR.

The proposed name 2-(hydroxymethyl)propane-1,3-diol is correct.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) includes three different methods for systematic names of alcohols. The preferred names are formed using the suffix ‘ol’ and (if necessary) the prefix ‘hydroxy’.

P-63.1.2 Systematic names of alcohols, phenols, enols, and ynols
Hydroxy compounds are named in three ways:
(1) substitutively, using the suffix ‘ol’ and the prefix ‘hydroxy’. The presence of several ‘ol’ characteristic groups is denoted by the numerical multiplying prefixes ‘di’, ‘tri’, etc.; the final letter ‘a’ in a multiplying prefix is elided before the suffix ‘ol’. Rule P-44 is applied when a principal chain or a senior ring system must be chosen. When there is a choice for numbering, the starting point and the direction of numbering of a compound are chosen so as to give lowest locants to the ‘ol’ suffixes;
(2) by functional class nomenclature and the class term ‘alcohol’;
(3) as assemblies of identical units by multiplicative nomenclature when the conditions for its use are fulfilled (see P-51.3)
Method (1) generates preferred IUPAC names. Names as assemblies of identical units, method (3), are preferred to those that are formed by simple substitution (P-51.1.5).

The first step is the selection of the principal chain, which will serve as the parent structure. The relevant general criteria for the choice of a principal chain in such compounds are:

greater number of groups that are expressed as suffix
longest chain
lower locants for groups that are expressed as suffix

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)

For the compound that is given in the question, the principal characteristic group is $\ce{-OH}$, which is expressed as suffix ‘ol’. Whichever way you look at it, the principal chain can only have two of these groups. Therefore, the principal chain is a propane; the parent structure including the substituents corresponding to the principal characteristic group is propane-1,3-diol.

The compound that is given in the question actually has a branched chain structure. Therefore, the next step is the identification and naming of side chains, which will be treated as prefixes. In this case, the resulting name is 2-methylpropane-1,3-diol.

The third $\ce{-OH}$ group cannot be expressed as suffix ‘ol’ since it is not situated on the principal chain. Therefore, it has to be indicated by the prefix ‘hydroxy’.

P-63.1.4 Substitutive nomenclature, prefix mode
Hydroxy groups are indicated by the prefix ‘hydroxy’ when:
(1) a group having priority for citation as the principal characteristic group is present; or
(2) a hydroxy group cannot be denoted by a suffix.

In this case, the hydroxy group is placed on the methyl side chain, which results in the compound substituent group ‘hydroxymethyl’.

P-29.4 COMPOUND SUBSTITUENT GROUPS
P-29.4.1 Compound substituted substituent groups
A compound substituted substituent group is formed by substituting one or more simple substituents into another simple substituent that is considered as the principal chain. The choice of the principal chain is fully discussed in Section P-44.3. The first criterion to be applied is that the principal chain is the longest chain; it will be applied in the following examples of acyclic compound substituent groups.
(…)

Parentheses are used around such compound prefixes in order to avoid ambiguity.

P-16.5.1 Parentheses (also called curves or round brackets)
P-16.5.1.1 Parentheses are used around compound (see P-29.1.2) and complex (see P-29.1.3) prefixes; after the multiplicative prefixes ‘bis’, ‘tris’, etc.; to enclose a multiplied parent compound consisting of a parent hydride and a substituent suffix or a multiplied substituent prefix even though preceded by a basic numerical prefix, such as ‘di’, ‘tri’, etc.

Therefore, the complete systematic name for the compound that is given in the question is 2-(hydroxymethyl)propane-1,3-diol.


text:
What's the chemical equation for the lithium iron phosphate battery
The title says it all, I'm searching for the chemical equation to the lithium iron phosphate battery.
I know that the cathode is made of $\ce{LiFePO4}$ and that upon discharging, it is transformed to $\ce{FePO4}$. The Anode is made of graphite.
So I think that the reaction on the anode is:
$\ce{LiFePO4 -> FePO4 + Li+ + e-}$
Is this correct? If so, what's happening on the anode?

Related: http://chemistry.stackexchange.com/questions/60437

Close, the cathode reaction is:
$$ \ce{LiFePO4 -> FePO4 + Li^+ + e^-}$$
The anode is:
$$\ce{Li^+ + C6 + e^- -> LiC6}$$
The best description I can give is that the lithium ion intercalates into the graphite and "steals" some electron density from the six-membered conjugated ring.

Just to clarify, we call the LFP/FP electrode the "cathode" regardless of whether it's charging/discharging. It's sloppy terminology but that's the way the it is.


text:
cross exchanging the coefficients and valences of cations and anions with each other in chemical formulas
Why are the coefficient and valences of iron and bromine cross exchanged with each other in the formula for ferric bromide, FeBr3?

The formulas for ionic compounds can be predicted from the charges of their ions. Cations and anions combine to form ionic compounds with a ratio of cations to anions that will produce a neutral (uncharged) unit. Generally, compounds are uncharged species.
Most binary ionic compounds are comprised of a metal and a nonmetal. In the most generic case, a metal cation with a charge of $+m$ and a nonmetal anion with a charge of $-n$ will combine in a $n:m$ ratio to balance charge. The example below, I am using $\ce{M}$ to represent a generic metal and $X$ to represent a generic nonmetal.
$$\ce{nM^{m+} + mX^{n-} -> M_n N_m}$$
This ratio guarantees charge balance. In the generic case the total charge of all cations in the formula unit is $+m\cdot n$, while the total charge of all the anions in the unit is $-n\cdot m$. These two values are equal and we get a compound. It is possible that $n:m$ is not the simplest whole number ratio and we can simplify. See some examples:

Calcium $\ce{Ca^2+}$ chloride $\ce{Cl-}$ gives $\ce{CaCl2}$.
Iron(III) $\ce{Fe^2+}$ sulfide $\ce{S^2-}$ gives $\ce{Fe2S3}$.
Magnesium $\ce{Mg^2+}$ oxide $\ce{O^2-}$ gives $\ce{Mg2O2}$, which can be simplified to $\ce{MgO}$.

