text:
Chirality of symmetrically substituted biphenyl
I am facing trouble with the following problem: is 2,6-dimethyl-2',6'-dinitro-1,1'-biphenyl chiral?

My idea is that there is a axis of symmetry along the line joining the two benzene molecules. Hence it should be achiral.
However my book states it otherwise.

Related: Why are substituted biphenyls optically active?, Racemization of biphenyl compounds, and How are the Cahn-Ingold-Prelog rules applied to chiral molecules without a stereogenic carbon center? i.e biphenyls and allenes

No, the biphenyl above is not chiral.
2,2′,6,6′-tetrasubstituted biphenyls are nonplanar at room temperature in order to avoid steric clashes between the substituents. This can in general give rise to chirality depending on the substituents, such as in this example.

However, this chirality is only true under certain conditions, namely:

The substituent groups must be sufficiently bulky in order to prevent rotation about the C–C single bond that connects the two rings. If this rotation occurs freely, then the two enantiomers will be able to interconvert, and chirality will not be observed (under typical conditions)

Both pairs of substituents on the phenyl rings must be non-identical; otherwise, the molecule will possess a plane of symmetry, rendering it achiral even if rotation about the C–C bond is restricted.

Here, since the two pairs of substituents on both rings are the same (i.e. two methyl groups on one ring, and two nitro groups on the other), the second criterion is not fulfilled. As a result, the molecule is not chiral.

Yes, the molecule should exhibit chirality, but this can likely only be observed below room temperature.
In the optimised1 molecular configuration the molecule cannot be superimposed on its mirror image. This is because of the out-of-plane rotation of the nitro groups.

The thermodynamic most favourable structure is of $C_2$ symmetry and has no mirror plane (see above).
Contrary to the previously presented arguments, an analysis of the Lewis-structure is misleading as it is massively misjudging the space requirement of the nitro moieties. This is a feature that is already present in 2,4,6-trinitrotoluene,2 as described in the abstract of Clarkson et al.:3

Two genuine energy minimum structures were found. In both structures the 4-nitro group is planar to the phenyl ring, while the 2,6-nitro groups are slightly out of plane with the phenyl ring due to steric interaction with the methyl group.

They later describe:

The two stable molecular structures of TNT, Fig. 1, are related by internal rotations of the 2 and 6-nitro groups and the methyl group.
[...]
Structure A is the more stable geometry, though the difference in energy is small ($0.650~\pu{kcal mol^−1}$ at B3LYP/6-311+G**, Table 4). Structure A displays near ideal $C_\mathrm{s}$ symmetry with one of the methyl hydrogen atoms perpendicular to the phenyl ring in the plane of $\sigma_\mathrm{h}$. The 2,6-nitro groups of A are non-planar and rotated into the same face of the phenyl ring, maximizing the number of van der Waals interactions to the methyl group.

The 2,6-phenyl moiety however is much larger than a methyl group, which is the reason, why the symmetry of the nitro groups is broken. Forcing this $C_\mathrm{s}$ symmetry will lead to a first order saddle point (transition state) on the potential energy surface which is only about $1.6~\pu{kJ mol^-1}$ higher in energy than the minimum structure.
At room temperature the inter-conversion between the enantiomers is likely to happen through this structure. I have not checked whether this state actually connects via an intrinsic reaction co-ordinate calculation, but from the visual investigation of the imaginary mode this seems very likely.

If you cool the system enough you should be able to identify two enantiomers, however, at room temperature interconversion should be too fast to observe.

It would be good to know which book you have used and in which context the statement was issued. From a purely theoretical point of view, I would agree that the molecule is chiral. From practical point of view I would not asses this property, since it is probably too hard to measure.

The suggested $C_\mathrm{2v}$ structures are either second- or fourth-order saddle points, depending on the rotation of the methyl groups. These structures are at least $15~\pu{kJ mol^-1}$ higher in energy than the former and no viable pathway for inter-conversion.

Footnotes

All structures have been optimised on the DF-M06L/def2-SVP level of theory with Gaussian 09 Rev. D01. Stationary points have been characterised by calculating the vibrational frequencies at the same level of theory.
W. Robert Carper, Larry P. Davis, Michael W. Extine, J. Phys. Chem. 1982, 86 (4), 459–462.
John Clarkson, W. Ewen Smith, David N. Batchelder, D. Alastair Smith, Alison M. Coats, J. Mol. Struc. 2003, 648 (3), 203-214.

The simplified point of view in a way that students should be expected to answer the question
It is not the axis of symmetry (which is a $C_2$ axis in this case) that is important but rather that the molecule additionally features planes of symmetry ($\sigma_\mathrm{v}$). There are two perpendicular planes of symmetry in your example compound which both contain the $C_2$ axis. Thus, the molecule’s point group is $C_\mathrm{2v}$ and it must be achiral.
Obviously, your textbook contains an error.
The practical point of view
See Mart’s answer. In a nutshell: the molecule would be chiral but racemic (rapidly interconverting) at room temperature due to the specific most stable orientation of the nitro groups.

I think the textbook is correct. The minimum energy structure only has $C_2$ symmetry, see my answer for details. I think for biphenyls and related compounds point groups are never obvious, tiny changes can easily break symmetry.

@Martin-マーチン Ah, the pitfalls in simple looking molecules.

I don't know about point groups and the axis of symmetry.can you please share a resource from where to know about them

@user471651 https://chemistry.stackexchange.com/q/43073

I do not believe the molecule to be chiral. While other answers correctly note that rotation about the central phenyl-phenyl bond is hindered (under reasonable conditions), the substituents on each individual ring are identical. For this reason, one cannot unambigously assign a chirality designator, such as $S$ or $P$.

This question wants you to check for axial chirality. However, this molecule doesn't possess axial chirality (even though in my haste to answer, I did fall into the trap of saying it did, my apologies.
I did not make not of the fact that the substitution pattern here is symmetrical, and because of the existence of this element of symmetry, the given compound should be achiral.
Axial chirality, which is a special case of chirality in which a molecule does not possess a "stereogenic center" (which is what is most commonly taught in introudctory organic chemistry courses) but an axis of chirality, i.e an axis about which a set of substituents is held in a spatial arrangement that is not superposable on its mirror image.
It is commonly observed in atropisomeric biaryl compounds wherein the rotation about the aryl-aryl bond is restricted. Owing to restricted rotation around the biaryl bond, the two stereoisomers cannot interconvert under ambient conditions. I wrote a lengthy post about chirality in the past; you might find instructive.
Whatever, I said about axial chirality (in my previous post) still holds true, it just isn't applicable in this case. My bad! sorry
Additionally, since OP asked about how to assign stereochemistry in the case of compounds possessing axial chirality, I'll put down a few words on that.
First draw a Newman projection along the axis of hindered rotation. Now assign the ortho, and meta substituents priorities based on Cahn–Ingold–Prelog rules. Now, starting with the substituent of highest priority in the closest ring and moving along the shortest path to the substituent of highest priority in the other ring (as illustrated below), the absolute configuration is assigned P for clockwise and M for counterclockwise. In the example shown, A has priority over B.

Now, try doing this for your given compound, and you run into trouble. You cannot unambiguously assign a configuration.

I would check the simplest definition of chirality before looking at potential chirality centers or point group representations:

Is the molecule's mirror image superimposable on the original?

If you redraw the biphenyl along any mirror plane, you will get the exact same image. That means it's mirror is superimposable on the original, and the molecule is achiral.
Even in difficult chem classes, I always check the simplest definition first. As long as you're confident of the stereochem, it's the quickest to test and hardest to mistake. I would only worry about elements of symmetry if you're in an upper level inorganic class or into that sort of thing.

You're assuming that biphenyl's twist is $90^{\circ}$. If it's less than that, the molecule is axially chiral... Chirality has a simple definition but that doesn't always translate to a simple scenario. That's why Elial's book is so damn thick.

In the trivial example of two perpendicular and symmetric biphenyls, simple works quite well. I wasn't suggesting point group theory isn't useful, but in my opinion the poster should lock down what chiral means before learning what a C2v axis is.

That, I agree with, especially since $C_{2v}$ is not the important symmetry element to look at.

C2v point group*, not axis. Don't have enough points to edit. Also meant phenyls* instead of biphenyls.


text:
Alcoholic potassium hydroxide and aqueous potassium hydroxide
I searched much about alcoholic KOH and aqueous KOH, but I was unable to understand it.
I mean, I want to know how ionisation of KOH in alcohol is different from that in water.
I also don't know what actually happens during its ionisation. Why is alcoholic KOH used in dehydrohalogenation of alkyl halides?

Could it be that there is some deprotonation of the alcohol in equilibrium, which would act as a better base?

Very related: http://chemistry.stackexchange.com/questions/15728/why-does-alcoholic-koh-prefer-elimination-whereas-aqueous-koh-prefers-substituti

Why is alcoholic KOH used in dehydrohalogenation of alkyl
  halides?

One possible explanation is lack of solubility of the alkyl halides in water. Thus, one switches to a less polar, presumably alkyl side chains containing alcohol. (H2O is formed in minute amounts from OH and the proton from the hydroxyl group of the alcohol and can thus be neglected.) Boiling points vs. reaction conditions can also be relevant.
KOH forms homogenous mixtures with H2 at any ratio, this may not be true for some alcohols (For the amounts needed for the reaction this should not be relevant). In terms of ionisation, one can suspect that K+ is not as well solvated in alcohols as in water because the alcohol molecules will have a harder time arranging optimally around the ion. Also, the solvent shell structure may be a lot less pronounced. For OH-, one basically expects the same. In alcohol, one might also see some ion pair formation (or rather retention?) for KOH.

Basically, alcohols can't stabilize the hydroxide ions as well, do potassium hydroxide is more basic. In addition, alcohol is a much better solvent for organic reactions. Those are the two prime factors for using alcoholic potassium hydroxide.

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text:
Chirality of spirocyclic compound
I am facing trouble with the following problem.
Is (R)-1,7-dioxaspiro[5.5]undecane chiral?

My idea is that the compound possess a plane of symmetry and thus should be achiral. However if we take its mirror image then the bond pointing below of the plane will become above the plane and thus it will not form a nonsuperimposable mirror image indicating that it is chiral. Certainly one of the approach is incorrect. I couldn't understand which one.

This is a great example of axial chirality.

Related: Axial or planar chirality in spiro[3.3]hepta-1,5-diene

The only place where a plane of symmetry may make sense would be vertically through the centre, through the spiro carbon. However, as the image by Tetrahydrocannabinol shows, these rings are not in the same plane (implied by the wedge and dash notation). Therefore, using that plane as a plane of symmetry would map each of the two rings into the void.
The only element of symmetry I can see in the compound is a $C_2$ axis (rotation by $180^\circ$). This axis would also bisect the spiro carbon, but it would map the oxygens onto each other. Having only a $C_2$ axis allows a compound to be chiral (point group $C_2$) and thus your compound is chiral.

Thank you for making it cleat that the compound does not possess any symmetry..However if we just take its mirror image about a plane which is perpendicular to the plane containing the compound,the two bonds with are pointing below and above the plane ( implied by the wedge and dash) will get interchanged but then also we would be able to perfectly superimpose over the original one..How do you look into this?.This would in fact make it achiral.

