Reactions of alkynes with sodium/ammonia vs sodium amide What happens when alkynes react with sodium in ammonia ($\ce{Na/NH3}$), or when they react with sodium amide ($\ce{NaNH2}$)?

Are these two sets of conditions different?

Sodium in ammonia ($\ce{Na/NH3}$) and sodium amide ($\ce{NaNH2}$) are very different sets of conditions, despite superficially looking similar!

Sodium metal itself, $\ce{Na}$, is a one-electron reducing agent – it is oxidised to $\ce{Na+}$ in the process. When it reduces neutral hydrocarbons, negatively charged species are formed, which pick up a proton from the ammonia solvent. Overall, $\mathbf{Na/NH_3}$ is a way of reducing some organic compounds (including alkynes).

Internal alkynes $\ce{R-C#C-R}$ can be reduced to trans-alkenes using $\ce{Na/NH3}$. Specifically, solvated electrons are created which add to the triple bond as shown. Ammonia acts as a proton source to protonate the carbanions formed. The vinyl anion intermediate 4 can interconvert between a cis and trans geometry, and the trans geometry is preferred as this minimises steric repulsion between $\ce{R^1}$ and $\ce{R^2}$. When this anion is protonated, the trans-alkene is formed. This is a nice synthetic procedure for the preparation of trans-alkenes, complementary to Lindlar hydrogenation which produces cis-alkenes from alkynes.

On the other hand, in sodium amide $\ce{NaNH2}$, sodium is already in the +1 oxidation state and is no longer a reducing agent. However, the amide ion $\ce{NH2-}$ is a very strong base (the $\mathrm pK_\mathrm a$ of ammonia, $\ce{NH3}$, is 38). So $\mathbf{NaNH_2}$ is a way of deprotonating some organic compounds (including terminal alkynes).

The proton attached to the terminal carbon in a terminal alkyne is acidic enough ($\mathrm pK_\mathrm a \sim 25$) to react with sodium amide, forming the corresponding carbanion 7. Such carbanions are good nucleophiles and can be used in many synthetic procedures involving nucleophiles, such as addition to a carbonyl compound, as illustrated here.

In the first case hydrogenation occurs(anti addition) whereas in the second case, acid base neutralization occurs. So they aren't identical.
According to Wikipedia,

$$\ce{2 Na + 2 NH3 -> 2 NaNH2 + H2}$$

This shouldn't occur "usually" in reactions, unless it is specified at very low temperature.
However, using FeCl3 as a catalyst one can produce sodamide and hydrogen from sodium metal and ammonia in appreciable amounts.


Why is the endo product the major product in a Diels-Alder reaction? I understand that the endo product is, at least in all cases in my textbook, the major product in a Diels-Alder reaction. However, I don't understand why this is the case.

The explanation in my textbook says that the endo product allows for more interaction between the diene and dienophile during the transition state.

I also found the following post, which suggests that secondary orbital interactions are the cause. Stereochemistry of product in Diels-Alder reaction

Then I found the following paper that indicates, at least to my understanding, that steric strain during the transition state is the main reason behind the endo rule (or at least for the example reaction in the article). https://www.ncbi.nlm.nih.gov/pubmed/24449044

So, I'm left wondering what is the reason for the endo rule?

What is the endo-selectivity of Diels-Alder reactions?
If the diene used in the Diels-Alder reaction has asymmetric substituents at the end carbons, and if the dienophile is unsymmetric, then two different isomers of the final adduct can form. The isomer, where the functional group(s) (usually carbonyl) on the alkene end up on the same side as the newly formed double bond, is called the endo-isomer [1]. The other one is the exo isomer. The naming of the isomers can be a little ambiguous in cases where there are multiple functional groups of different types on the starting alkene. Usually the electron-withdrawing group is considered when identifying which isomer is endo and which is exo.
 [2]
The experiments tell us that, in general, the endo-product is the one which is favoured kinetically i.e. it forms faster. This is not a rule as such, because there is no way to predict to what extent the endo product will be favoured. (In many cases, the exo product is the thermodynamic product i.e. more stable, so allowing the reaction to equilibriate will result in higher yield of exo product)
Why does this happen?
It's difficult to answer this question definitively, because there is still a lot of active research going on in the area. I will try to summarise the main effects which have been proposed as possible causes of the endo-selectivity.