We need to be careful to only think this way for ionic compounds. Covalent compounds can be more complex, and their are often multiple binary compounds based on varying oxidation state. Additionally, for covalent compounds, we do not simiplify the coefficients to the lowest whole number ratio, as the formula unit describes discrete molecules with fixed composition.
For example, $\ce{C4H8}$ (butene) is a different compound from $\ce{C2H4}$ (ethylene).


text:
How can I dissolve nickel(II) oxide?
To insulate $\ce{Pt}$ wire heaters, I plated them with $\ce{Ni}$. After switching on the heater for the first time, green $\ce{NiO}$ is formed which is insulating and protects the $\ce{Pt}$ from shorting. After using the heaters for a week or so, the $\ce{Ni}$ slowly falls away and the $\ce{Pt}$ heater shorts and cuts.
What is the best method of stripping away the balance $\ce{NiO}$ from the used up $\ce{Pt}$ heater?

It took me 10 seconds to find the Wikipedia page for NiO. It's a basic oxide.

Wikipedia says, its solubility in H2O in water is negligible. While it also says it's soluble in KCN. But KCN's safe use a bit of concern.

Can $\ce{HCl(aq)}$ dissolve it?

@DMHO http://link.springer.com/article/10.1023/A:1020634014278

As mentioned by @Mockingbird, nickel oxide is somewhat soluble in potassium cyanide which will dissolve the oxide to form nickel cyanide, $\ce{Ni(CN)2}$ and potassium tetracyanonickelate, $\ce{K2Ni(CN)4}$  but potassium cyanide is very, very toxic.
Do not use it until and unless you are a professional chemist.


text:
What is a highest oxide of element?
When I go through oxides of period 3 elements in periodic table I came across a term "highest oxides" and when I searched I didn't found any appreciative answer for my question.
What are highest oxides and why they are called so?

In particule accelerator the remove all electrons of atoms. So you can make $\ce{Au}^{79+}$ http://www.rhichome.bnl.gov/RHIC/Runs/ :)

Highest oxide of an element is the oxide compound of the element which has highest possible value of oxidation state of the element to which the oxygen is bonded. Particularly, the highest oxide of period 3 is chlorine heptoxide, $\ce{Cl2O7}$ in which chlorine's oxidation state is +VII.
In general, the highest oxide of transition metal which has oxidation value of + VIII is ruthenium tetroxide, $\ce{RuO4}$ and osmium tetroxide, $\ce{OsO4}$.


text:
Can Substances on Titan be Reacted at a Thermodynamic Gain?
Titan, a moon of Saturn, has readily available water ice ($\ce{H2O}$) and methane ($\ce{CH4}$) on its surface, at a frigid $\pu{93.7 K}$.
It would be significant if these (or possibly, other available hydrocarbons) could be reacted in a self-sustaining way, releasing usable energy.  I presume the best way of doing this would be to react by the simple following two step process:

Split water into hydrogen and oxygen:
$$\ce{2 H2O → 4 H2 + O2}$$
Burn oxygen and methane:
$$\ce{CH4 + 2 O2 → CO2 + 2 H2O}$$

Keeping in mind the starting temperature, does this sequence of reactions operate at a net energy release (that is, is it theoretically possible to use this reaction to power itself, with energy leftover)?  If not, could other hydrocarbons* found on Titan fit the bill?
*Methane, ethane, propane, and benzene exist. Others may as well.

The thermodynamic problem is not so much with the hydrocarbons reacting with O2 as it is with forming O2 from water.  That reaction is energetically uphill.  A big hill.  In fact water is a waste product of your second reaction.  The reason the second reaction is energetically favorable is due to the high stability of CO2 and water relative to the reactants, methane and oxygen.
Would it make sense that you should be able to then convert the water from your second reaction into O2 and H2 in a way that would consume less energy than it took to make the water in the first place?  That is almost equivalent to what you are proposing by starting out with the water already present on Titan.
The prohibitive energetics of your reactions apply on Earth as well as Titan.  You need a process to split the water to burn with hydrocarbons that will consume less energy than the energy you will get back from burning the hydrocarbons.  You could conceive of using solar power as the energy source for converting water into O2 and H2.  Or other energy sources.  But that wasn't your question, and you definitely won't get the energy back that you would require via your second reaction.  And good luck with my solar energy idea on Titan ;)

So, just to be sure I understand your answer: The energy released by reaction 2 is not enough to offset reaction 1, even if the water from reaction 2 is used. Is the story similar for other hydrocarbons?

Exactly.  And yes, by all the same arguments it is true for all hydrocarbons.  In fact, methane would be the most energetically favorable for that oxidation reaction because it is the most reduced hydrocarbon.

https://en.wikipedia.org/wiki/Syngas


text:
Is there a difference between potassium manganate and permanganate?
Is potassium manganate the same as potassium permanganate?
Also, I have seen potassium manganate written as potassium manganate (VII) or potassium manganate (VI).
What do the roman numerals stand for?