@aniline You would be mapping an oxygen onto a carbon. That would not be the same thing.

Oh yes so if in the first ring instead of oxygen there would have been carbon then the compound would be achiral.

@aniline Absolutely correct! =)

The compound is chiral as there is no element of symmetry. The two rings are more or less perpendicular to each other. Something like this:

Since it is disymmetric it is chiral.

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text:
Esterfication of 2,2,2-triphenylethanoic acid

Explain the formation of (1,1,1-triphenylmethyl)methyl ether when a solution of 2,2,2-triphenylethanoic acid in concentrated sulfuric acid is poured in methanol.

I cannot figure out where is one carbon atom lost. Has it escaped as $\ce{CO2}$? Or is there something more strange happening.
I wouldn't call this reaction an esterification. There is no ester functional group in the final compound.
I guess there's some special role of the three phenyl groups but I'm unable to come up with any mechanism. I hope dehydration is involved as one of the reaction steps.

Similar thing as here

Note that the name “2,2,2-triphenylethanoic acid” is not wrong but the preferred IUPAC name is triphenylacetic acid.

Disclaimer: The question has pretty much been answered in the comments. I'm just making it substantial.

The triphenylacetic acid is dehydrated in the presence of concentrated sulfuric acid to give an acylium ion, namely triphenylacetylium cation, which is stabilized by resonance.
Then, the triphenylacetylium cation undergoes decarbonylation (removal of carbon monoxide) to give triphenylmethyl cation, which is stabilized by the three phenyl groups.
The oxygen in methanol acts as a nucleophile to attack the carbocation, and is deprotonated to give our desired product triphenylmethyl methyl ether.
It is noted that this reaction should be done in a well-ventilated area because of the evolution of toxic carbon monoxide gas.

The answer is great, but it could benefit more from explaining why the regular esterification does not take place. Is it the minor product? Or does it not take place at all? why?

Esterification doesn't happen because of steric hindrance - its intermediate has much bigger functional group then acylium. Also, decarbonylation is irreversible and carbocation heavily stabilised.

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text:
Relative O-O bond lengths in O2X2 molecules (X = H, F, Cl)
I am facing problem in the following question:

What is the order of bond length of $\ce{O-O}$ in the following
  compounds:
$\ce{H2O2, O2F2, O2Cl2}$

In my view since fluorine is more electronegative than oxygen it will attract the bonded electrons towards itself resulting in low bond length in case of $\ce{O2F2}$ than in case of $\ce{O2Cl2}$. However if the same concept is applied to $\ce{H2O2}$ it should have the greatest $\ce{O-O}$ bond length however my book states it otherwise.

Fluorine is more electronegative, so it pulls the oxygen nuclei away from the center of the molecule, elongating the bond.

You're right and your reasoning is OK. Also don't be surprised with mistakes in textbooks @khaverim It may be counter-intuitive but it does shorten the bond.

this can simply be explained using bent's rule as well

First things first: the shorter a bond, the higher the multiple bond character. I.e. a double bond is shorter than a single bond, and a bond with a bond order of $1.5$ is in the middle of the two. Thus, the shorter the bonds in $\ce{X2O2}$, the higher the corresponding double bond character.
The key to understanding why the bond in $\ce{O2F2}$ has a very high double bond character is the gauche effect. The effect has its name because of the observation that $\ce{O2F2}$ prefers a gauche configuration rather than the expected (due to sterics) anti configuration. The reason is electronic in nature: a lone pair of oxygen A can interact with the $\sigma^*(\ce{O_B-F})$ orbital. This lowers the energy of the populated orbital at the expense of the unpopulated one. Since we have an interaction along $\ce{O_A-O_B}$, this increases the double bond character of the $\ce{O-O}$ bond. At the same time, since we are using the $\sigma^*$ for stabilisation, the $\ce{O-F}$ bond decreases in bond order — it becomes less than a single bond. In Lewis resonance depictions, this can be explained by the following mesomeric structures:
$$\ce{F-O-O-F <-> F-O^+=O\bond{...}F-}$$
This effect is strongest in $\ce{O2F2}$ since fluorine is more electronegative than oxygen and therefore the $\sigma^*$ orbital has a much higher oxygen contribution.
In hydrogen peroxide, a similar mechanism is at play that also contributes to reducing the bond length and increasing the bond order. In $\ce{Cl2O2}$, the effect is weakest.

At the moment, I have studied just the basics of molecular orbital theory, not much in detail. How plausible is Bent's rule as an explanation for shortening of O-O bond in $O_2F_2$?

@AumkaarPranav I keep forgetting most things I know about Bent’s rule but afaicr Bent’s rule predicts a lone pair to receive the highest possible s contribution which would mean it would do a poor job in predicting the difference between $\ce{H2O2}$ and $\ce{O2F2}$. Take that with a grain of salt, however

It was explained to me that between the two bonds that an $O$ atom forms, $\ce{O-F}$ would have more p nature (due to Bent's rule, that is, $O$ would direct more p nature towards a more electonegative atom), which would make the $\ce{O-O}$ bond richer in s-character and hence would be of shorter length.

First, let me confirm your answer. The table given below shows a summary of optimised structures of $\ce{XOOX}$ compounds; experimental data when available is in parentheses. As a reference the bond length in $\ce{O2}$ is $121\ \mathrm{pm}$.
Here you can clearly see that $\ce{O2F2}$ has the shortest $\ce{O-O}$ bond length of the three compounds in consideration in the original question. In fact, the fact that it is so close to $\ce{O=O}$ bond length in the dioxygen is suggestive of high double bond character between the two oxygen atoms in dioxygen difluoride. Here's another table summarising the bond order calculations:

Let's study the MO diagram proposed in the paper I am using as a reference for this answer.

Here, the orbitals of this $C_2$ system have been generated by considering
the interaction of the frontier orbitals of $\ce{X ... X}$ with those of
the $\ce{O2}$ molecule. In the case of $\ce{H...H}$, they are simply the
a + b combinations of the $1s$ functions, and for the halogens they are a combination of their valence p functions.
Thus, 1a is derived from the $2p \sigma_g$ function of $\ce{O2}$ and is
$\ce{X–O–O–X}$ bonding. Similarly, $1b$ and $2a$ come from the $2p\pi_u$ functions of $\ce{O2}$.
The higher electronegativity of fluorine and bromine leads to better interaction with the quite low lying $\ce{2p\pi_u}$ orbitals. In this scheme, we also note that the energies of 1b and 2a in these $\ce{XOOX}$ molecules decrease with the electronegativity of X.
The orbitals $2b$ and $3a$ derive from the bonding overlap
of $\ce{X...X}$ functions with the $\ce{O2}$ $2p\pi_g$ orbitals.
Now, addressing the 2b and 3a functions in $\ce{FOOF}$, we note that they are similar to those in $\ce{HOOH}$. However, the high electronegativity of fluorine results in transfer of electron density from oxygen and considerable stabilisation of this $\ce{O–O}$ anti-bonding function.
The 3b and 4a functions in FOOF and BrOOBr are halogen based, non-bonding ‘lone pairs’. The higher electronegativity of fluorine, as outlined above, leads to greater bonding interaction between the $\ce{F}$ $2p$ and $\ce{O2}$ and $2p \pi_u$ functions resulting in the low energy of $2a$ and $1b$.
It similarly leads to the relatively low energy of the corresponding anti-bonding functions which are $4b$ and $5a$ in $\ce{FOOF}$.  These are $\ce{F–O}$ anti-bonding but $\ce{O–O}$ bonding. These are the highest occupied orbitals in $\ce{FOOF}$ with the result that the $\ce{F–O}$ bond order is considerably reduced at the cost of the $\ce{O–O}$ bond.
In $\ce{FOOF}$ there is effectively a 4-orbital, 8-electron interaction
between the p orbitals of $\ce{F...F}$ and the $2p\pi_u$ functions
of $\ce{O2}$ resulting in weak $\ce{F–O}$ bonds and little net transfer of
charge.
This repulsive interaction is responsible for the long $\ce{F–O}$ bond. There is some bonding interaction with the $2p\sigma_g$ and $2p\pi_g$ functions of $\ce{O2}$. The latter results in a $\ce{O -> F}$ charge transfer and a degree of strengthening of the $\ce{O–O}$ bonds.
In the resonance scheme proposed below, I would, based on this, call attention to resonance form 3.
Anyway, I have made an attempt to summarise the salient features of the discussing in the paper cited below in the reference section. The effects that I have called attention to here, are significant for fluorine but considerably weaker for chlorine, bromine etcetera.

Reference:
1] Bonding in mixed halogen and hydrogen peroxides, Adam J. Bridgeman and    Joanne Rothery, J. Chem. Soc., Dalton Trans., 1999, 4077-4082 (link)


text:
Reaction of water and methane
Methane is a non-polar compound, i.e. it shows no bipolar movement but even after that it reacts with water which is a polar compound and polar and non-polar compounds do not react. Why does this reaction take place, then:

Q: The reaction of methane and water is a way to prepare hydrogen according to
$$\ce{CH4 (g) + H2O -> CO2 (g) + H2 (g)}$$
...