Secondary orbital interaction

The interaction between the frontier molecular orbitals at the points where the new $\sigma$-bonds form are the first-order interactions. However, there are other interactions possible between the diene and dienophile, there are called second-order interactions, or secondary orbital interactions (SOI). This explanation was proposed by Woodward and Hoffmann, in the book The conservation of orbital symmetry[3].
The following diagram will be useful:
 [4]
As you can see, in the transition state leading to the endo-product, the carbonyl part of the alkene can interact with the back of the diene, and the interaction is favourable, as the symmetry matches. In the transition state leading to the exo-product, this interaction would not be available, as the carbonyl group is away from the diene. For a long time, SOI was considered as the right answer, and was extensively invoked in explaining not just Diels-Alder reaction, but other types of cycloaddition reactions as well.
However, there are a lot of evidences, both experimental and computational, which calls into question the validity of SOI. There is a good paper from Salvatella et.al. summarising them: "Do  Secondary  Orbital Interactions  Really  Exist?" Acc. Chem. Res., 2000, 33, 658.
For example, the dimerization of butadiene was shown to have slight endo preference.
 [5]
This was frequently used as an example of the SOI effect. However, computational studies at RHF and CASSCF level have found that the reaction proceeds via a stepwise mechanism. [6] If the mechanism is stepwise, not concerted, then it does not go through the constrained transition state which is required for the secondary orbital interactions to come into existence. So, we cannot invoke SOI for this endo-preference.

Steric effects

In certain Diels-Alder reactions, especially those involving the cyclopentadiene, steric effects have been proposed.
 [5]
Experimental data shows that when $\ce{R}$ = $\ce{CH3}$ and $\ce{X}$ = $\ce{CHO}$ the endo product is formed in $17.0%$ yield, i.e. the exo product now dominates. [7] This indicates that the steric repulsion of cyclopentadiene ring with $\ce{CH3}$ and the  is greater than that with $\ce{CHO}$, which is consistent with $\ce{CH3}$ being bulkier.
These observations can also be explained by SOI, but the trends in the endo-selectivity for a range of $\ce{R}$ and $\ce{X}$ groups seem to fit better to the steric argument.[5]
In a more recent study by Fernandez and Bickelhaupt, the selectivity was found to arise from activation strain, i.e. steric strain as the reactants approach the transition state.[12] They used $\mathrm{M05{\text -}2X/def2{\text -}TZVPP}$ to model the reaction of cyclopentadiene and maleic anhydride.

The above image, taken from their paper, shows the full reaction coordinate, with ZPE-corrected energies in kcal/mol and distances in Å. As you can see, the exo-reactant complex (exo-RC) is higher in energy than the endo one, suggesting that the strain comes into existence even before the TS. The extra strain in the exo-pathway is thought to arise from the steric repulsion between the methylene of cyclopentadiene and the oxygen of the anhydride.

Hydrogen bonds

Hydrogen bond-like interactions between C-H in alkene and the double bond has also been proposed as a cause of endo-selectivity in particular cases, such as the reaction of cyclopropene and butadiene.[8] Here, both steric effects and H-bond like interactions are thought to be present.
 [5]
This is somewhat different than the SOI interaction, because in SOI, only orbital overlap (delocalization) is considered. H-bond interactions are a sum of orbital overlap and electrostatic attraction.

Electrostatic forces

Electrostatic forces can be used to explain the unusual exo-selectivity of the furan+cyclopropenone addition.

In the exo-TS, there is a favourable electrostatic interaction between the electronegative O of furan and the electropositive carbonyl carbon of cyclopropenone. This makes the exo-isomer the preferred one. The exo-TS is about 1.81 kcal/mol lower in energy than the endo-TS, according to computational studies. [9]
The usual endo-selectivity of other Diels-Alder reactions can also be explained by the electrostatic repulsion between slightly electropositive terminal C-H of the diene, and the electropositive carbon of the electron-withdrawing group on dienophile:
 [5]

Solvent effects

Sustmann and others carried out computational studies on Diels-Alder reactions in both gas phase and with dielectric-continuum solvation.[10] They found that in the gas phase, the exo-TS is lower in energy i.e. kinetically favoured. Using a solvation model resulted in the endo-TS being the favoured one.
They suggested that solvent effect might be the cause of endo-selectivity. The endo-TS is generally more polar than the exo-TS, so it is stabilised to a greater extent in solvent. It is also observed that the preference for endo-TS increases the more polar the solvent is, which lends support to this argument. The gas phase exo-preference would also disprove the role of SOI effects, because they should be present regardless of solvation.
However, computational studies like these have to be considered carefully, because they were mostly carried out using HF or B3LYP methods, both of which severely underestimate dispersion forces (which are necessary to describe SOI). So, the results might be artificially skewed in the direction of exo-TS. [11]
Conclusion
All of this means that we are not really sure if there is a single cause that can explain the endo-selectivity in Diels-Alder reactions.
In an extensive experimental-computational study , Paddon-Row et. al.(2020) studied reactions of isotope labelled butadienes with mono- or di-substituted alkenes.[11] Surprisingly, it was found that mono-substituted alkenes (such as methyl acrylate) did not have a strong endo-preference when the reaction was done in benzene solution. Methyl acrylate and butadiene, for example gave 50:50 exo-endo product yield. Di-substituted alkenes showed higher endo-selectivity. In their analysis they suggested that SOI effects are important in some cases, whereas steric effects are forcing the selectivity in some other cases.
So, it might just be that different effects are working in different reactions, and somehow the end result is that endo-product is favoured in almost all cases. Then it might not be useful to look for one particular blanket cause of the selectivity, rather, each reaction has to be explained uniquely.