The roman numerals describe the oxidation state of the manganese. See http://chemistry.stackexchange.com/questions/62333 and http://chemistry.stackexchange.com/questions/59777 for other examples.

I'm voting to close this question as off-topic because of what Klaus said.

No, they are somewhat different. $\ce{K2MnO4}$ is potassium manganate and $\ce{KMnO4}$ is potassium permanganate. Notice the two potassium for each $\ce{MnO4}$.
The Roman numerals mark the oxydation state. So, manganate will have a (VI) and permanganate will have a (VII).

There are two different nomenclature concepts at work here. One includes potassium manganate and potassium permanganate and the other uses potassium manganate(VII) and potassium manganate(VI). These two do not mix, and each of the two has an unambiguous name for each compound. Thus, to answer your title question, potassium permanganate and potassium manganate are not the same.
The more easily explained difference is that between manganate(VI) and manganate(VII) (the cation is irrelevant; as always). The Roman numerals represent oxidation states as they always do. In both cases, we are dealing with $\ce{MnO4^n-}$ and the oxidation state numeral tells us the exact charge. If you do your maths correctly, you will arrive at the following:
$$\begin{align}\text{manganate(VI)} && \ce{\overset{{+VI}}{Mn}\overset{-II}{O}_4^n-} & \Longrightarrow &&\ce{MnO4^2-}&& \Longrightarrow &\ce{K2MnO4}\\
\text{manganate(VII)} && \ce{\overset{{+VII}}{Mn}\overset{-II}{O}_4^n-} & \Longrightarrow &&\ce{MnO4-}&& \Longrightarrow &\ce{KMnO4}\end{align}$$
The two substances differ by the number of potassium counterions present in the crystal; potassium manganate(VI) has two per manganate(VI) anion while potassium manganate(VII) has one.
For reasons not entirely known to me, it was decided once that manganate(VI) — $\ce{MnO4^2-}$ — should be the ‘standard manganate’. It fits in well if the analogy to sulphate ($\ce{SO4^2-}$) and chromate ($\ce{CrO4^2-}$) which all have the $\mathrm{+VI}$ oxidation state and same charge; but that’s the last difference. -ate names are not commonly used for the highest oxidation state only (e.g. chlorate $\ce{ClO3-}$) and they do not refer to a certain oxidation state (e.g. phosphate with the oxidation state $\mathrm{+V}$). Whatever the reasons were, manganate was decided to be $\ce{MnO4^2-}$, i.e. the manganate(VI) you just heard about above.
The other ‘manganate’-like anion, manganate(VII) from above, is in a higher oxidation state. At least this part of the nomenclature is consistent: the higher oxidation state is marked by the prefix per. Thus, permanganate must have a higher oxidation state than manganate.
Summary: Manganate always maps to manganate(VI) and vice-versa; permanganate always maps to manganate(VII) and vice-versa.


text:
Identification of Sir Raman's handheld scope
I read an article somewhere around 5-10 years ago that talked a little bit about the life of Sir C. V. Raman. The focus was on the effect of Raman spectroscopy but included an interesting little tidbit.
Sir Raman apparently used to carry a small handheld scope of some description which allowed him to examine an object and only see the scattered light. He would always keep it in his pocket so that he could pick a flower or look at an everyday object and see how it changed. The article included a photo of him using it. I think I even remember seeing a video of him using it. As I recall, it was a small tube, about a hands length in size, and I think it was made from brass.
What was that tube, how did it work, and is there a resource for me to get one of my own?

I think you are referring to a handheld prism based spectroscope with an eye-view/eyepiece. Using such spectroscope one does not observe Raman signal but only observe the dispersed light coming from an object giving information about the absorption characteristics of the object or emission characteristics (for light sources).
The following is such device (personal photo).

Here is a ray diagram for understanding the working principle. White light is the light entering the device. Yellow lines are normal. (Other ways to achieve this are possible, hence this ray diagram is only an example.) On the right, is the image taken from the eyepiece of the device using a cell phone. Sun light has continuous spectra, while white fluorescent lamp (Hg vapor based) shows distinct lines. (Ray diagram is drawn by me. Images shown on the right are captured by me.)

Such device works using prism(s) to achieve dispersion. The amount of dispersion provided by the spectroscope is a parameter governed by the prism used. Higher the dispersion, more fine grained are wavelengths of light dispersed and in such cases one may need to move the eye or (prism) to see full range of spectrum.
Prism based spectroscopes are frequently used for studying gems and diamonds. White light which passes through such samples undergoes absorption (usually by the minor impurities, which imparts color) and the dispersed light gives this information.
You can easily make such portable spectroscope using combination of small optical quality glass prisms. (Or you can just buy, available on Amazon and some companies dealing with optics sell these.)

About observing Raman signal using eyes, it is very difficult since Raman signal is very weak, but it is possible. As per my knowledge no commercial product exists which can do that.

Please make sure you use a picture that does not violate copyright. Please substitute the offending picture. On  yourgemologist.com it is explicitly stated: None of this site can be downloaded for posting on another website or server for any reason. It will be a violation of the copyright for anyone to copy, duplicate, distribute, and/or re-print this material in any format or any medium without written permission.

I was rather taken aback at the idea of visible eye Raman given the low number of photons, but as it's not my area of specialty, didn't feel comfortable questioning it. Useful link : http://www.yourgemologist.com/Spectroscope/spectroscope.html and the diffraction info is coming back to me from undergrad physics. I think this is it, yes. Thank you!