Polar and non-polar liquids do not mix, but that doesn't tell you anything about their ability to react, which depends on the thermodynamics and kinetics of the reaction. The reaction shown is in the gaseous state.
The reaction of methane gas with gaseous water, which you show, is a common reaction called Methane reformation. Methane gas is heated with steam (gaseous water) to produce hydrogen (and $\ce{CO2}$).
As shown in that link, the reaction takes place in two steps:
$$\begin{align}\ce{CH4_{(g)} + H2O_{(g)} &->  CO_{(g)}  + H2_{(g)}} &&\text{(Endothermic)}\\
\ce{CO_{(g)} + H2O_{(g)} &-> CO2_{(g)} + H2_{(g)}} && \text{(Exothermic)}\end{align}$$
It appears it probably requires some heating to get this reaction to proceed (the first step is endothermic), I'll edit when I have a chance to look up the thermodynamic properties and see if we can glean what drives the reaction, or if it requires catalysis to avoid the energy barrier.
Side-note: by "bipolar movement" I think you mean "dipole moment"
Edit: According to this source, the first step of the reaction happens at ~3–20 atmospheres of pressure. The second step uses a catalyst to make the reaction more energetically favorable.
The thermodynamic variables for the individual species (at $25~\mathrm{^\circ C}$) in the net reaction are:
$$
\begin{array}{c|cc}
 & \Delta S^\circ \left( \frac{\mathrm{kJ}}{\mathrm{K\cdot mol}} \right) & \Delta H_\mathrm{f}^\circ \left( \frac{\mathrm{kJ}}{\mathrm{mol}} \right) \\
\hline
\ce{CH4_{(g)}} & 0.1862 & -74.81 \\
\ce{H2O_{(g)}} & 0.1887 & -241.8 \\
\ce{CO2_{(g)}} & 0.2136 & -393.5 \\
\ce{H2_{(g)}} & 0.1306 & 0 \\
\end{array}
$$
The balanced net reaction is:
$$ \ce{CH4_{(g)} + 2 H2O_{(g)} -> CO2_{(g)} + 4 H2_{(g)}}$$
Therefore:
$$
\begin{array}{l|ccc}
 & \Delta S^\circ \left( \frac{\mathrm{kJ}}{\mathrm{K\cdot mol}} \right) & \Delta H^\circ \left( \frac{\mathrm{kJ}}{\mathrm{mol}} \right)  \\
\hline
\text{Reactants} & 0.5636 & -558.41 \\
\text{Products} & 0.7576 & -393.5 \\
\hline
\text{Net Reaction} & 0.2136 & 164.9 \\
\end{array}
$$
Right away, it's apparent that the reaction is entropically favorable ($S_{\text{rxn}} > 0$), but enthalpically unfavorable ($H_{\text{rxn}} > 0$) at room temperature. To find out if the reaction is spontaneous, consider:
$$
\Delta G = \Delta H -T\Delta S
$$
The reaction is spontaneous when $\Delta G < 0$. Because $\Delta S$ is so close to $0$ ($0.194~\mathrm{kJ/(K\,mol)}$), and $\Delta H$ is so high ($164.91~\mathrm{kJ/mol}$), we should expect already that this reaction is non-spontaneous at room temperature. This is consistent with the fact that methane doesn't readily decompose in moist air, as any cow pasture will tell you.  But we can check:
$$\Delta G = 164.91\ \frac{\mathrm{kJ}}{\mathrm{mol}}-298.15~\mathrm{K}\cdot0.194~\frac{\mathrm{kJ}}{\mathrm{K\cdot mol}}=107.07\ \frac{\mathrm{kJ}}{\mathrm{mol}}$$
Quite non-spontaneous. However, since the overall reaction is entropically favorable, we can use $\Delta S$ to overpower the $\Delta H$, by jacking up $T$. Since the industrial reaction takes place between $700$ and $1000~\mathrm{^\circ C}$, let's plug in $700~\mathrm{^\circ C}\ (973.15~\mathrm{K})$ and check that the reaction is indeed spontaneous:
$$\Delta G = 164.91\ \frac{\mathrm{kJ}}{\mathrm{mol}}-973.15~\mathrm{K}\cdot0.194~\frac{\mathrm{kJ}}{\mathrm{K\cdot mol}}=-23.85 \frac{\mathrm{kJ}}{\mathrm{mol}}$$
Voilà! A spontaneous reaction!
Do note, however, that this calculation:

ignores any temperature-dependence of the thermodynamic variables;
ignores the role of pressure.

Tldr: This reaction is not energetically favorable at room temperature. However, the reaction is entropically favorable. If you heat it up enough (i.e. $\gtrapprox 700~\mathrm{^\circ C}$), the reaction will proceed spontaneously.
Source for thermodynamic values: Oxtoby, Principles of Modern Chemistry Seventh Edition

Heating is probably a good idea; the first reaction is usually done at temperatures $700$$-$$1100 \ ^\circ \mathrm{C}$ while the other one at $360$. If temperatures are lower (around $250$) possibly with catalysts (copper and zinc oxides, aluminum) one might expect competing reactions: $$\ce{CO + 2H2 -> CH3OH,\ CO2 + 3H2 -> CH3OH + H2O}.$$ It would be interesting to see the pressures involved.

I edited my answer with an update about the pressures required. (3-20atm)

For the record, the carbon monoxide-water mixture is know as syngas.

CO-H2 mix not water. Also catalysts change activation energies and mechanisms not overall energetics.

"This is consistent with the fact that methane doesn't readily decompose in moist air, as any cow pasture will tell you." The cow pasture won't tell you that by smell. Methane is odorless; the smell cones from small amounts if more complex compounds.


text:
Why can enolizable aldehydes undergo Cannizzaro reactions?
Why does 2-methylpropanal undergo a Cannizzaro reaction even though it has an $\alpha$-hydrogen?

It would best be called a Cannizzaro-like reaction, since Cannizzaro's are specifically defined as being non-enolizable.

There is never an either–or in chemistry; it is typically a both instead. In the case of a hydroxide ion approaching an aldehyde with an α-hydrogen, it can both deprotonate the hydrogen creating the corresponding enolate and attack the carbonyl nucleophilicly to generate the required intermediate for a Cannizzaro reaction.
Most of the time, if there are two (or more) possible pathways and neither of them is intrinsically less hindered due to a strong reason, both pathways will be taken.

So what is the major product ? The one formed due to Cannizaro or Aldol ?  Is there any way to predict this in general or are such results mostly usually only experimental ?

@UtkarshGupta You definitely can’t predict it a priori. It will also depend a lot on the reaction conditions, some making it lean more towards Cannizzaro, others more towards aldol. Thus, the results are experimental (and researchers subsequently attempt to rationalise them).

I have to agree with Jan's answer here, that the final results will always be experimental. But, having not found any conclusive answer on the internet anywhere, and that this result comes up often, I believe a theoretical answer would help, so here it is:
The average textbook will commonly write:

2-methylpropanal undergoes the Cannizaro reaction even though it has an alpha hydrogen atom.

Now, what they mean to say, as Jan already pointed, is that Cannizaro is a major product (and not the only product). Why is it a major product? Let us see:
(I hope you are familiar with both aldol and Cannizaro mechanisms clearly. If not, click those links first)
First off, just because there is an alpha hydrogen atom here, does not mean that 2-methylpropanal cannot undergo Cannizaro reaction. In fact, it can very well undergo the Cannizaro reaction, and you would face no problem in drawing out the corresponding products for it. Try it out!
However, there will be a problem if you do aldol condensation with 2-methylpropanal. Have a look at the first two steps:

Step A is fine. The enolate is slightly unstable due to being a third degree carbanion, but it's not the problem. The problem actually is step B. The nucleophilic site here is a 3 degree carbanion and that presents severe steric hindrance here.
Hence, due to step B being sterically hindered, the aldol product ends up being a minor product, giving the major product title to the Cannizaro product instead.

"The enolate is slightly unstable due to being a third degree carbanion": the enolate ion will have most of the negative charge on the oxygen (another resonance structure which you haven't shown here), so I would argue that the resonance structure you have shown being a three degree carbanion would make negligible difference to the stability.

@AshishAhuja There is an important mistake you have done. Basicity is a thermodynamic concept whereas nucleophilicity is a kinetic concept. Stability plays less role here. The main thing is that steric hindrance makes it difficult to form aldol product, so aldol reaction will be slower. So cannizaro product is definitely KCP. But I can't say about TCP

@Ritil I never said that aldol reaction here will not be slow; I agree that due to steric hindrance it will be difficult to form aldol pdt. I just pointed out that a certain line in the answer (which I have quoted in my original comment) is not logical.

I would like to join this conversation and hopefully codify some of the ideas presented. First, the Cannizzaro reaction is generally conducted under more vigorous conditions than that are required for an aldol reaction or aldol condensation (when water is lost to form an enone or enal). I would hazard to guess that Cannizzaro studied only aromatic aldehydes. But if isobutyraldehyde ($\bf{1}$) and base afford disproportionation products isobutyric acid ($\bf{6}$) and isobutyl alcohol ($\bf{9}$), then it is safe to say it is a Cannizzaro reaction (if it quacks like a duck and walks like a duck...).
Assuming the reaction is being conducted in aqueous $\ce{NaOH}$, the enolate $\bf{2}$ of isobutyraldehyde $(\bf{1})$ is formed reversibly. The aldol product $\bf{4}$ may form to some extent via alkoxide $\bf{3}$ (follow the red arrows.) Aldol $\bf{4}$ cannot eliminate water under the reaction conditions because the α-position has no hydrogens available for elimination. Aldol $\bf{4}$ is a known compound prepared under other conditions (acid catalysis with Dowex-50). If it were subjected to the reaction conditions under consideration, it would undergo retro-aldolization to afford isobutyraldehyde (blue arrows) [continued].

So isobutyraldehyde cannot escape its ultimate fate --- the irreversible Cannizzaro reaction. While hydroxide adds reversibly to the carbonyl group of isobutyraldehyde to provide alkoxide $\bf{5}$, transfer of hydride to isobutyraldehyde is irreversible (red arrows; its all downhill from here!). The bis-alkoxide $\bf{5}$ has been suggested as the active hydride transfer species. The immediate products are isobutyric acid ($\bf{6}$) and the alkoxide $\bf{7}$. Proton exchange as shown for economy of space but more likely via the agency of aqueous $\ce{NaOH}$, gives the respective carboxylate $\bf{8}$ and isobutyl alcohol $\bf{9}$. If these two species are independently subjected to the reaction conditions, isobutyraldehyde will not be formed. Moreover, treatment of aldol $\bf{4}$ under the reaction conditions will also channel itself into the anion of isobutyric acid ($\bf{8}$) and isobutyl alcohol ($\bf{9}$).
Diagram $\bf{10}$ represents a possible transition state for the hydride transfer. A concerted mechanism obviates the questionable existence of hydride in aqueous solution.

2-Methyl propanal with only one alpha-H can undergo Aldol reaction, the next step condensation to eliminate water molecule requires another alpha-H which is not available in case.
While Aldol is a reversible reaction, much of the chances to get is backwards exist.
On the other hand, if hydroxide attacks aldehyde group, leading a way to Cannizzaro: the step-II, hydride transfer to another aldehyde is 'irreversible'.
Verdict: Aldehydes with just one alpha-H do not prefer Aldol; they rather proceed to Cannizzaro.

There still is another alpha hydrogen from where another alkene could form, and a condensation could still occur. So, I disagree with your first point. Secondly, could you clarify your meaning with "hydride transfer to another aldehyde is 'irreversible'." If it is irreversible, so what? Do you mean to imply that the concentration of all Cannizaro products will increase with time (because they are irreversibly formed)? Going by this logic, why can't we say that all reactions - even those with major aldol - will eventually arrive at Cannizaro as the major product? Thanks!

Referring to Jans answer one might ask for a reason that in the case of 2-methyl-propanal the canizzaro reaction gets favoured over the aldol reaction in comparison to other $\alpha$-deprotonizable aldehydes.
Take a look at the intermediate of the aldol reaction: You will get a trifold substituted carbo-anion, which is quite unfavourable in terms of stability. (Talking basic conditions as these are also required for canizzaro reactions)

No, if you put the charge on the oxygen in line with the most favoured resonance depiction, you have a trisubstituted double bond which is rather stable.

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


text:
Averaging of dipolar coupling in solution due to tumbling
The angular part of dipolar splitting depends on $\theta$ as:
$$3\cos^2\theta - 1$$
In the liquid state or in solution molecular tumbling occurs which causes the NMR signal to become an average of all $\theta$ which is expressed mathematically as:
$$\int_0^\pi(3\cos^2\theta-1)\sin\theta\,\mathrm{d}\theta$$
where $\sin\theta$ is described, in my the book that I'm reading, as a weighting factor to take into account the probability of finding the system with an angle $\theta$
It is this weighting factor that confuses me. Surely all angles are equally likely? It looks to me like the $\theta$ part of the triple integral one would do to integrate over a sphere using polar coordinates (notice the limits on the integral). However, even if this is correct, I am still finding it difficult to picture this volume element $\sin\theta\,\mathrm{d}\theta$.
Please help me to rationalise why the weighting factor is there and also to visualise this $\sin\theta\,\mathrm{d}\theta$ volume element.