References
(1) Organic Chemistry, Clayden, Greeves, Warren, 2nd ed., Oxford University Press, p.884
(2) Image source: https://employees.csbsju.edu/cschaller/Reactivity/pericyclic/peri%20endo.htm
(3) The conservation of orbital symmetry, R B Woodward, R Hoffmann,   Weinheim : Verlag Chemie GmbH, 1971
(4) Image source: https://www.cpp.edu/~psbeauchamp/pdf/316_MO_DA_rxn.pdf
(5) Image source: Acc. Chem. Res., 2000, 33, 658
(6) Y. Li and K. N. Houk, J. Am. Chem. Soc., 1996, 118(4), 880–885
(7) Y. Kobuke, T. Fueno, J. Furukawa, J. Am. Chem. Soc., 1970, 92, 6548-6553
(8) M. Sodupe,R. Rios,V. Branchadell, J. Am. Chem. Soc., 1997, 119, 4232-4238
(9) S. M. Bachrach, J. Org. Chem., 1995, 60, 4395-4398
(10) T.  Karcher,  W.  Sicking, J. Sauer and R. Sustmann, Tetrahedron Lett., 1992, 33, 8027–8030
(11) W. J. Lording, T. Fallon, M. S. Sherburn, M. N. Paddon-Row, Chem. Sci., 2020, 11, 11915–11926
(12) I. Fernández, F. M. Bickelhaupt, J. Comput. Chem., 2014, 35, 371– 376. DOI: 10.1002/jcc.23500

From what I know, there reason why the endo product is preferentially formed is still not clear and subject to further enlightenment. But it seems to arise from more efficient interactions between the pi electrons of the dienophile and the pi electrons of the developing double bond in the diene during the transition state. In this website you can the both transition states in 3d

http://www.chemtube3d.com/DAendo_vs_exo,cyclopentadiene_and_maleic_anhydride.html


Determine products of NH4NO3 hydrolysis What are the products of the following hydrolysis reaction? How do you know?

$$\ce{NH4NO3 + H2O <=> ?}$$

It will never happen. You can not hydrolyze a salt to any strong acid (nitric acid in your case) or base because the reverse reaction is predominant.

The products will be $\ce{HNO3 + NH4OH}$.

This is because a salt hydrolysis reaction always produces an acid and a base. The $\ce{H+}$ in $\ce{H2O}$ will bond with $\ce{NO3-}$ to form $\ce{HNO3}$, a strong acid. The remaining $\ce{OH-}$ in $\ce{H2O}$ will bond with $\ce{NH4+}$ to form $\ce{NH4OH}$, a weak base.


Is “amount of substance” the same thing as “number of moles”? Background:
Amount of substance is a fundamental physical quantity which has mole (mol) as it's SI unit. Therefore all expressions for amount of substance should have the unit mole on simplification.
If $A$ is the amount of substance (normally in moles), $m$ is the mass of the substance in a particular unit (normally in grams), $M$ is the mass per unit amount of substance (normally in grams/mole), then
\begin{align}
A &= \frac{m}{M}\tag1\label1
\end{align}
If the substance is a molecule then $M$ is known as molar mass of the substance.
But in my textbook and in many websites on the internet, I have encountered the phrase number of moles. I think it refers to the amount of substance or maybe the numerical part in the amount of substance. The formulas given there were strange.
If $n$ is the number of moles, $m$ is the mass of the substance in grams and $x$ is the atomic weight (for atoms of elements) or molecular weight (for molecules of elements and compounds), then,
\begin{align}
n &= \frac{m}{x}\tag2\label2
\end{align}
This equation is not dimensionally correct if I am right.
My questions:
First of all I would like to ask whether amount of substance and number of moles refer to the same thing, or is it that amount of substance has a unit along with a numerical value whereas number of moles does not have a unit and represents the numerical value in the magnitude of the amount of substance.
Secondly, between equations $\eqref1$ and $\eqref2$, which one is completely correct (both in meaning and dimension)?
Thirdly, since in some place I have encountered gram atomic/molecular mass in place of molar mass, I would like to know what are the differences between both and do they have the same units or different units?

"Amount of substance" is one of the seven base quantities of the SI [1]. Its dimension is noted with the symbol N, and its base unit is the mole (mol). Similarly "length" is another base quantity with dimension noted L and with base unit the meter (m).

"Number of moles of [substance X]" rigorously means the numerical value (thus dimensionless) associated with the corresponding "amount of [substance X]" expressed in moles, but from my experience it is often misused for "amount of [substance X]" (thus of the dimension of an "amount of substance"). - Think about how you would respond to the question: "How many meters are there between your house and the nearest bakery?" You could answer "35" or "35 meters", both would be grammatically correct; but strictly speaking the question expects just a numerical answer.