You mean dispersion (instead of diffraction).

@ankit7540 I knew that. I was just testing to see if you knew that ;) actually I think I mistyped that because I was looking at the diffraction grate type rather than the prism type and was drooling over buying one. Thank you for the very thoughtful answer.

Sir Martyn Poliakoff did the video about the spectroscope on his Periodic Videos channel on youtube.
To my knowledge, he presented the original instrument but did not mention where to obtain a replica.


text:
Ultramarine blue colour absorbing agent

Ultramarine blue ($\ce{Na2Al2SiO8 + Na2S}$, a sulfide-silicate complex containing sodium and aluminium) is insoluble when mixed in water and gives a blue color appearance to the water solution.
Which chemical (from my understanding, it should be acid) can be used to make the blue color disappear in water (absorb the coloring agent)?

Possibly $\ce{H2SO4}$ will absorb the sodium, releasing $\ce{H2S}$. I want to understand basically

What is the coloring agent (is it sodium only)
Which chemicals can I think about to remove the color (absorption of coloring agents)

No single component of ultramarine blue gives its blue color. The whole chemical by itself is blue. The sodium itself is not blue. Nor does $\ce{H2SO4}$ "absorb" the blue color. It just reacts with ultramarine blue to form new compounds which are not blue. You make it sound like a physical change where in fact it is a chemical change.

Thanks.Rephrasing - Will H2SO4 make the blue disappear. I just want experts opinion before actually trying it in the lab

"Dilute $\ce{HCl}$, $\ce{HNO3}$, and $\ce{H2SO4}$ rapidly destroy the blue color, producing hydrogen sulfide ($\ce{H2S}$) in the process.", so yes.


text:
Reaction of propylene oxide with ammonia or hydrazoic acid
I am supposed to identify the products A and B in the following scheme:

The three options I have been given are:

I think that since $\ce{HN3}$ is a stronger nucleophile than $\ce{NH3}$, B should form via an SN2 mechanism. So, cleavage should take place in such a way that $\ce{OH}$ group remains with the carbon attached to the $\ce{CH3}$ group. Hence I think option 1 is the answer, but I am not sure about it.

Related: Regioselectivity of acid-catalyzed ring-opening of epoxides

In fact, these are at most chemoselective reactions, so there will probably be a mixture of all possible products.
Ignoring stereochemistry and assuming that we want to know the 'most probable' products, as far as I can tell the main difference is that $\ce{NH3}$ is a rather nucleophilic basic compound, whereas $\ce{HN3}$ is an acid and not very nucleophilic. We are not given the solvent, so we cannot know what side reactions can happen, or what solvation effects we have. So we may assume an 'inert' solvent is used.
Within these hypotheses, I'd say that $\ce{NH3}$ will react by $\mathrm{S_N}$2, attacking mostly on the least substituted side ($\ce{CH2}$), whereas $\ce{HN3}$ will protonate the oxygen first, and the opening by the azide anion will happen by $\mathrm{S_N}$1, i.e. attacking the most stable carbocation (the one with the positive charge on the most substituted carbon. So for me, the 'correct' answer would be 2.


text:
Suggestions for 200-600 nm range transmitting polymers/materials
I am a student of physics who is doing research related to Cherenkov measurements. For that purpose, I need suggestions for some materials that have the following properties:

$80\%$ transparency in the range of $200$-$600$ $\mathrm{nm}$
Low scintillation light produced from electrons
Relatively lower density
A high refractive index

Some materials in this regard that I have tested are PMMA, TPX, NAS, LURAN, TOPAX, and Plexiglass. I want to test more materials to see if something does better; what polymers or other materials might I try to achieve these properties?

Photons with wavelength around 200 nm are highly damaging to almost any kind of organic matter (not only do the photons get absorbed, they trigger molecular decomposition rather than just being turned into heat). Depending on the photon flux around this wavelength, your parts could degrade quickly. Inorganic materials are a must.

No plastic of any practically useful thickness$^\dagger$ has $80\%$ transparency down to $200~\mathrm{nm}$. Period.
Below is a UV transmission chart for various materials from a marketing brochure published by BrandTech, a manufacturer of disposable UV/Vis cuvettes (click image to enlarge):

The chart only goes down to $210~\mathrm{nm}$; extrapolating the trend, not even quartz is $80\%$ transparent at $200~\mathrm{nm}$, but it's probably a decent choice for your purposes as long as it sufficiently meets your other criteria.
Beyond that, fluorite $\left(\ce{CaF2}\right)$ is the only specific material I know of with better UV transparency; it appears to have better than $90\%$ transparency down to around $200~\mathrm{nm}$ (figure is from ThorLabs online catalog):

A chart at that ThorLabs link indicates two additional candidate materials: $\ce{MgF2}$ and sapphire. As with quartz, the choice among these will boil down to how well they meet your other criteria.

$^\dagger$ As a caveat to the above: If a very thin plastic film ($\leq 100~\mu \mathrm{m}$ or so) is acceptable, a specialty material might give sufficient transmittance. You would have to convince the manufacturer of such a material to make it for you, however, which would probably not be trivial, or cheap.

Ah thanks a lot, really appreciate the material insight you gave as well as the insight into the transmission rates.


text:
Reactivity of meta and para xylene towards nitration
Is meta xylene more reactive towards nitration or is it para xylene?
I would say it is meta xylene as it has two methyl groups meta to each other. Since both are activating groups, there would be a greater electron density at the ortho and para positions with respect to both the groups. This would make meta xylene more susceptible to nitration (EAS) as compared to para xylene.