The $\sin(\theta)\,\mathrm{d}\theta$ comes from the surface element of a sphere of unit radius where $\theta$ is the polar angle measured from the north pole, as it were, to the equator. The surface element is $\mathrm{d}s = \sin(\theta)\,\mathrm{d}\theta\,\mathrm{d}\phi$ where $\phi$ is the azimuthal angle measured is if longitude, i.e around the equator. The integral is
$$\frac{1}{4\pi}\iint f(\theta)\sin(\theta)\,\mathrm{d}\theta\,\mathrm{d}\phi$$
where $f(\theta)$ is the function you want to average.
The sketch shows the idea


text:
Which is the most acidic proton of (3Z)-3-ethylidenecyclopent-1-ene-1-carbaldehyde?

At first glance, I thought the most acidic protons would be the ones in red due to their proximity to the carbonyl. However, the answer key has the hydrogens in green as the most acidic. Can someone explain how that is? (I'm guessing resonance, but I need a more concrete explanation.)

If you draw curly arrows for the deprotonations, it becomes clear that only the pathway in green is productive due to the ability to form an extended enolate.
If you deprotonated the green protons, the resulting negative charge is able to be delocalised through the alkene and onto the carbonyl, the product of which is essentially an extended enolate. Due to this stabilisation, the pka of the green protons is relatively low (the resulting conjugate base is stabilised).
Deprotonation of the red proton however doesnt yield a charge which may be delocalised onto the carbonyl ( you would end up with a 5 valent carbon if you draw the arrows ). What you essentially end up with in this case then is an allyl anion (which itself is conjugated to the exocyclic alkene), which although stabilised by resonance isn't quite as favourable as in the green case (the pka of the red protons is higher).

This answer would be more helpful with arrow-formalisms drawn out.

@CPak This is homework-type question Not did OK not drawing it.

Think about keto-enol tautomerization; there is a resonance structure where the alpha hydrogen is right next to a positively charged carbon atom. Here its the same except that the resonance expands through the conjugated system and there is a resonance structute where the green hydrogens are next to a carbocation. Generally, if this happens the hydrogens become acidic due to hyperconjugation effect.


text:
Strongest negative inductive effect group between trimethylammonium, ammonium and dimethylsulfonium groups

Which of the following group exerts the strongest -I effect?

$\ce{-N(CH3)3+}$
$\ce{-NH3+}$
$\ce{-S(CH3)2+}$
$\ce{-F}$

My idea is that positive species exerts more -I effect than neutral species. thus, 4) is neglected. Moreover, as nitrogen and sulfur have almost the same electronegativity 3) has two methyl groups and 2) none so 2) should exert more -I effect. However, the answer key says it to 1). Why is this?

Methyl groups are great at stabilising carbenium ions via an inductive effect — which should actually be considered a resonance effect — known as hyperconjugation. This effect, which is actually due to neighbouring $\ce{C-H}$ bonds, not indiscriminately due to neighbouring carbon atoms, is electronic in nature. The cationic carbon has an empty p orbital, which can line up with the $\sigma_{\ce{C-H}}$ orbitals of the neighbouring carbon. These two orbitals can then interact, stabilising the $\sigma_{\ce{C-H}}$ orbital and destablising the empty p orbital for an overall energy gain of the system. It can be understood in resonance terms by the following mesomeric structures:
$$\ce{H-CR2-CR2+ <-> H+\bond{...}CR2=CR2}\tag{1}$$
For this effect to be helpful, it is vital for the positive charge to derive from an empty p orbital.
In ammonium derivatives, this is not the case. Nitrogen is $\mathrm{sp^3}$ configured with four bonds to neighbouring atoms; the positive charge stems only from an extraneous proton in nitrogen’s nucleus; it cannot be attributed to any specific (empty) orbital. In fact, all valence and bonding orbitals are fully occupied; only antibonding and far-removed orbitals are vacant. Therefore, hyperconjugation cannot serve to stabilize ammonium cations.
Instead, we need to base our discussion on truly inductive effects based solely on the electronegativity of the bonding partners. I assume that this is the analysis your answer key wants you to perform. The first step is to eliminate fluorine which is not a cation. The second step is to distinguish between sulfur and nitrogen — the latter has a higher electronegativity and thus features a less stable cation.
The final step is then comparing an $\ce{-NH3+}$ group to an $\ce{-NMe3+}$ group. Here, we need to compare secondary effects: the electronegativity of carbon is higher than that of hydrogen. Thus, a nitrogen bound to three hydrogens experiences a greater partial negative charge from these three bonds than a corresponding nitrogen bound to three carbons. Therefore, the $\ce{-NH3+}$ cation is slightly more stabilised thanks to hydrogen’s lower electronegativity — and thus $\ce{-NMe3+}$ exhibits the greater -I effect.

Thanks for the answer.here you have used the electro negativity factor to determine which of the two nitro group will have more $-I$ effect .but why instead we should not consider the inductive effect of the substituents.methyl group has more $+I$ effect than hydrogen.so it would  increase electron density of nitrogen and thus nitrogen will become more negative.so both approaches give dissimilar results.how to effectively determine which approach I'd better in certain situations than in others.

@aniline Well yes, a $\ce{CH3}$ group has more electron density than a hydrogen atom. However, that is intrinsically due to carbon’s electronegativity and it cannot serve well to shift additional electron density on to the nitrogen. The typically stated $+I$ effect of methyl groups is hyperconjugation which is actually a $+M$ effect.

@Jan If we compare the $-I$ effect of $\ce{O+(CH3)2}$, then this will have the strongest negative inductive effect?

@Apurvium We should.

@Jan Does this mean that hyperconjugation,inductive and resonance effect are same thing?

The inductive effect can be quantitatively measured by the Hammett equation
$$\sigma(\ce{X}) = \mathrm{p}K_\mathrm{a}(\ce{H}) - \mathrm{p}K_\mathrm{a}(\ce{X})$$
where $\mathrm{p}K_\mathrm{a}(\ce{X})$ refers to the $\mathrm{p}K_\mathrm{a}$ of $\ce{X}$-substituted benzoic acid and $\mathrm{p}K_\mathrm{a}(\ce{H})$ to the $\mathrm{p}K_\mathrm{a}$ of unsubstituted benzoic acid. A more electron-withdrawing group will lead to a more acidic carboxylic acid group, a lower $\mathrm{p}K_\mathrm{a}(\ce{X})$, and hence a more positive $\sigma$ coefficient.
Values for a wide range of substituents in the meta and para positions have been tabulated in Table I of Ref. 1. Also included are Swain and Lupton's $F$ ($F$ for field) parameters which are designed to remove the influence of resonance effects on the original Hammett coefficients $\sigma$ (given in the same table; see Ref. 2 for original discussion).
$$\begin{array}{ccccc} \hline
\text{Entry number} & \text{Substituent} & \sigma_\mathrm{m} & \sigma_\mathrm{p} & F \\ \hline
\text{main text, p 167} & \ce{NMe3+} & 0.88 & 0.82 & 0.89 \\
58 & \ce{NH3+} & 0.86 & 0.60 & 0.92 \\
241 & \ce{SMe2+} & 1.00 & 0.90 & 0.98 \\
15 & \ce{F} & 0.34 & 0.06 & 0.45 \\ \hline
\end{array}$$
So, perhaps the answer should be $\ce{SMe2+}$. I don't have a good explanation for why, given that sulfur is less electronegative than nitrogen. I also do not think that the difference between $\ce{NMe3+}$ and $\ce{NH3+}$ can solely be attributed to the inductive effect of the methyl groups, as it is quite possible that there are solvation or hydrogen bonding effects that affect the $\mathrm{p}K_\mathrm{a}$ values. The bottom line is that I don't think this can be rationalised as simply as the question seems to want.
References

Hansch, C.; Leo, A.; Taft, R. W. A survey of Hammett substituent constants and resonance and field parameters. Chem. Rev. 1991, 91 (2), 165–195. DOI: 10.1021/cr00002a004. (there is a non-paywall copy here)
Gardner Swain, C.; Lupton, E. C. Field and resonance components of substituent effects. J. Am. Chem. Soc. 1968, 90 (16), 4328–4337. DOI: 10.1021/ja01018a024.

As you said sulphur is less electronegative than nitrogen.then. why not $\ce{NH_3^+}$ has got more -I effect.meanwhile it has got no methyl groups as well.thanks.

"I don't have a good explanation for why".

Considering that the $\mathrm{p}K_\mathrm{a}$ values of anilinium and benzoic acid differ only by about $0.6$, I think great caution needs to be applied when pressing $\ce{NH3+}$ into the scheme above.

@orthocresol In the answer above yours, Jan has stated that +I of CH3 group is just a result of hyperconjugation, and CH3 won't show +I wherever hyperconjugation is not possible. However in your answer, you haven't treated inductive effect like that. It would be really helpful if you could confirm what Jan had stated.

@AumkaarPranav my answer only gives empirical measurements. I purposely refrain from trying to give some sort of reasoning, because I don't want to say something I'm not totally sure about. I understand it's very tempting to search for an explanation that uses the familiar heuristics of inductive effect, hyperconjugation, ... (humans like to understand things, of course!) but I'm afraid that this answer is not where you will find it. What I will say is that I do find Jan's explanation to be believable, but I will stop short of "confirming" it, because I am not an expert on this.

$\ce{-NH3+}$ should be the answer. Your thinking about the group having positive charge being more electron withdrawing is correct. Sulfur being less electronegative than nitrogen, will not withdraw electron more strongly, hence will have less $-I$ effect. Now the two nitrogen groups remain. The $\ce{-NMe3+}$ group has three methyl groups donating their electron density to the $\ce{N}$ atom, hence fulfilling some of its electronic needs. So it will have less $-I$ effect than $\ce{-NH3+}$. I hope the thought process is clear.


text:
Effect of solvent in the reaction of alkyl halide with nitrite ion
$$\ce{RX + NaNO2 -> ?}$$
Is there a difference in the outcome of the reaction if the solvent is changed from an alcohol (e.g. EtOH) to a polar aprotic solvent such as DMSO or DMF?
I have two books - one book mentions an alcoholic solvent gives $\ce{RONO}$ as product; the other mentions $\ce{RNO_2}$ is the product in DMSO/DMF. Can anyone please confirm?

Checked. It answers the question title, not the content.

Let me first clear out the difference between reaction of alkyl halide with $\ce{AgNO2}$ and $\ce{NaNO2/KNO2}$
The synthesis of nitro compounds from reaction of alkyl halides with alkali metal nitrites using DMSO/DMF as medium is known as the Victor meyer reation($\ce{AgNO2}$) or Kornblum reaction($\ce{NaNO2/KNO2}$). The products of these reaction are generally mixture of nitroalkanes and alkyl nitrites. But the main difference between them is that the % yield of nitroalkane is higher on reaction with $\ce{AgNO2}$ than that of $\ce{NaNO2/KNO2}$.Whereas % yield of alkyl nitrite is higher on reaction with $\ce{NaNO2/KNO2}$ than that of $\ce{AgNO2}$. The required mechanisms for the reactions are..