Note that according to the SI, a number of entities (molecules, atoms,..) is dimensionless and the unit of a dimensionless quantity is simply the number one, 1 [1]. My way of reconciling these two views (an "amount of substance" of dimension N and of unit the mole vs. a number of entities, dimensionless and of unit 1) is that the "amount of substance" refers to a combination of an amount (a number) and a given type of entity of a given substance, whereas a number of entity simply refers to the number itself. Why not consider an amount of substance as a dimensionless quantity? I think it is a reminiscence of history, at the time when chemists had to measure amount of substance without knowing how this substance was structured (but I can be wrong).

Regarding the equations, I agree with this other answer by @desc (equation 1 is correct, equation 2 isn't). Importantly: "atomic weight" is dimensionless (it is a ratio between a mass and a reference mass) and should not to be mistaken with "atomic mass". [2]

Finally, regarding the units "gram-atom" or "gram-molecule", these are outdated units which, according to SI were connected with "atomic weights" and "molecular weights" [2]:

  Following the discovery of the fundamental laws of chemistry, units
  called, for example, "gram-atom" and "gram-molecule", were used to
  specify amounts of chemical elements or compounds. These units had a
  direct connection with "atomic weights" and "molecular weights", which
  are in fact relative masses. "Atomic weights" were originally referred
  to the atomic weight of oxygen, by general agreement taken as 16. But
  whereas physicists separated the isotopes in a mass spectrometer and
  attributed the value 16 to one of the isotopes of oxygen, chemists
  attributed the same value to the (slightly variable) mixture of
  isotopes 16, 17 and 18, which was for them the naturally occurring
  element oxygen. Finally an agreement between the International Union
  of Pure and Applied Physics (IUPAP) and the International Union of
  Pure and Applied Chemistry (IUPAC) brought this duality to an end in
  1959/60. Physicists and chemists have ever since agreed to assign the
  value 12, exactly, to the so-called atomic weight of the isotope of
  carbon with mass number 12 (carbon 12, $^{12}$C), correctly called the
  relative atomic mass $A_r$($^{12}$C). The unified scale thus obtained gives
  the relative atomic and molecular masses, also known as the atomic and
  molecular weights, respectively.
  [2]

https://www.bipm.org/en/publications/si-brochure/chapter1.html
https://www.bipm.org/en/publications/si-brochure/mole.html

Your question is mainly semantic. The amount of a substance and number of moles are both unitless quantities (and the same thing). A mole is simply defined as "Avogadro number" of something. That is, 1 mole of (say) carbon atoms is about $6.02\times10^{23}$ carbon atoms. Both have a dimension of 1 (the number one); a mole is conveniently represented as a unit of measurement but doesn't actually hold any physical meaning. Using moles (and derivative quantities, such as molar mass which is the mass of "Avogadro number" items of something) is only a way to avoid extremely large or small numbers.

Another important point is that the variables in an equation only have the meaning you attach to them, so though conventions as to the meaning of each letter in your equations do exist I will answer to your definitions.

Equation (1) is correct - dividing the mass of a sample with its molar mass will give the number of moles.

Equation (2) is incorrect - diving the mass of a sample with the weight of a single item (atom or molecule) will give the number of items, not the number of moles.

Two notes:

Common convention is that $n$, and not $A$, is the number of moles. So the most conventionally correct equation is $$n = {m \over M}$$
where $n=[\mathrm{mol}]$, $m=[\mathrm g]$ and $M=\left[\mathrm{g \over mol}\right]$ (brackets are used to denote the unit of measurement of quantities).
The reason the second equation usually works is a double error - the atomic "weight" is usually given in number of nuclear particles (protons and neutrons) and not in actual weight units. This number is very close to the molar weight in $\mathrm{g \over mole}$ (on purpose, the definition of a mole is meant to be this way), but these are different quantities. Many chemistry books tend to confuse them. The actual atomic or molecular weight (in grams) of a single particle is very small and it is not likely you will actually use it much.


Bonding in the phosphate ion I'm looking for an explanation of the bonding in the phosphate (PO43−) ion:

(Image courtesy of Wikipedia)

Phosphorus (15P) - being the fifteenth element - has fifteen electrons, five valence electrons and the following electron configuration:

1s2
2s2 2p6
3s2 3p3

This being so, by what means does it form its bonds? Naively, one might expect it to form 3 single covalent bonds and be happy. But it doesn't, it forms five bonds, presumably using each of its five electrons. My first question is - why would it do this? It seems to me that this would result in the Phosphorus atom having ten electrons in its outer shell (four from the double bond with the oxygen and two each from the single bonds with the other three oxygens).

Apart from the oxygen with a double-bond, we have three oxygens forming single bonds. Does this not leave them one electron short?

Finally, why are the oxygens shown as carrying a negative charge?

First of all, let me state the obvious: Phosphorus is awesome.
After we got that out of the way we can focus on why.

There are many different modifications of phosphorus in nature. With increasing thermodynamic stability they are $$\ce{P_{white} -> P_{red} -> P_{violet} -> P_{black}}.$$

Apart from this there are many low molecular different allotropes, like $\ce{P4 (white)}$, $\ce{P6}$, $\ce{P8}$, $\ce{P10}$, $\ce{P12}$. And because that is not enough, there are chain like polymeric allotropes, too.