Yes, see here, p. 188.


text:
Adiabatic compressibility factor of ethyl alcohol
I am looking for a source where I can find the adiabatic compressibility factor of ethyl alcohol at approximately 15 degrees Celsius. I have found the compressibility factor at 20 degrees. Is there maybe a way to convert this to the desirable temperature?

Maybe in Perry's :)

I could calculate values using REFPROP, NIST Standard Reference Database 23, Version 9. What temperature and pressure do you need? Exactly 15 °C and 1 bar?

The Isothermal compressibility at 20 degrees C is given in the CRC.  And there is an article "Adiabatic compressibility and structure of aqueous solutions of ethyl alcohol" that includes measurements of pure ethyl alcohol in AIP.  It is free online, "If you have an individual subscription, a subscription provided by one of AIP's Member Societies, have claimed access to a Conference Proceeding, or have made an individual purchase".  I don't know if either of these solutions would be helpful or easier than what has already been suggested.

@Loong 15.5 degrees and 1 bar, thank you very much!!

A calculation using the NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9, and assuming a temperature of $T=15.5\ \mathrm{^\circ C}$ and a pressure of $p=1\ \mathrm{bar}$, yields an isothermal compressibility of ethanol of
$$\varkappa_T=-\frac1V\left(\frac{\partial V}{\partial p}\right)_T=1.0795\times10^{-9}\ \mathrm{Pa^{-1}}$$
and an adiabatic (or isentropic) compressibility of
$$\varkappa_S=-\frac1V\left(\frac{\partial V}{\partial p}\right)_S=9.1749\times10^{-10}\ \mathrm{Pa^{-1}}$$
The number of digits is the default setting of the software and does not necessarily correspond to the number of significant digits.
By way of comparison, the results for a temperature of $T=20\ \mathrm{^\circ C}$ and a pressure of $p=1\ \mathrm{bar}$ are
$$\varkappa_T=1.1130\times10^{-9}\ \mathrm{Pa^{-1}}$$
and
$$\varkappa_S=9.4729\times10^{-10}\ \mathrm{Pa^{-1}}$$
The data in “Pressure and Temperature Dependence of Liquid Density”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL include a value of $\varkappa_T=1.119\times10^{-9}\ \mathrm{Pa^{-1}}$ for $T=20\ \mathrm{^\circ C}$. Therefore, it seems a likely supposition that the corresponding calculation result of $\varkappa_T=1.1130\times10^{-9}\ \mathrm{Pa^{-1}}$ is reasonable but the number of digits is excessive.


text:
Buffers: what does the strong base react with?
I am a bit confused on buffers:
If I have a solution with a weak acid (say $\ce{NH4}$) and its conjugate weak base ($\ce{NH3}$), I have a buffer solution (in $\ce{H2O}$).

$$\ce{NH4 <--> NH3 + H+}$$

But now If I add a strong base to my solution ($\ce{OH-}$ for example) I get that my pH will go up but I don't understand the dynamics of it.

Will my added $\ce{OH-}$ react with the weak acid $\ce{NH4}$ (more $\ce{NH3}$ produced)?

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

OR will my added $\ce{OH-}$ react with the $\ce{H+}$ produced by my weak acid (Less $\ce{H+}$ produced)?

$$\ce{NH3 + H+ (products of $\ce{NH4}$) + OH- -> NH3 + H2O}$$
I understand that both cases up end up with the pH going up but but I want to be sure to understand with what exactly my $\ce{OH-}$ molecule reacts once added to my solution.
Is it the $\mathrm{1^{st}}$ reaction which means more $\ce{NH3}$ produced?
Or is it the $\mathrm{2^{nd}}$ reaction which means less $\ce{H+}$ produced?

The reaction NH$_4^+$ $\Leftrightarrow$ NH$_3$ + H$^+$ is in a dynamic equilibrium that means the reaction will happen in both directions all the time and the acid dissociation constant will tell you how much on average you have all components in solution so by that you can also calculate the pH value. So of course since you have NH$_4^+$ in solution as well as H+ both can in principle react with the OH$^-$ in providing the proton to form water. The question is, which of the two reactions happens more. This will be determined by the two equilibrium constants
$K_{1} = \frac{[NH3][H2O]}{[NH4+][OH-]}$
$K_2   = \frac{[NH3][H2O]}{[NH3][H+]}$
You see the only difference are the concentrations of the educts, so how large are these? Suppose you prepare a 0.05M NH$_4$Cl solution. After a short time there will be thermodynamic equilibrium and by using the $pK_\mathrm{a}=9.25$ value for NH$_4$Cl you can calculate the respective concentrations of NH$_4^+$ and NH$_3$ in equilibrium:
$K_a=\frac{[H+][NH_3]}{[NH_4^+]}$
By using $[H^+]=[NH_3]$ we get
$10^{-9.25}=\frac{[x][x]}{0.05-[x]}$
gives $[x]=5 \cdot 10^{-6} M$, so actually not much has dissociated. So you can be nearly completely sure that the reaction takes place with the NH$_4$ rather than protons or hydronium ions. And that is what a buffer is for! The pH should not change even when you add $OH^-$ until you reach the stability window of the buffer, that is when the NH$_4^+$ concentration gets too low compared to the $H^+$ concentration. Free protons will also never exist in solution, they can nearly always be associated to some water molecule.


text:
Prove that a 10-Degree Temperature Increase Doubles the Rate Constant (k), when the Activation Energy is Approximately 50 kJ/mol
I read that increasing the temperature by $10~^\circ\mathrm C$ will double the rate constant ($k$), when the activation energy for the reaction is relatively close to $50~\mathrm{kJ/mol}$.
However, no matter how hard I try to wrap my head around it (through graphical examples that require plotting the natural logarithm of $k$ vs the inverse of temperature), I just can't grasp this so called "rule of thumb".
As a result, I tried proving this statement through mathematical expressions:
(Note that the activation energy was evaluated for $50\,000$ because it needs to be expressed in $\mathrm{J/mol}$ to be used in Arrhenius' equation)
Prove that...