Now as for the reaction of $\ce{NaNO2/KNO2}$ with alcohol, there is no involvement of alkyl halide. Alkyl nitrite can also be prepared by adding concentrated $\ce{H2SO4}$ to the aqueous solution of $\ce{NaNO2/KNO2}$ and alcohol. The mechanism of this reaction is

Source books:

The Nitro Group in Organic Synthesis By Noboru Ono
Organic Chemistry of Explosives By Jai Prakash Agrawal, Robert Hodgson
Sourcebook of Advanced Organic Laboratory Preparations By Stanley R. Sandler, Wolf Karo


text:
When finding hybridisation of atoms, why are double bonds and triple bonds only considered once?

To figure out what hybridization an atom has is to just count the number of atoms bonded to it and the number of lone pairs. Double and triple bonds still count as being only bonded to one atom.   (Chem.Libretexts)

Why does this rule hold? Why are double and triple bonds counted only once?
Wouldn't a double bond have the equivalent number of electrons as 2 lone pairs?
And if orbitals are all about electrons, and there are three times as many electrons in a triple bond, shouldn't it count three times?

The hybridised orbitals are almost always involved in $\sigma$-bonding, and in housing the lone pair(s) of electrons. This helps us determine the geometry of the molecule.
Consider acetylene, $\ce{C_2H_2}$.
The carbon is $\mathrm{sp}$-hybridised. So, it has 2 $\sigma$-bonds - one to a hydrogen, one to another carbon. All this is in the third shell. So, a total of 4 orbitals (1 $\mathrm{s}$, 3 $\mathrm{p}$) undergo hybridisation and form 2 $\mathrm{sp}$-orbitals and 2 $\mathrm{p}$-orbitals. The $\pi$-bonding occurs due to these unhybridised $\mathrm{p}$-orbitals.
Hope this clears your doubt!
PS: The only exception to the first statement that I know is https://en.wikipedia.org/wiki/Aryne.

The rule fails pretty quickly. For example, it almost systematically fails for anything in the third period and higher or for most oxygen species. However, it is indeed valuable for carbon and nitrogen centres.
If you would count double bonds twice and triple bonds three times, you would always, as a rule (carbenium ions not considered) arrive at $\mathrm{sp^3}$ for carbon. However, the idea of teaching students hybridisation is not only to explain how bonding can actually occur in carbon (which would have a $[\ce{Ne}]\,\mathrm{2s^2\,2p^2}$ in ground state) but also how double and triple bonds differ from single bonds.
In general, double and triple bonds are considered to derive from an underlying σ bond to which a number of π bonds (one or two) have been added. These π bonds are fundamentally different from σ bonds; the bonding orbitals of the latter are always directed towards the bond partner while the principal direction of the formers’ bonding orbitals is perpendicular to the bond axis. This perpendicular arrangement is achieved by forming the π bonds using unhybridised p orbitals, while σ bonds are formed with $\mathrm{sp}^n$ hybrid orbitals.
Therefore, you are actually counting the underlying σ bonds (and lone pairs, where applicable) rather than indiscriminately counting bonds. If you do that, you are ensured to always have sufficient p orbitals remaining unhybridised to form the necessary multiple bonds.


text:
Why do methylamine and ethylamine have nearly the same basicity constant?
It is known that ethyl has a stronger inductive effect than methyl.
Its positive inductive effect is greater than methyl.
Then why does ethylamine has a $\mathrm{p}K_\mathrm{b}$ of $3.35$ and methylamine has the $\mathrm{p}K_\mathrm{b}$ of $3.34$
($\mathrm{p}K_\mathrm{b}$ values in water)
Shouldn't the $\mathrm{p}K_\mathrm{b}$ of ethylamine be lower than methylamine?

http://chemistry.stackexchange.com/questions/24447/which-base-is-stronger-methylamine-or-propylamine

Ethylamine has a pKb slightly greater than that of methylamine.
As you are saying

It is known that ethyl is better than methyl for inductive effect. It's positive inductive effect is greater than methyl.

is absolutely true. But the basicity in aqueous medium is not only determined by +I effect but also solvation of substituted ammonium cation.
Even though +I effect in ethylamine is more than that in methylamine, ethylamine  is more bulkier and sterically hinders the formation of intermolecular H−bonds with water.

The difference of methyl’s and ethyl’s inductive effects is very small.
Likewise, the difference in $\mathrm{p}K_\mathrm{b}$ values is very small.
Both are in the magnitude of practically identical or within experimenal error. As such, be happy that the theory holds true up to the point that the values are similar and don’t expect it to explain every last digit behind the decimal dot.

But i believe that  in aqueous medium we'll consider the h-bonding with water


text:
Reaction with Urushibara Catalyst
I'm not a chemist, nor an organic chemistry student, but I do a lot of research. I have a question about the end of this synthesis.

Dissolve 5g pure phenylnitropropene in 50ml Ethanol and add to Ni solution.2 Now slowly add 3ml conc. HCl3 and 1 gram shredded aluminum w/ manual stirring. Aluminum will slowly dissolve with a more vigorous effervescence of hydrogen than the first step. Maintaining good stirring with a glass stirring rod is essential in beginning. Attempted magnetic stirring will result in frustration because nickel is ferromagnetic and will stick to stirbar preventing surface area exposure necessary for reduction. After aluminum is dissolved add three more milliliters HCl and one more gram Al. Repeat adding acid and Al until 10 grams Al and about 30ml HCl has been added. Aluminum reacts slowly. Expect addition to take about six hours, longer if temp falls below 50°C. Constant stirring towards end is not necessary, just give mix a good stir occasionally.

My understanding is that the aluminum is being added to hydrochloric acid in order to create hydrogen gas, which will adsorb to the U-Ni-A catalyst. This then allows for the hydrogenation of that double bond in P2NP, AND the reduction of the nitro group to an amine group simultaneously? Or does the aluminum serve some other purpose than to just create hydrogen gas? Or is that not what's happening at all?
If I'm completely off I'd love a step by step of what's going on here, or some key words/processes for me to look up.


text:
Differentiating random and ordered sequential mechanisms in enzyme kinetics
Can random and ordered sequential mechanisms be descern by Lineweaver-Burk plot? Why or why not? Because for orederd sequential mechanism I got different LB plot depending on which substrate is held constant (when A, which binds first to the enzyme, is held constant lines for different concentrations of A in Lineweaver-Burk plot have same intercept 1/Vmax). And how are they differed based on product inhibition studies?


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

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

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

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

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

You can place the ring structures anywhere in your document.

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

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

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

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

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

Finally, you can add any substituents as required.
(Also note that for unsubstituted ferrocene in the gas phase, the cyclopentadienyl rings are actually eclipsed.)


text:
Reaction of Metallic Tin with NH4Cl and treatment with Sulfur
Metallic $\ce{Sn}$ is reacted with $\ce{NH_4Cl}$ and the resulting solution is treated with Sulfur.

What is the gaseous product?
What is the color of final precipitate?
The above resulting solution is added to $\ce{AuCl3}$, resulting in a purple    coloured solution. What is the composition of the solution?

I do realise the answer to (3) will be the purple of Cassius i.e colloidal gold in $\ce{Sn(OH)2}$ (as per Vogel's Qualitative Inorganic Analysis textbook fifth edition). But I can't figure out the answer to the first two questions (I tried Googling with no fruitful search).

Are you sure that the solution is treated with sulfur and not sulfide?

Yes, I am not getting the purpose of treatment of sulfur.

Yeah I know; me neither am sure of the purpose. But this what the question in my question paper said.

Note: this is an incomplete answer because question is unclear till now

Metallic $\ce{Sn}$ is reacted with $\ce{NH4Cl}$..

$$\ce{2NH4Cl + Sn → SnCl2 (white ppt.) + 2NH3 (both aq and g.) + H2 ^}$$
Ammonium chloride react with tin to produce tin(II) chloride, ammonia
and hydrogen. (source)
Some tin(II) chloride will react with aqueous ammonia to form tin (II) hydroxide and amm. chloride.(source)
$$\ce{SnCl2 + 2(NH3•H2O) or NH3(aq) → Sn(OH)2 or SnO.H2O(white ppt.)  + 2NH4Cl (white ppt.)}$$

..and the resulting solution is treated with Sulfur.

I am suspecting that is hydrogen sulfide and not sulfur.(Are you sure?)
$$\ce{SnCl2 + H2S → SnS(dark brown ppt.) + HCl}$$
Tin(II) chloride react with hydrogen sulfide to produce tin(II) sulfide and hydrogen chloride. Tin(II) chloride - concentrated solution. Hydrogen sulfide - saturated solution. (source)
So, answer to your questions:

What is the gaseous product?

Hydrogen, Some ammonia gas.

What is the color of final precipitate?

Stannous sulfide ($\ce{SnS}$)- dark brown ppt.

The above resulting solution is added to $\ce{AuCl3}$, resulting in a purple coloured solution. What is the composition of the solution?

$$\ce{2[AuCl4]- + 4Sn^2+ + 2H2O -> [2Au + Sn(OH)2)](purple ppt.) +
3Sn^4+ + 2H+ + 8Cl-}$$
(Vogel)
Note: $\ce{AuCl3}$ exist as $\ce{AuCl4-}$ ions in solution.

The answer says that the gaseous products are not just Hydrogen but Hydrogen and Ammonia

@UtkarshGupta Since ammonia is very soluble, some of it is present in aqueous form in the solution. Some may have passed as gas but most of the ammonia is in aqueous form. I will edit the answer though but are you sure it was sulfur and not hydrogen sulfide?

I am pretty sure the question mentioned Sulfur (I have it in my hand right now). There is not much I can do about it. Thanks for your help :)

Ok, I will see what I can do..


text:
Most stable resonating structure of 1-nitro-4-nitrosobenzene
The most stable resonating structure of 1-nitro-4-nitrosobenzene is:

In my view since in each case we have 4 charges and except in case 1 all have complete octets, so 1 is neglected. However, I am facing difficulty in eliminating other two in order to get the desired answer. I noticed that in case 3 it is in conjugation (alternating single and double bonds).

The answer is not that difficult. You should have a certain number of rules to follow in determining the most stable resonance (not "resonating") structure.
It probably starts with check all for octets (which you correctly did). 1 has ten electrons on the nitroso nitrogen, so that rules it out.
It probably then goes on to ask you to check for the least number of formal charges (which, judging by your description, you probably did).
The next step should probably say something about the negative charges being on the most electronegative elements. Electronegative atoms like O, F... (and N to a slightly lesser extent) are quite happy to take on a negative charge whereas carbon will not quite be so pleased.
Here, resonance forms 2 and 4 have the negative charge on carbon, whereas structure 3 has it on oxygen. 3 is therefore the most stable resonance structure.
Incidentally, all of the resonance forms 2 through 4 have complete conjugation throughout the molecule so that's not a factor. The $\mathrm{sp^2}$-hybridised anionic carbon can take part in conjugation equally well. Conjugation does not require alternating single and double bonds - it only requires an extended, continuous, $\pi$ system. Generally this means that all atoms along the chain must have unhybridised p orbitals parallel to each other.