Apart from this it is possible to formulate many different cations and anions, that are derived from the above molecular structures. Just to name a few, there are $\ce{P3+, P5+, P7+, P9+}$ mainly observed in gaseous phase an $\ce{P^{3-},P2^{4-}, P3^{5-}, P4^{2-}, P7-,...,}$ usually in combination with alkali metals. Most amazingly it may form anionic polymer strains of the general form $\ce{[P7-]_{\infty}, [P15-]_{\infty}}$.

But now for the most important part, phosphorus has stable oxidation states in compounds, ranging from $\mathrm{-III}$ to $\mathrm{+V}$. Here are a few examples:
$$\ce{\overset{-III}{P}H3,\ \overset{-II}{P_2}H4,\ [\overset{-I}{P}H]_{n},\ \overset{\mathrm{\pm0}}{P4},\ H3\overset{\mathrm{+I}}{P}O2,\ H4\overset{\mathrm{+II}}{P2}O4,\ H3\overset{\mathrm{+III}}{P}O3,\ H2\overset{\mathrm{+IV}}{P2}O6,\ H3\overset{\mathrm{+V}}{P}O4}$$
While dealing with these compounds it is usually completely unnecessary to describe bonding with hybrid orbitals.

In case of phosphane $\ce{PH3}$ it would be wrong. Assuming $\ce{{}^{sp^3}P}$ one would expect $\angle(\ce{H-P-H})\approx109^\circ$, while it is found to be $\angle(\ce{H-P-H})=93.5^\circ$, which is almost the same angle as the $\ce{p}$ orbitals are having towards each other.

In general your assumption is correct, that it is possible to form only three covalent bonds to reach a stable configuration. And that will most likely be the case when phosphorus forms compounds with more electropositive elements.

Now dealing with oxygen, means dealing with a much more electronegative element, i.e. $\ce{En(O)}\approx3.4$, $\ce{En(P)}\approx2.2$. That also means that bonds are much more polarised towards the oxygen.

Analysing the phosphate anion $\ce{PO4^{3-}}$ it is crucial to recognise its symmetry, which is tetrahedral $T_\mathrm{d}$. In this arrangement it is perfectly safe (but not at all necessary) to describe phosphorus as $\ce{sp^3}$ hybridised.

A Natural Bond Orbital analysis (BP86/def2-TZVPP) reveals that there are 4 equal $\ce{P-O}~\sigma$ single bonds and each oxygen has three lone pairs. The contribution of the $\ce{d}$ orbitals to bonding is well below $1\%$ and can be interpreted as numerical noise (use as polarisation functions) of the DFT method.
\begin{array}{rlrr}\hline
       &             &\mathrm{\%P (hyb)}  &\mathrm{\%O (hyb)}\\\hline
3\times&\ce{Bd(O-P)} & 24 (\ce{sp^3}) & 76 (\mathrm{sp^{2.3}})\\
       &\ce{Lp(O)}   &                & 100 (\mathrm{sp^{0.4}})\\
2\times&\ce{Lp(O)}   &                & 100 (\ce{p})\\\hline
\end{array}

This is consistent with the partial charges, i.e. $q(\ce{P})=2.2$, $q(\ce{O})=-1.3$. Therefore a more accurate Lewis formula is with charge separation.

The corresponding NBO reflect the bonding picture one would expect given all details from above. It should be noted, that natural bond orbitals are a linear combination of the canonical orbitals and do not have a physical meaningful energy eigenvalue. The two top rows represent the $\ce{p}$ lone pair orbitals of oxygen, the third row represents the $\ce{sp^{0\!.4}}$ lone pair orbitals, the fourth row gives the $\sigma$ bonding orbitals. (The last row is the orientation of the molecule, core orbitals are not displayed.)

The corresponding canonical orbitals which have a physical meaningful eigenvalue are delocalised over the whole molecule, hence they are not providing a simple understandable bonding picture. While NBO fail to respect the symmetry point group, canonical orbitals are constructed to obey this principle. (Here shown from highest energy, top, to lowest energy, bottom, core orbitals not displayed.)

I happen to look at this site just now.
The reason is all a matter of hydridization.
Look at the electron configuration of Phosphate and that of the next element that use next orbital i.e. K.
The hybridization would bring in 3s, 3p, and 4s together. To complete fill them, number of electrons or bonds required would therefore be 5.
I guess that there are 5 sp3 hybrid orbitals.
Because of electron negativity of Phosphorus and oxygen, there are negative charge on the oxygen atoms that are singly bonded, while the doubly bonded oxygen, being closer could have insignificantly charged negatively.


Electrophilic substitution of an indole as opposed to benzofuran and benzothiophene Regioselectivity in electrophilic aromatic substitution of benzofuran and indole

This response as to the regioselectivity of the electrophilic aromatic substitution of indole left me with a few more questions than it did answers.