A rise of temperature of $10\ \mathrm{^\circ C}$ will double the rate constant, when the activation energy for the reaction is approximately $50\ \mathrm{kJ/mol}$.

Proof starts here -----------------------------
For:  $\Delta T = T_2 - T_1$ ($T_2$ is the final temperature, and $T_1$ the initial temperature)
When: $\Delta T = 10 \rightarrow T_2 - T_1 = 10$
It follows that: The ratio between $k_2$ (at $T_2$) and $k_1$ (at $T_1$) will equal $2$, owing to the doubling of the rate constant ($k$):
$$
{k_2\over k_1} = 2
$$
Employing Arrhenius' equation:
$$
{k_2\over k_1} = {A\mathrm e^{\left({-E_\mathrm a\over RT_2}\right)}\over A\mathrm e^{\left({-E_\mathrm a\over RT_1}\right)}}
       = \mathrm e^{\left[{-E_\mathrm a\over R}\left({1\over T_2} – {1\over T_1}\right)\right]}
       = 2
$$
Or simply:
$$
{k_2\over k_1} = \exp{\!\left({-E_\mathrm a\over R} \cdot \Delta{1\over T} \right)} = 2
$$
So:
$$
\exp{\!\left({-E_\mathrm a\over R} \cdot \Delta{1\over T} \right)} = 2
$$
This might be the part where I messed up -------------------------
$\Delta{1\over T}$ can be set in terms of $\Delta T$:
$$
\Delta{1\over T} = \left({1\over T_2}-{1\over T_1}\right)
       = \left({T_1\over T_1T_2}-{T_2\over T_1T_2}\right)
       = \left({T_1-T_2\over T_1T_2}\right)
       = \left({-\Delta T\over T_1T_2}\right)
$$
Since Arrhenius' equation above holds true for $\Delta T = 10$ ...
And since $\Delta{1\over T}$ can be expressed as $-\Delta T \over T_1T_2$ ...
Then, the rate constant ($k$) will double when:
$$
\Delta{1\over T} = {-10\over T_1T_2}
$$
Substituting into Arrhenius' equation:
$$
\exp{\!\left({-E_\mathrm a\over R}\cdot{-10\over T_1T_2}\right)} = 2
$$
Simplifying:
$$
\begin{align}
\exp{\left({10E_\mathrm a\over RT_1T_2}\right)} &= 2 \\
{10E_\mathrm a\over RT_1T_2} &= \ln{2} \\
{1\over T_1 T_2} &= {R\cdot\ln{2}\over 10E_\mathrm a} \\
T_1 T_2 &= {10E_\mathrm a\over R\cdot\ln{2}}
\end{align}
$$
From $T_2 - T_1 = 10$ we can solve for $T_2$ as $T_2 = T_1 + 10$. Substituting above:
$$
\begin{align}
T_1 (T_1 + 10) &= {10E_\mathrm a\over R\cdot\ln{2}} \\
T_1^2 + 10T_1 - {10E_\mathrm a\over R\cdot\ln{2}} &= 0
\end{align}
$$
For $E_\mathrm a = 50\,000$:
$$
T_1^2 + 10T_1 - {500\,000\over R\cdot\ln{2}} = 0
$$
Solving the quadratic equation:
$$
\begin{align}
T_1 &= 290~\mathrm K \\
T_2 &= 300~\mathrm K \\
\Delta T &= 10~\mathrm K
\end{align}
$$
I don't think this proof is correct because substituting any value for activation energy ($E_\mathrm a$) onto the quadratic equation will produce different values for $T_1$, but that doesn't tell me anything.
What did I do wrong?
Can this be proved any other way?

see this earlier answer.

As you have shown, the rule of thumb applies to reactions occurring at approximately room temperature.

I have read that this is theoretical for some reactions and can be held as a guideline, it is by no means however valid for every type of reaction. One should be very carefull to use this idea to generalize to all reactions.

Your method and your mathematics seem perfectly fine, and your calculated result is correct. You are also quite correct that the result will change depending on the value of $E_\mathrm a$ that you choose.
The problem you're running into is the assumption that the stated rule of thumb holds exactly, regardless of temperature. It doesn't.