Side note: "Most stable" resonance structure is a slightly sloppy description. None of these resonance structures actually exist and the true structure is a linear combination of all of these resonance structures. It should more properly be called the "greatest resonance contributor" or "greatest contributor to the resonance hybrid".


text:
Which dihydroindenyl cation is the most stabilised?
I am facing difficulty in the following problem.
The carbocation which is most stabilised among the following is …

In my view c) should be most stabilised because even if one hydrogen is abstracted it will help in restoring the aromatic character. However my book gives the answer as a).

"extent of delocalisation"?

@EashaanGodbole oh yes,I did not think about that.a) is more delocalised than c).thanks.

@aniline You may answer your own question. However, we are looking for long answers that provide some explanation and context. Do not just give a one-line answer; explain why your answer is right, ideally with citations.

The carbocation a) is mostly stabilised because it has maximum resonance structures which leads to maximum delocalisation of the charge.

None of the others has so many resonance structures.Therefore a) is most stable.

You could improve this self-answer by adding that c) is only diallylic while b) is monoallylic and d) is nonstabilised with the corresponding resonance structures ;) But happy hat.


text:
Most stable cation
I am facing difficulty in the following problem:

Which of the following cations is the most stable?

If we take the number of hyper conjugative structures into consideration then a) should be most stable. It is known that $\ce{-OH}$ is a $\ce{-I}$ group. So if it is present nearer to the carbonation it must destabilise it.Near the carbonation $\ce{-I}$ effect should be more effective than $\ce{+M}$ effect. However my textbook gives the answer as c). Is there something that I am missing out? Thanks.

@aniline think about the lone pairs on the oxygen atom and how they can contribute

@getafix Yes it's true that $OH$ group will also exert $+M$ effect.But since it is very close to the carbonation its inductive effect should dominate over the mesomeric effect.Moreover mesomeric effect should be neglected since it will bring about a positive charge on more electronegative oxygen.

That's what make (c) more stable

@getafix I couldn't follow.as I have pointed out in the comment above how can be judge that in a fight between inductive and mesomeric effect how can we say that mesomeric effect will win when it has a destabilising effect as well.

What does a carbocation look like? How is the p orbital positioned. How are the lone pairs on oxygen atom positioned? Can electrons transfer from oxygen to the vacant p orbital ? How does the inductive effect work? Think about these questions. I'm afraid I cannot write up a complete explanation right now

(c) is not a carbocation per se. There is a better resonance structure.

Please at least promote or improve this question instead of down voting or voting to close so that the requisite correct explanation could be brought forward.My attempts are quite clear in the question and comments above.

It is difficult to arrive at a strict definition of a carbocation, and that is perhaps what makes their chemistry interesting!
I am of the opinion that (a) is the only bona-fide carbocation of all four - the rest have potential interactions, either through sigma framework or through pi electrons with -O atoms (nearby).
(c) looks like a protonated aldehyde in disguise - the protonated aldehyde resonance form (if you like) will be the greatest contributor primarily due to enthalpic considerations - you technically have a distorted C=O bond (think: why is it distorted?)
It is hard to find such an arrangement in any of the other structures, and hence (c) should be most stable. That effectively answers the question, but you may also want to note that (b) is a protonated epoxide in disguise (think bridged bromonium ions vs carbocations), whereas (d) looks like a substituted oxetane.
We could expect the chemistry of (b), (c) and (d) to be vastly different from (a), and also from each other.


text:
Why is phosphorous acid more acidic than phosphoric acid?
Phosphoric acid has basicity of 3 i.e it can loose 3 $\ce{H+}$ while phosphorous acid has basicity of 2. Why is phosphorous acid more acidic than phosphoric acid? Acidity refers to the ability to liberate protons. Phosphoric acid liberates more protons than phosphorous acid (as the basicity of phosphoric acid is 3 and that of phosphorous acid is 2). Why then is phosphorous acid more acidic than phosphoric acid?

Consider that proticity is not the only factor governing acidity. For example, hydrochloric acid is considered to be strongly acidic, but it only has the capacity to donate one proton. At the same time, carbonic acid is considered to be weakly acidic, though it has two protons available for donation.

@T.Kent does carbonic acid really exists ? ^^

I'm pretty sure it is because phosphorous acid is more polar than phosphoric acid. This is because in the phosphoric acid there are more OHs and an O, so it balances out more towards the center.
In phosphorous acid, there is a H directly bonded to the central phosphorus. Hydrogen doesn't pull as much on electrons as oxygens do; hence, that part of the molecule becomes more positive, causing a stronger dipole moment than that in phosphoric acid.

You’re using a wrong definition of acidity. Acidity and its adjective acidic do not mean ‘number of protons liberated eventually.’ Instead, acidity is well defined as the equilibrium constant $K_\mathrm{a}$ of the corresponding first deprotonation reaction $(1)$:
$$\begin{align}\ce{H_nA &<=> H_{($n-1$)}A- + H+}\tag{1}\\[0.6em]
K_\mathrm{a} &= \frac{[\ce{H_{($n-1$)}A-}][\ce{H+}]}{[\ce{H_nA}]}\tag{2}\end{align}$$
For phosphoric acid, the corresponding deprotonation is reaction $(3)$ and the equilibrium constant is as written; for phosphorous acid, see $(4)$.
$$\begin{align}\ce{H3PO4 &<=> H2PO4- + H+} &&K_\mathrm{a} = 7.11 \times 10^{-3}\tag{3}\\[0.8em]
\ce{H3PO3 &<=> H2PO3- + H+} && K_\mathrm{a} = 5.01 \times 10^{-2}\tag{4}\end{align}$$
Phosphorous acid has the higher acidity constant corresponding to a lower $\mathrm{p}K_\mathrm{a}$ value and is thus more acidic than phosphoric acid.


text:
Which is the strongest base among the given anilines?
I am facing difficulty in the following problem.

In my view N,N-dimethylaniline 2 should be the strongest base because it has two methyl groups which increases electron density of N and hence it can donate it easily than in other cases. However I am not sure about it.

Related: Rates of solvolysis of p-dimethylamino benzyl halides

In my opinion, option 4 should be the answer. This is because too much crowding near the nitrogen atom will throw it out of the plane, as a result of the which the lone pair of electrons of nitrogen atom get localized and are better available for accepting the protons. Moreover, presence of $\ce{-CH3}$ groups also increase basicity due to +I effect.

In my view your answer is correct. Among all the compound, 3rd one is the least basic, due to the presence of $\ce{-N)2}$ group which withdraws the electron density of nitrogen and decrease donating capacity.
While comparing $1$ and $2$, 2nd compound is more basic due to the presence of 2 $\ce{-CH3}$ groups as they are electron withdrawing group makes the lone pair of $\ce{N}$  more available for donation.
Now while comparing $4$ and $2$, 4th compound is the less basic, even though $\ce{-CH3}$ groups is present the -I effect of $\ce{-NO2}$ group reduce the the electron donating effect of methyl .

That was what I had thought.but the correct answer is what is posted in the answer above.


text:
Arranging the phenol derivative in decreasing order of acidity
I am facing difficulty in the following problem.

What is the decreasing order of acidity of the following phenol derivatives?

In my view benzene-1,2-diol 1 should be most acidic because $-\ce{OH}$ is a strong $-I$ group thereby it will stabilise the negative charge on oxygen formed after removal of a proton.
$\ce{OCH3}$ is a $+R$ and $-I$ group. Near the carbanion $-I$ effect should be more effective. Hence the second most acidic should be 2-methoxyphenol 2. However the answer key gives it as 4-methoxyphenol 3.

The acidity order should be 1 > 2 > 3 as they have pKa values of 9.48, 9.98, and 10.21 respectively.
+R effect of $\ce{-OCH3}$ is more significant than +R effect of $\ce{-OH}$
In case of 1, -I effect works at -ortho position and also +R effect works at -ortho position.
In case of 2, -I effect works at -ortho position and also +R effect works at -ortho position. Net effect is +R effect as +R effect is dominant than -I effect.
In case of 3, -I effect works at -para position and also +R effect works at -para position. Net effect is +R effect as +R effect is dominant than -I effect. But this +R effect is more dominant than in case 2 because -I effect is more dominant at -ortho position than at -para position but+R effect of equally effective at both -ortho and -para positions.
As a result of which, acidity is reduced to greater extent in the third compound compared to the second compound.

How do you know that +R effect is more dominant in case b).

If your assumption is taken as granted then would be the relative acidic strength of o-,m-,p-methoxy benzoic acid.this accordingly will be then m->o- > p-.

Yes. It is not I am saying for granted, it is general organic chemistry. Also +R effect is not meaningful at meso position.

Answer to the first comment, I have very clearly explained in my answer. Please read the answer very carefully

With regards to paragraphing, you need to press Enter twice to make a line break.

Hey Resorcinol, I wanted to know how to decide that a 5 memebered ring due to a H-bond was being formed or not in compounds A and B, as that would have reduced their acidity?


text:
Which has more inductive effect: an aldehyde or a carboxylic acid?
A chemistry reference book that I have been reading gives the order of inductive effect ($-I$) as follows:

While the Wikipedia page on inductive effect gives the following order:
$\ce{-NH3+} > \ce{-NO2} > \ce{-SO2R} > \ce{-CN} > \ce{-SO3H} > \ce{-CHO} > \ce{-CO} > \ce{-COOH} > \ce{-COCl}> \ce{-CONH2} >
\ce{-F} > \ce{-Cl} > \ce{-Br} > \ce{-I} > \ce{-OR} > \ce{-OH} > \ce{-NH2} > \ce{-C6H5} >\ce{-CH=CH2} >\ce{-H}$
The reference book has $\ce{COOH}$ ahead of $\ce{CHO}$ but Wikipedia has it the other way around. Which source is correct?

And where will ester ( O-CO-CH3 ) lie in this order?

I also find it strange that the acyl chloride COCl is considered less electron-withdrawing than aldehydes (CHO) and ketones (CO) in the Wikipedia list.

I believe the Wiki page is more accurate, because the Hammett Equation substituent constants (para) indicate -CHO should be more "electron withdrawing" than -COOH, but there is little reason to use such lists as "absolute" - I think they are simply good guidelines for reactivity.
I looked at the data in this link: https://www.bluffton.edu/homepages/facstaff/bergerd/classes/CEM311/handouts/sigmas.pdf
However, substituent constants are not absolute proof of "inductive effect" and it is difficult to conclude from such data that a particular substituent is more "inductively" electron withdrawing than another. To do that, we would need a model (or a set of compounds, equivalently) where we can achieve separability of sigma and pi electrons.

$\ce{COOH}$ will show more -i effect than $\ce{CHO}$ because $\ce{COOH}$ hybridisation considering free radical as a sigma bond will be $\ce{sp^2}$ and both oxygen will also attract electron in angle 120° so that re will be more positive than aldehyde therefore it will attract electron more and show more -i effect.

CHO group will have more -I effect than COOH because OH group in COOH group will decrease the charge on the carbonyl C which reduces it's withdrawing effect.

Shouldn't decreasing the charge on C increase the electron-withdrawing effect (on nearby substituents)?

Note that electron density is enhanced at the carbonyl carbon in carboxylic acid by resonance effect ( lone pair on oxygen is in conjugation with pi bond on C-O bond ) . Hence, the R effect increases electron density at carbon in carboxylic acid while there is no R effect in aldehyde .
And , we know that ;
The more electron rich is carbon , the less electronegative it is and the less electronegative it is, the less negative inductive effect it shows .
Hence , aldehyde group would show more I effect than carboxylic acid.