In figure 3b, does the pi bond in the bottom half of the benzene on the left get pushed to the nitrogen, allowing for stabilization across the benzene ring in addition to the resonance structures of the nitrogen cation? And if so, and this is the case of a viable nitrogen cation contributing to the stabilization of a valid resonance structure, is it true that all (or most) polycyclic compounds that undergo electrophilic substitution at a location on the nitrogen-containing ring that result in the temporary resonance structure of a nitrogen cation along with stabilization across a benzene ring will be more stable than substitution at a location that would only provide the resonance across the benzene (e.g. without the possible nitrogen cation, had electrophilic substitution occurred elsewhere)?

Furthermore, does this imply that in the electrophilic aromatic substitution pattern of benzothiophene, seeing as how sulfur is less electrophilic than both oxygen and nitrogen, that we would see a similar substitution pattern at the 3-position, as opposed to the substitution at the 2-position that we would see (And that is, to my puny little brain, comprehensible) on benzofuran?

because when electrophile react on 3 position more resonance structures are posible than the 2 position and thus it show stability.Thats why we say that position 3 is more good for indol to attach electrophile.


How to judge accuracy and precision of measurement devices? I am unsure about the accuracy and/or precision of measuring instruments.
For example, if we have a graduated cylinder, a pipette and a beaker, how and why can we assert them that they can be used with accuracy, precision, or both?

Beakers, Erlenmeyer flasks, dropping funnels and the like are not volumetric instruments. They are not precisely calibrated, and their scales serve only as approximate guides. Therefore, the beaker that is mentioned in the question is right out.

Graduated cylinders, volumetric flasks, bulb pipettes, graduated pipettes, and burettes are volumetric instruments. They are calibrated; however, like any other measuring instrument, they only have a limited accuracy and precision.

The accuracy and precision do not only depend on the type of the instrument (e.g. graduated cylinder or pipette). They also depend on the dimensions, graduating divisions, and accuracy class of the instrument, which are usually internationally standardized.

In principle, volumetric instruments with a small diameter have a better precision because the same difference of the liquid level (i.e. the meniscus) corresponds to a smaller difference in volume. Therefore, typical long and thin pipettes are usually better than short and thick graduated cylinders. For example:

A standardized graduated pipette (Class A or AS) with a nominal capacity of 10 ml and an overall length of about 360 mm has a maximum permissible error of ±0.05 ml.
A standardized graduated measuring cylinder (Class A) with a nominal capacity of 10 ml and a maximum overall height of 140 mm (tall form) has a maximum permissible error of ±0.1 ml.
If you use an even larger graduated measuring cylinder (Class A) with a nominal capacity of 100 ml and fill it just to the 10 ml mark, the maximum permissible error is ±0.5 ml.

These errors represent the maximum permissible error at any point on the scale, and also the maximum permissible difference between the errors at any two points.

Since the same sources of error are, naturally, inherent both in calibration and use, a volumetric instrument should have a good precision in order to be able to be calibrated with a good accuracy.

However, note that graduated pipettes are usually calibrated ‘to deliver’ (TD, Ex), whereas graduated measuring cylinders are calibrated ‘to contain’ (TC, In). Hence, graduatted pipettes and graduated measuring cylinders may not be interchangeable at will.

The term precision means two different things. The number 1.23456789 has more "precision" than the number 1.23. That is, there is a mathematical (and computer science) meaning that is different than the one used by physical scientists.

In physical science, precision means the property of reproducibility of a value or result. If I am precise, I get the same answer every time (or nearly so). If I am accurate, I get the correct answer. Here is one example: I have a machine which counts people going through the door. Whenever someone walks through, it adds 1.1 to its total. This machine is precise but not very accurate. I modify it and now it adds between 0.95 and 1.05 to the count each time someone walks thru. It is definitely not more precise, but may or may not be more accurate.


Nomenclature of common naming of alkynes vs alkenes; both -ene? Is there a difference between common naming of alkenes vs alkynes? Why is Ethylene the common name for an alkene and acetylene for an alkyne? Isn’t acet- the common form of naming ethyl groups anyway so why is ethyl used in a common name?

The official IUPAC name for ethyl with a double bond is ethene, but this is a relatively recent change.
Before the official name (as an exception) was ethylene. I’m not sure about propene/propylene but am fairly confident the official IUPAC name for propyl double bond is propene and not propylene. However, these compounds are often referred to with the unofficial name as per tradition.

For alkynes, the IUPAC name is, for ethyl, the name is ethyne.

Don’t remember to specify the location of the double/triple bond (ex: 2 - ethene)

For alkenes, the formula is $\ce{C_nH2_{n+2}}$ and for alkynes is $\ce{C_nH_{2n}}$.


Enthalpy of neutralisation for weak acids and weak bases at standard state I recently read about why for weak acids and bases the enthalpy of neutralisation is less as compared to the strong ones. This was because the acids and bases being weak, they dissociate incompletely and even in this they absorb energy.