I would set up the problem as follows. Consider the ratio $\rho$ between the reaction occurring at some Kelvin temperature $T$, and at $T+10$:
$$
\rho = {e^{-E_\mathrm a\over R\,\left(T+10\right)}\over
        e^{-E_\mathrm a\over RT}}
     = e^{{E_\mathrm a\over R}\left({1\over T} - {1\over T+10}\right)}
     = e^{{E_\mathrm a\over R}\cdot {10\over{T^2+10T}}}
$$
Plotting $\rho$ as a function of $T$ for $E_\mathrm a = 50~\mathrm{kJ\over mol}$, it becomes immediately clear that the rule of thumb does not hold over a very wide temperature range:

In fact, it only holds strictly for one specific temperature, which is why your math led you to a single temperature as your answer. By my calculations, using a value of $8.3144598~\mathrm{J\over mol\, K}$ for $R$, that temperature is $289.56~K$, which is equal to your $T_1=290~K$ to within three significant figures.
The gray box in the figure marks where the rule holds to within $20\%$ – that is, where $\rho$ falls between $1.6$ and $2.4$. This corresponds to a temperature range of $257~\mathrm K$ to $353~\mathrm K$.


text:
When a molecule loses hydrogen atoms, does it become oxidised?
I had the following question on a bio test covering cell respiration and glycolysis, but I'm not sure if I agree with the answer.

When a molecule loses hydrogen atoms (as opposed to hydrogen ions), it becomes
B) oxidised

According to my textbook (Principles of Life, 2nd Edition), oxidation is defined as

Relative loss of electrons in a chemical reaction; either outright removal to form an ion, or the sharing of electrons with substances having a greater affinity for them, such as oxygen. Most oxidations, including biologial ones, are associated with the liberation of energy.

Hydrogen atoms have one proton and one electron in their typical state. If a molecule is losing hydrogen atoms, then its net charge wouldn't change. For example, if a (hypothetical) molecule had 7 protons and 6 electrons, it would have a charge of $+1$. If it then lost a hydrogen atom, it would now have 6 protons and 5 electrons, which still yields a net charge of $+1$. Am I missing some critical piece of information, or is this still considered oxidation?

This phrasing is only true for organic molecules. If for example, sodium hydride loses hydrogen, the sodium ion will get reduced. But since you seem to come from a biochemical background, this simplification is okay since you will be dealing with organic molecules primarily.
The idea behind that statement is that hydrogen atoms in biomolecules are typically only bound to carbon, nitrogen, oxygen and sulfur. Hydrogen is less electronegative than all these elements and therefore any $\ce{X-H}$ bond will be polarised towards $\ce{X}$; the non-hydrogen atom.
When determining oxidation states, bonds are formally cleaved heterolytically in such a way that the electronegative partner gets both electrons. Then, the electrons on the formal atomic ions created this way are counted and subtracted from the number the compound should have. Bonds between the same element are cleaved homolytically and the same procedure applied. Thus, if we take ethene ($\ce{C2H4}$, structure see below) the $\ce{C-H}$ bonding electrons are formally attributed to carbon entirely while the $\ce{C=C}$ double bond is split equally. Doing this, each carbon atom ends up with six electrons, formally, and therefore a formal charge of $2-$ or an oxidation state of $\mathrm{-II}$. If we did the same for ethyne ($\ce{C2H2}$) — a compound with two hydrogen atoms removed — we arrive at five formal electrons and thus an oxidation state of $\mathrm{-I}$. Therefore, going from ethene to ethyne is an oxidation.
$$\begin{align}\ce{H2C=&CH2} & \ce{H-C&#C-H}\\
\text{ethe}&\text{ne} &\text{eth}&\text{yne}\end{align}$$
Having gotten the theory out of the way, what does this mean for the simplified rule? Well, as I mentioned any $\ce{X-H}$ bond is polarised away from the hydrogen and thus effectively adds an additional electron to $\ce{X}$. Carbon, which is the most common bonding partner of hydrogen, is less electronegative than all the other typical elements of biomolecules, and thus will formally lose electrons to them. When removing hydrogen from a compound chemically, we will always remove at least one $\ce{C-H}$ bond in favour of a $\ce{C-C, C-O, C-N}$ or $\ce{C-S}$ bond. Thus, the carbon atom will formally lose electrons and we will have an oxidation.

This can also be rationalised in another way to determine oxidation states: the additive method. Starting from a neutral compound, assign each hydrogen atom a $\mathrm{+I}$, each oxygen $\mathrm{-II}$. Add that up and invert the sign; this is the oxidation state of all the carbons (assuming a molecule of the formula $\ce{C_mH_nO_x}$). Divide by the number of carbons to get an average oxidation state of carbon.
$$\chi(\ce{C}) = -\frac{n - 2x}{m} = \frac{2x - n}{m}$$
As you see, each hydrogen atom in a compound enters into the formula with a negative sign. Thus, removing a hydrogen must increase the average oxidation state of the carbons. Unfortunately, there is no trivial way to extend this formula to compounds containing oxygen and nitrogen or oxygen and sulfur.

If a molecule gains or looses electrons, it is reduced or oxidized, respectively.  If a hydrogen atom leaves a molecule, taking it's single, formerly shared electron with it, then the molecule has lost an electron and is thus oxidized.  Your argument sort of assumes that formal charge alone determines a molecule's oxidation state.  So, sorry, the test answer:

B) oxidized

is correct ;)


text:
What is the variation of atmospheric pressure at large heights?
In an approximation (though not a very accurate one), temperature can be assumed constant throughout the atmosphere. The Earth's gravitational field can also be assumed constant (a better assumption).
For these approximations, what is the calculated pressure at the top of Mt. Everest ($8848\ \mathrm m$), given that the pressure at the top of Mont Blanc ($4807\ \mathrm m$) is $4.32\times10^4\ \mathrm{Pa}$?