Can anyone do a constructive criticism that what is wrong in my answer ?


text:
Which occurs first: nitration or oxidation of aniline by concentrated nitric acid?

Which of the cases is possible or both or none?

Would definitely oxidise. Protect the amine but reacting with acyl chloride to make abide to make the aromatic ring less activated.

None, oxidative oligomerisation happens.

Concentrated $\ce{HNO3}$ is ‘only’ around $70~\%$. To add a nitro functionality by electrophilic aromatic substitution, typically the $\ce{NO2+}$ cation is required which is formed only by strong dehydrating agents such as sulfuric acid:
$$\ce{O2N-OH <<=>[\ce{H2SO4}][-\ce{HSO4-}] O2N-OH2+ ->[][-\ce{H2O}] O=\overset{+}{N}=O}\tag{1}$$
Concentrated nitric acid is not strong enough to dehydrate iself. Thus, nitration will typically not be observed.

Not really, nitration doesn't even need any acid in some variants.

So oxidation would occur  first?  @Jan,Melanie Shebel


text:
Why does sulfonation of aniline occur para rather than ortho?
According to a textbook by K. L. Chugh:

Sulfonation [of aniline] is a reversible reaction. The p-isomer, being more stable, does not get desulfonated easily. Hence, p-sulfanilic acid is the major product.

I am curious as to why the ortho isomer (2-aminobenzenesulfonic acid) is not more thermodynamically stable. There is a possibility of intramolecular hydrogen bonding (in a six-membered ring):

Why is the para isomer (4-aminobenzenesulfonic acid) favoured?

The sulfonyl group is kind of bulky due to the size of the sulfur, so steric hindrance is probably making it energetically disfavored for the group to be placed ortho. It's unlikely you can see this easily from a model or a drawing. If you have access to computational resources, you might be able to see the issue.


text:
Arranging different oxygen-containing organic compounds by decreasing acidity
I am facing difficulty in the following problem.

Arrange the following compound in decreasing order of acidity.

It is clear that conjugate base of (Q) is more stable.hence it is most acidic.in R) there is much steric hinderence so my view is that it should be less acidic.however I am not sure about it.

And this is the type of question for which Professor Mayr made us chemistry students rote memorise typical $\mathrm{p}K_\mathrm{a}$ values. Here are some.
\begin{array}{lc}\hline
\text{compound type} & \text{typical }\mathrm{p}K_\mathrm{a}\\ \hline
\text{alcohol} & \approx 15 \\
\text{ketone} & \approx 20\\
\text{carboxylic acid} & \approx 5\phantom{0}\\
\text{secondary }\ce{C-H} & \approx 45\\ \hline\end{array}
Therefore, the order is $\mathrm{Q > P > R > S}$

why the downvote? I would agree with the answer. Carboxylic acids are undoubtedly more acidic than others here. A hydrogen attached to an Oxygen(in OH) should be more acidic than a H attached to a C (in ketone) and there is no reason for an ether to be acidic.


text:
Obtaining 100% pure metal
While reading my book everywhere I found that 99.9% pure metal is obtained. Is it possible by any process that we get 100% pure metal on industrial level?

Nothing is 100% pure.

99.9% pure is 100% pure within certain tolerance. What tolerance do you want?

Typical industrial silicon wafers for the fabrication of integrated circuits are 99.9999999 % pure.

related http://chemistry.stackexchange.com/questions/2870/what-is-a-pure-substance

The only way to get a "100% pure" sample of an element is by mass spectroscopy, or something similar, which acts on each atom (or ion, actually) one at a time.
So you could get a sample of pure rubidium of a few dozen atoms. Caveat: "100 % pure" applies only to those atoms actually observed... there could be other elements in the vicinity that have not been observed.
Practically, though, there is no way to prevent a single atom of hydrogen, for example, from eventually contaminating your "100 %" pure sample.

True and false. There is no absolute vacuum, so any cluster would absorb some gas on their surface sooner or later even inside the ms chamber

Yes, that's what I wrote.

When we start talking about Fineness of metals we can deal with some misunderstanding. We generally have to make a distinction between absolute purity and the common metals basis purity. The first refers to all the impurities (all the elements of the periodic table) the second refers  only to metals. On industrial level is not possible to reach 100% absolute purity, because this means that the whole process must be carried on in ultra-clean ultra-vacuum environment. We can however reach very high purity. Usually the purity is measured in Nines below you can find a table that summarize the purity grades:
\begin{array}{lllll}
Nines & Purity \%  & Total Parts & Matrix  \space ppm & Impurity \space ppm \\
1N    & 90\%       & 1,000,000   & 900,000    & 100,000      \\
2N    & 99\%       & 1,000,000   & 990,000    & 10,000       \\
3N    & 99.9\%     & 1,000,000   & 999,000    & 1,000        \\
4N    & 99.99\%    & 1,000,000   & 999,900    & 100          \\
5N    & 99.999\%   & 1,000,000   & 999,990    & 10           \\
6N    & 99.9999\%  & 1,000,000   & 999,999    & 1            \\
7N    & 99.99999\% & 1,000,000   & 999,999.9  & 0.1          \\
-     & 100\%      & 1,000,000   & 1,000,000  & 0
\end{array}
In the past Royal Canadian Mint used to mint a 5 nines gold coin and that was something very unusual, the purest gold ever produced was 999.999—six nines gold that was claimed to be refined by the Perth Mint in 1957. At sigma aldrich you can buy 99.999% trace metals basis so is not the absolute purity. If you are interest in the topic this article could be helpful Schulze, K.K. JOM (1981) 33: 33. doi:10.1007/BF03354422 while a general discussion can be found here.


text:
Validation of _atom_site_frac_*: Range outside of [0,1)?
I have the following cif file
loop_
_atom_site_label
_atom_site_type_symbol
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
...
Al2 Al -0.35936(1) -0.08768(9) -0.24188(8)
Al3 Al -0.27044(1) -0.35380(9) -0.13044(8)
F1 F -0.5320(2) -0.2015(2) -0.3794(2)
F2 F -0.3800(2) -0.2437(2) -0.1891(2)
F3 F -0.0350(2) 0.0944(2) 0.1528(2)
...
N3 N 1.2782(3) 0.7260(3) 0.3726(3)
...
C2 C 1.0165(4) 0.6764(4) 0.2267(3)
C3 C 1.1053(4) 0.6806(4) 0.3576(3)
...
H1 H 1.015(4) 0.603(4) 0.147(3)
...
H6 H 1.068(6) 0.773(6) 0.220(5)
H7 H 1.100(5) 0.761(5) 0.441(4)
...
H10 H 1.303(5) 0.653(4) 0.319(4)
...
H12 H 1.070(5) 0.574(5) 0.364(4)
...
H16 H 1.328(6) 0.749(6) 0.471(5)
H17 H 1.313(5) 0.801(4) 0.348(4)

Of interest is the fact that many of these "fractions" of the cell length are greater than 1, and many are negative. To my sensibilities these atoms lie outside the unit cell.
Should the IUCr definition form CIF dictionary of _atom_site_fract_x state that the permitted range is 0.0 -> 1.0, but forgot?
Should these atoms be discarded, should they be mod'd into the unit cell, or should they be interpreted in some other manner?

I couldn't find in the reference that the range is permitted only from 0 to 1.

@Martin-マーチン: It isn't in the reference. The question is whether it should be.

Ah, thanks! I completely missed that reading it, sorry.

Everything should be ok. The first Al atom in the list (not shown above) is arbitrarily set at (0,0,0) so you could just add a constants to this (and same to all other atoms of course) if you don't like negative values etc. But if you leave things as they are and work out xyz coords using the known transformation matrices and the unit cell params given, you should have the structure.  As it happens, in this structure the unit cell boundary cuts through the molecule. The unit cell is only the repeat pattern, it has no physical significance per se .

@porphyrin: If you put this as an answer I'll mark it as accepted. As an aside, I suppose that chemists feel uncomfortable putting a slice through (say) a benzene ring and that if you want a conventional unit cell with a slice just mod it in?

I completely agree with @porphyrin comment. This is an example of crystal data which formatting does not entirely meet IUCr requirements, that's all. This happens a lot, and in most cases it's no big deal. This particular set of coordinates is just not uniform within a single cell. Different crystallographic software uses different means to "reassemble" atoms "correctly" and write changes to *.cif, e.g. Olex2 uses Standardise function for that:


text:
A metal, M, forms two fluorides containing 40.43% F and 50.44% F by mass. Identify the metal

A metal, $\ce{M}$, forms two fluorides containing 40.43 % $\ce{F}$ and 50.44 % $\ce{F}$ by mass. Identify the metal.

I get that you can find out how much of the metal reacts with 1 mole of fluorine atoms in the 40.43 % one. This can be calculated to be 27.99 g.
Then from that you can work out that 28.49 g of fluorine reacts with the metal in the 50.44 % one and that that is about 1.5 moles of $\ce{F}$.
I'm a bit lost after that. How might you continue from here?
So the law of multiple proportions conveys that the fluorides contain fluorine in a 1.5:1 or 3:2 ratio. This means the possibilities are
$\ce{MF2}$ and $\ce{MF3}$, or
$\ce{MF4}$ and $\ce{MF6}$
Since there is twice as much fluorine, the mass of $\ce{M}$ in $\ce{MF2}$ must be 56 g. The next one is 112 g. If there are no errors in what was said, the metal must be iron or cadmium. What makes one a better candidate over the other?

Assuming that your calculation is correct, which oxidation states sound better? Fe-(II)/(III) or Cd-(IV)/(VI)? You'll agree that's OK for iron, what about cadmium? Have one look into the periodic table, and you know the answer.

I'd like to thank you all for your help. I understand that Cd(VI) or even Cd(IV) is quite unlikely, making Fe the better answer.

You need to use the general concepts of the law of definite proportions and the law of multiple proportions to solve this:
Sample 1 is 59.57% $\ce{M}$ and 40.43% $\ce{F}$.
Sample 2 is 49.56% $\ce{M}$ and 50.44% $\ce{F}$.
Assume $\ce{M_{x}F_{y}}$ for sample 1 and $\ce{M_{a}F_{b}}$ for compound 2.
Let $m$ by the molar mass of $\ce{M}$. We have
$$\frac{19.00y}{mx + 19.00y}=\frac{40.43}{100}$$
and
$$\frac{19.00b}{ma + 19.00b}=\frac{50.44}{100}$$
We can rearrange the equations to:
$$m\frac{x}{y}=28.0$$
and
$$m\frac{a}{b}=18.7$$
Note that $$\frac{28.0}{18.7}\approx\frac{3}{2}$$
so we can start by trying:
$$m\frac{3}{1} = 28.0$$
and
$$m\frac{2}{1} = 18.7$$
This is weird because the compounds would be $\ce{M3F}$ and $\ce{M2F}$.
Fortunately, as long as the proportions work, it's fine, so divide both fractions by 6:
$$m\frac{1}{2} = 28.0$$
and
$$m\frac{1}{3} = 18.7$$
This gives $m=56.0$, fairly close to the molar mass of iron which is $55.9$.
Cadmium is a bad candidate, since that would be $\ce{CdF4}$ and $\ce{CdF6}$.