I have two questions regarding this:

At standard state, for a solute, we take the hypothetical case of unit molality behaving as one which is infinitely diluted, then shouldn’t even weak acids or bases act as strong acids/bases (because they are, at least in theory assumed to behave as if they were infinitely diluted)?
It's mentioned that during dissociation they (weak acids/bases) absorb heat. Wouldn't even strong ones absorb at least some heat to effect their dissociation.

I find the first question to be more important in terms of conceptual value.

Q1, well you are correct to a certain level. standard state of a solute in a solution of any specified solvent is the state having solution under pressure exactly equal to standard pressure 1 bar,composition of solute equal to 'a standard composition', solution must be an 'ideal solution (perfect solution)', and finally solution must show 'properties of an infinitely diluted solution'. here standard composition refers to either, standard amount concentration (1 mol/dm3), standard molality (1 mol/kg) or standard mole fraction (1). note that temperature is not a part of definition of standard state. this is a hypothetical state for any temperature we use or what standard composition you may chose. here you have chosen standard molality,its ok.(note that we can prove that for an ideal solution showing properties of infinite dilution at a given pressure and temperature molar enthalpy of solute is a constant what ever its composition may be.plug into somewhere that defines you exactly the terms 'perfect solution',infinite dilution and properties of infinite dilution to understand this, its easy). infinite dilution means amount concentration or equivalently any other composition term of solute approaching zero. however acid  composition is molality =1mol/kg as you prefer or any other standard composition, but yet it is not infinitely diluted although it shows properties of infinite dilution. key thing here to understand is that infinite dilution is one thing and showing properties of infinite dilution is another. this is not the only hypothetical state you are dealing with, standard state of a gas existing in pure form , as a solute or in a mixture is the pure form of gas at pressure 1 bar and gas is ideal or equivalently compressibility factor =1. going back to your question, we can talk about strong and weak acids or bases. in fact Ka and Kb values are based on this.  Q2; strong and weak are relative terms. the better word is strength. even strong acids or bases absorbs heat .


Why is methyl iodide more dense than ethyl iodide, even though the latter has more carbon atoms? It is understood that more carbon atoms present in a hydrocarbon, the greater the density of the hydrocarbon.

But then why does $\mathrm{CH_{3}I}$ have a greater density than $\mathrm{C_{2}H_{5}I}$?

For the same reason why a bucket of sand will weigh more than a bucket of rocks.
Smaller molecules will result in a more dense, thus more heavy substance.


Is dehydrating copper(II) sulphate pentahydrate a chemical or physical change? Is removing the waters of hydration from $\ce{CuSO4 * 5H2O}$ a chemical or physical change? Sorry for the trivial question. I get varying answers across the net.

The traditional answer is that it's a chemical change when bond-making and bond-breaking is involved.  The material certainly changes appearance, and heating copper sulphate somehow releases bound water.

That said, bonding in metal-aquo complexes runs the gamut from very weak to exceedingly strong - exchange rates span 15 orders of magnitude (Fig. 1 in Jan Reedijk, Metal-Ligand Exchange Kinetics in Platinum and Ruthenium Complexes. Platinum Metals Rev. 2008, 52, 2–11. DOI: 10.1595/147106708X255987), and where do you draw the line between a "physical" electrostatic interaction (in, say, $\ce{Na(OH_2)_6^+}$) and a chemical bond (as seen in $\ce{Cr(OH_2)_6^3+}$?

The definition of a chemical reaction: We call changes that change bonding form of atoms, which results changes in the structure and properties of matter, is called a chemical reaction.
So this reaction isn't a chemical reaction. Copper (II) Sulphite is still Copper (II) Sulphite and water was'nt in bond with Salt it was trapped in crystal system of salt
P.S: SO3 is Sulphite and SO4 is Sulphate


can I mix. trisodium phosphate, potassium hydroxide, and sodium hydroxide together? Can I mix trisodium phosphate (tsp), potassium hydroxide, and sodium hydroxide ( or sodium hypochlorite ) and not cause a bad reaction?  And will this make for a good degreaser and cleaning chemical for me ?  I own a pressure washing business.  I do fast food resturanta, gas stations

I advise against using potassium or sodium hydroxide in homemade cleaning agents. You could easily end up with a mixture that could severely burn eyes (up to blindness), skin or lungs.

Also, grease from fast food restaurants is fundamentally different from gas station grease. The former consists most probably of fat while the latter is a mineral oil product.

Your home-made degreaser, with a hydroxide concentration that could be handled without complete personal protective equipment, might or might not have some effect on fat due to saponification but for removing mineral oil based grease I would not expect it to be effective.

Potassium hydroxide, and sodium hydroxide tend to produce heat when you dissolve them in water, tbs is very peaceful. So if you dump a kilo of NaOH in a bucked of water it might shoot up the ceiling and burn your skin if it splashes on you. If you add it slowly you will be OK. If you want to make a liter of the solution start by adding a table spoon of NaOH and look what happens. You can increase the rate of addition as you get more comfortable. I would personally add 100 g of NaOH to 1 liter in 10 portions over 5 minutes, but that is because I know what I am doing. You will need to stir solution with a right tool (at least don't use your hand or aluminum spoon).