$T = 0.0\ \mathrm{^\circ C}$
$\rm g = 9.81\ \mathrm{m\ s^{-2}}$
$\ce{O2}$: $M = 32\ \mathrm{g/mol}$, percentage composition of atmosphere =
  $21\ \%$
$\ce{N2}$: $M = 28\ \mathrm{g/mol}$, percentage composition of atmosphere =
  $79\ \%$

I thought about using $pV=n\mathbb{R}T$ for this question – perhaps using the difference in the volume of air above the different heights, but I couldn't come to an answer.

Over that small change in altitude just assume that pressure is linear with altitude, use the gas law, then look up the answer on line to see how close you were ;)

@airhuff We use MathJax to format mathematical expressions and equations. I have looked through your suggested edits and noticed that you use HTML for this. While your effort is much appreciated, most of it won't last long since it will be changed by other users. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

@Martin-マーチン, very helpful.  Since it worked, I thought the markup must be the same.  Big wrong there!  My apologies for all those edits and appreciation for the guidance.

From the given composition of the atmosphere you can calculate the mean molecular mass $M$ of air.
Since Earth's gravitational field and temperature are assumed to be constant you can use the barometric formula
$$p(h_1) = p(h_0)~\mathrm{e}^{- \frac{M g}{R T} \Delta h}$$
to calculate the pressure at the top of Mt. Everest.


text:
Why is N2 enthalpy zero, entropy 191.5, yet G is listed as zero and not [-T*0.1915]?
In standard tables, $\ce{N2}$ at STP has an enthalpy formation ($\Delta H_\mathrm f$) listed as $0\ \mathrm{kJ/mol}$, entropy ($S$) as $0.1915\ \mathrm{kJ/K}$, and free energy of formation ($\Delta G_\mathrm f$) as $0\ \mathrm{kJ/mol}$.
If we use the std eq.: $\Delta G = \Delta H-T\cdot\Delta S$, I don't get zero with those numbers, obviously. (assuming STP and all that) But, $\Delta G$ is zero for nitrogen?
My initial thought is that $0.1915\ \mathrm{kJ/K}$ is not $\Delta S$ as required by the equation, but just $S$, so I need to figure out $\Delta S$ for $\ce{N2}$ and use that instead. Okay, that would make sense, but it doesn't follow for this example, which I see everywhere:
$$\ce{N2 + 3H2 -> 2NH3}$$
$\Delta H$ is calculated for the reaction using the appropriate $\Delta H_\mathrm f$
$\Delta S$ is calculated using the appropriate $S$ values (So $0.1915\ \mathrm{kJ/K}$ for $\ce{N2}$, not $0$)
Then they use these reaction $H$ and $S$ to get $\Delta G$
However, if I just use the $\Delta G$ values found in databases like the CRC Handbook (so $0$ for $\ce{N2}$), I don't get the same answer using $\Delta G_\mathrm r=\Delta G_\text{products}-\Delta G_\text{reactants}$
This is driving me nuts, I know I must be missing something simple here.

Enthalpies of formation and Free energies of formation are expressed relative to the standard state of 25 C and 1 bar.  So, in the standard state, they are taken to be zero.  The entropy value you cited is the absolute entropy, relative to absolute zero.  If you wanted the entropy of formation of nitrogen relative to the standard state of 25C and 1 bar, that would be 0.

Okay, but why then do all textbook examples following the above reaction use 191.5 for dinitrogen when the reaction is at stp?

Can you give an example of how you don't get the same answer for the standard free energy change of the reaction; because it shouldn't matter if the formation bases differ, since the values for the change associated with the reaction are independent of the formation bases.

Chester, I still do not understand your temperature reference. The value, 191.5  J/mol-K is listed at 298 K rather than 0 K  link. From..."Table 18.1. Standard Molar Entropies of Selected Substances at 298 K." Am I interpreting this incorrectly? An example with numbers to follow.

No.  I meant that the 191.5 is at 298, but it doesn't have to be consistent with the enthalpy and the free energy.  Only the changes as a result of the reaction need to be consistent.  Please provide the example of the inconsistency that you found for the changes.

Got it, thanks! Using $\ce{\Delta H}$ and $\ce{\Delta S}$ values to get $\ce{\Delta G}$, and then G values calculated from $\ce{\Delta H}$ and S to get $\ce{\Delta G}$ I was able to get -32.7 kJ for both methods. So, when determining $\ce{\Delta G}$, it is okay to use G values obtained from S and not $\ce{\Delta S}$, because the change is the same? This is what confused me...$\ce{G}$ for N2 was not 0, but rather -116.8. I see now that it is because S was used and not $\ce{\Delta S}$, but seems to be okay because the change is the same as if we were using calculated $\ce{\Delta S}$ values.

Yes.  It's the deltas that matter.


text:
Mustard oil test for distinguishing between primary aliphatic and primary aromatic amines?
Can the mustard oil test be used for distinction between primary aliphatic and primary aromatic amines?
An alkyl isothiocyanate is formed when mustard oil is reacted with a primary aliphatic amine. But does mustard oil give a product with aromatic primary amines so as to make a considerable distinction?

No, aromatic primary amines cannot be distinguished from aliphatic amines using mustard oil test because they both give isothiocyanide and the same byproducts (mercury(II) sulfide and hydrochloric acid).


text:
Which factor drives the stability in chelated complexes?
Which factor drives the stability in chelated complexes?

Stability, I guess, is supposed to be associated with a release of energy. Does something like this happen in chelated complexes?

Related: What's so special about chelation?

@Loong But I have a different query

Well, your post doesn't point out the difference.

@Mithoron what about the release in energy?

You need to edit your post, pointing out how is it exactly and in significant way different.