Why do you say that the compounds would be $\ce{M3F}$ and $\ce{M2F}$?  I see that those are the proportions by mass, but not by mole.  Is that what you meant and if so, why is it weird?  Also I don't see where you would have lost a significant figure anywhere.  Your choice of rounding for F warrants rounding to 4 in any final answer.  Aside from those questions I think it is a + answer.

So I don't see where you would have lost a sig fig, but I do agree with Li Zhi  that if you were limited to 3, the rounding should not have happened at the step that yielded 27.99 and 18.67

I see now that you should have lost one in the summation (subtration) that yielded 18.67.

Also @ Zhe you divided both fractions by 6.  18.67 not 16.67.

Thanks @JosephHirsch. Fixed that and the rounding issues while I was at it.

you still get 28.00 since 1900.-768.17 retains 4 sig figs to yield 1132, but 18.7 is correct.  (though it doesn't matter) This, by the way will give 6*(28.00-18.7) or 55.8 as your calculated m.

A for algebra, F for chemistry. You add nothing new to what the OP already said.

If you set it up this way you retain 4 sig figs in your result, but 55.96
Sample 1 is 59.57% $\ce{M}$ and 40.43% $\ce{F}$.
Sample 2 is 49.56% $\ce{M}$ and 50.44% $\ce{F}$.
Assume $\ce{M_{x}F_{y}}$ for sample 1 and $\ce{M_{a}F_{b}}$ for compound 2.
Let $m$ by the molar mass of $\ce{M}$. We have
$$\frac{mx}{19.00y}=\frac{1-0.4043}{.4043}=\frac{0.5957}{0.4043}$$
and
$$\frac{ma}{19.00b}=\frac{1-0.5044}{0.5044}=\frac{0.4956}{0.5044}$$
We can rearrange the equations to:
$$m\frac{x}{y}=19.00*\frac{0.5957}{0.4043}$$
and
$$m\frac{a}{b}=19.00*\frac{0.4956}{0.5044}$$

OK, I see it.  The question now is "what makes one metal a better candidate than the other."  Anyway, if you use the setup I showed, avoiding the subtraction step that costs a sig fig, you get 55.96.  So your real "accepted" answer is that CdF4 and CdF6 are not good candidates, while obviously FeF2 and FeF3 are as they match the normal oxidation numbers for the metal.

Another approach would be to assume that one of the compounds is of the type $\ce{MF_n}$ with $n$ a small integer. Using a spreadsheet (or just building a table) you use the small integer multiples of $18.9984$ (say across the top 6 of the columns $n=1,2,3,4,5,6$ — hexavalent compounds are fairly common, higher valences are possible but much less likely. For the rows' first column you choose the values of the % of F. This allows you to calculate the mass (which we can (but don't have to) assume is the atomic weight (atomic mass)) of the metal). That calculation is that the mass of the F atoms is the respective % of that row, so the total mass is the Fs mass ÷ %F. Subtract the mass of Fs from this and you get the mass of the M. (Yes, you could have done the same thing using the $100-\%\ \ce{F}$ values, maybe even simpler, IDK).
Then, you look at the entries, and see two (and only two) pairs that are "interesting" - meaning they're close to one another. Under 2F ($37.9968$) is $55.984$ and under 3F is $56.001$ and under 4F is $111.970$ and under 6F is $112.002$. These suggest that Fe at $55.845$ or Cd at $112.414$ are good matches. In fact if you compute the difference between their theoretical %s and the measured values, the sum of the square of the differences are $\ce{Fe} = 0.41$ and $\ce{Cd} = 0.00$, which indicates that statistically Cd is the "better" answer. You then should compare the values and determine if the difference between them is meaningful given the measured results. Since the %s were given to 4 places, I think that Cd is the "more correct" answer, the error for Fe is simply too large, IMHO.
This approach has another advantage: you can inspect the table to see if there are any values which are small multiples of an actual atomic weight of a metal (atomic mass). This would allow you to evaluate the assumption that the formula is $\ce{MF_n}$ and not $\ce{M2F_n}$ or $\ce{M3F_n}$, depending on how good your mental arithmetic is.

So, you're proposing $\ce{CdF6}$? $\ce{CdF4}$ instead of $\ce{FeF2}$ and $\ce{FeF3}$? Also, I mentioned the issue with rounding. I was just too lazy to fix it after the fact...

Where did Zhe claim he should have rounded "as an intermediate step"?  He claimed that his final.
Actually, in manipulating the formula, he has an intermediate step where he takes 1900. - 768.17.  Even though the 1900 has 4 sig figs, when subtracting or adding in an intermediate step, you have to round immediately (in this case to the 1's place).  You can't carry 1131.83 to the end because it is a summation so it should have been rounded to 1132, but it only changes the final answer to 28.00

In the other manipulation we have 1900. - (50.44)19.00.  Once again this summation requires immediate rounding to the 1's place in this case giving 942 and 18.7  Again, you have to round to place immediately after summation.  In this case you lose 1 sig fig.  This gives an m=56.9 using his final step which is pretty far off. (or 56.1 and 56.0 if you use 19.00 to get the the denominators).

That should be  56.1 and 56.00 above^^

You know, if you had registered you would probably have had enough reputation to comment by now …


text:
Stability of diaminocyclohexa-2,4-dienyl cations
I am facing trouble in the following question

What is the order of stability of these compounds

In my view the stability must be greater for that compound which is able to disperse the positive charge most effectively. For that reason, 4,6-diaminocyclohexa-2,4-dien-1-ylium (R) should be most stable because the $\ce{-NH2}$ is ortho and para in that. The next most stable should be 2,5-diaminocyclohexa-2,4-dien-1-ylium (Q) followed by 3,5-diaminocyclohexa-2,4-dien-1-ylium (P). However the answer key gives it to be (P) > (Q) > (R). I couldn't understand where I am making mistake.

You've asked a lot of questions just like this one.  The lesson to be learned is that you need to draw out all of the resonance structures for each ion.  Then you can make arguments about their relative stability.

Maybe we really should start marking your questions as dupes …

Hint: You do not have an aromatic system in any of the three structures. I assume you thought you had because you used ortho and para.

@Jan I am aware of the Huckel's rule.I used ortho and para to denote positions relative to the carbocation.

@aniline It’s not even about Hückel’s rule which only relates to the number of electrons. You don’t even have a conjugated system.

The general idea in solving such questions should be to draw out the resonance structures as suggested by @ron

It is clear from the picture above that $P)$ has maximum resonance structures followed by $Q)$ and then $R)$ .Same
 is the order of their stability.

Now, you're on the right track, but note that there is 1 more resonance structure that you can draw for both P and Q.


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How uniform is the distribution of elements in the universe?
I was wondering about element distribution in the universe, specifically if any inconsistency in the relative distribution of elements would have any significant effects on the chemistry of a galaxy. How likely is it that there could be significant differences in element quantities from one galaxy to another, so that certain common, or 'important' chemical compounds or reactions might not be possible/probable.
For example could a galaxy not have the proper relative quantities of elements used in organic chemistry to allow the sort of organic chemistry we see occurring in many places in our own local space (comets and cosmic dust)?

Related: Is the relative natural abundance of isotopes of an element the same everywhere?

Take a look at : https://www.youtube.com/watch?v=KlBG_A4Djp4
It explains the s-process

@QuantumAMERICCINO I really enjoyed that, and while I don't have the expertise to answer my own question, it makes it clear to me that any variation in elements probably lies entirely with the composition of star types in a given space.  Perhaps the question is better suited for a astronomy focused exchange?


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Polarity index vs. Dipole moment
I'm looking to find whether water or methanol is more polar, and I'm getting conflicting answers.
My textbook says "the polarity of a bond is quantified by the size of its dipole moment." According to this table, at 20 °C, the dipole moments of methanol and water are 2.87 D and 1.87 D. This implies to me that methanol is more polar.
But there seems to be another measure called "polarity index," which I can't seem to find a definition for online. This table says the polarity indices of methanol and water are 10.2 and 5.1, which implies to me that actually water is more polar.
So what's the difference between the dipole moment and the polarity index? How is the latter defined? And is methanol or water more polar?

Related: What is the order of the following compounds based on polarity?

From Roger E. Schirmer "Modern Methods of Pharmaceutical Analysis":

Solvents are generally ranked by polarity, but polarity is not a uniquely defined physical property of a substance. Hence the relative polarity of a solvent will be somewhat dependent on the method used to measure it....
Solvent polarity is a complex function of many parameter in addition to adsorption energy. A more recent ranking of solvents by Snyder is based on a combination of parameter such as dipole moment, proton acceptor or donor properties, and dispersion force solvent...
Snyder's polarity index ranks solvents according to a complex theoretical summation of these properties. As a rule, the higher the polarity index, the more polar the solvent.

Snyder's paper was published in   Journal of Chromatography A, Volume 92, Issue 2, 22 May 1974, Pages 223-230 . Take a look at it to see how the index was calculated.
There are different types of polarity indices. And each of them has different parameters and ways to calculate the polarity.
Unlike polarity, dipole moment is a physical property. Polarity indices often take dipole moments as a parameters when calculating the polarity of solvents.  The reason why there are polarity indices is because dipole moments alone couldn't explain the nature and interactions of solvents.
Part of your confusion stems from the repetition of the word "polar" to describe different phenomena. There is polarity of bonds and there is polarity of solvents. They are different things.

I know bond polarities and molecule polarities are different things, but then what is the dipole moment of an entire molecule measuring?

Badyal YS, Saboungi M-L, Price DL, Shastri SD, Haeffner DR, Soper AK. Electron distribution in water. The Journal of Chemical Physics. 2000;112(21):9206-8 gives the dipole moment of water 2.90 +-0.6D, what means that water is more polar then methanol.


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Why is the carbon-carbon bond in hexachloroethane shorter than in ethane?
How can we explain the shorter carbon-carbon bond length in $\ce{(CCl3)2}$ ($1.49 \,\mathrm{\mathring{A}}$) than that of $\ce{(CH3)2}$ ($1.534 \,\mathrm{\mathring{A}}$)?

One reasoning would be applying Bent's rule:

In ethane the $\ce{C-H}$ bonds are approximately composed from $\ce{61\% sp^{3.3} C}$ and $\ce{39\% s H}$ orbitals.
That results that the $\ce{C-C}$ bond is composed from $\ce{sp^{2.3} C}$ orbitals.
In hexachloroethane the $\ce{C-Cl}$ bonds are composed from $\ce{48\% sp^{3.7} C}$ and $\ce{52\% sp^{4.8} Cl}$ orbitals.
That results that the $\ce{C-C}$ bond is composed from $\ce{sp^{1.8} C}$ orbitals.

Higher $\ce{s}$-character in bonds usually means that the optimal overlap of these orbitals is at shorter inter-nuclear distances. Since the $\ce{s}$-character in the $\ce{C-C}$ bond in hexachloroethane is much higher than in ethane, the bond length is expected to be shorter.
(Values obtained from an Natural Bond Orbital Analysis on the DF-BP86/def2-SVP level of theory.)