NaOH and KOH dust are nasty. You will need eye protection, gloves,      labcoat and a shower nearby in case of emergency.

This solution will absorb CO2 and lose activity, so you need to seal it.

tldr They are not going to blow up but you shouldn't add them too fast to each other.


Is it possible to have negative bond orders? In the MO theory, is it possible to have a molecule with bond order less than 0? The least bond order I have come across as of now is 0 (for di-atomic noble gases), indicating that the molecule cannot exist and the atoms get pulled apart.

Is it possible to create a scenario (at least theoretically) where we deal with a negative bond order by manipulating the system's conditions or any of that sort?

No, it's not possible, theoretically either.

If you see the Molecular orbital theory, maximum number of anti bonding electrons = number of bonding electrons.

And bond order is (Number of bonding electrons - number of anti bonding electron)/2

So, the minimum value that you can get is 0 and not negative, as anti bonding electrons can't exceed the number of bonding electrons.

I don't think that bond order can be zero. Think of bond order as the number of bonds an atom is forming in the molecule. So, least, it can form no bond . There would be no physical significance of an atom forming negative number of bonds. That would be absurd.


Buffer Solution - Sodium Benzoate A buffer with a pH of $\mathrm{4.79}$ contains $\mathrm{ 0.39\ M}$ of sodium benzoate and $\mathrm{0.10\ M}$ of benzoic acid. What is the concentration of $\mathrm{[\ce{H} ]}$ in the solution after the addition of $\mathrm{0.058\ mol}$ of $\ce{HCl}$ to a final volume of $\mathrm{1.6\ L}$?

My thought is that
$$\mathrm{pH=pK_a+Log_{10}(\frac{[base]}{[acid]})}$$

When we added $\mathrm{0.058\ mol}$ of $\ce{HCl}$, the concentration of both acid and base decreased by $\frac{.058}{1.6}$, so I take that ratio and apply it.  I found $\mathrm{pK_a}$ by just applying what's given.  Not sure where I am going wrong.

In that equation, the acid and base are a conjugate acid/base pair.  We assume HCl quantitatively dissociates and protonates benzoates.  The concentration of the acid increases, and the base decreases.

Empirically, the pH should go down in this case.  Without seeing your specific calculation, we can't comment where you are going wrong, but the proper flow should look something like:

$$\mathrm{pH=pK_a+Log_{10}(\frac{[base]-[Added H]}{[acid]+[AddedH]})}$$

$$\mathrm{pH=pK_a+Log_{10}(\frac{0.39M - 0.036M}{0.10M + 0.036M})}$$

$$\mathrm{pH=pK_a+Log_{10}(\frac{0.354M}{0.136M})}$$
From your calculated $\mathrm{pKa = 4.20}$, you should get there pretty quickly.

Just applying the difference in the log values of the changed ratios, I get  $\mathrm{pH = 4.61}$

The concentration of [H] ion will be $2.45x10^{-5}$ if the pH was 4.61.


Understanding strong acid titrated with weak base graph I don’t understand why the conductivity doesn’t change after the equivalence point. To my understanding, at the equivalence point there is $\ce{NH3}$, $\ce{NH4+}$ and $\ce{Cl-}$ in the solution. As more $\ce{NH3}$ is added it only partially dissociates, thus conductivity stays the same. But doesn’t the concentration of $\ce{Cl-}$ ions also contribute to the conductivity, and its conductivity decreases are more aqueous $\ce{NH3}$ is added as it becomes more dilute? Where is the issue in my understanding. Thanks for the help.

You are missing an important point. Conductometric titration is not just plotting the conductance after adding the titrant. One cannot plot conductance as it is, because you are significantly diluting the solution.  You have to apply a correction to take dilution into account. Once you apply a correction, the curve will remain flat because further addition of ammonia will not enhance conductivity.
See for example,

Conductometric titrations
Throughout a titration the volume of the solution is always increasing, unless the
conductance is corrected for this effect, non linear titration curves result. The
correction can be accomplished by multiplying the observed conductance either
by total volume (V+V') or by the factor (V+ V')/V, where V is the initial volume
of solution and V' is the total volume of the reagent added. The correction
presupposes that the conductivity is a linear function of dilution, this is true only
to a first approximation.

After equivalence point you add more and more NH3 solution (remember NH4OH does not exist). So you add a negligible amount of ions. As a consequence, the conductance does not change significantly. This is what your diagram shows.


Where does the Fe(III) ion go? Iron(II) sulphate solution is added to a test-tube. Following this, potassium manganate solution is added gradually to the test-tube as well. The observation seen is that the green solution turns pink at first, then potassium manganate turns colourless as manganese(VII) is oxidised to manganese(II).

But Fe(II) is oxidised to Fe(III) ions as well, which are brown in colour, so shouldn’t the solution turn brown instead of colourless?

I have seen these observations experimentally in the lab but still did not understand. I know I’m probably thinking about it the wrong way so any help would be much appreciated.

In solution, the iron(III) will be hexaaqua iron(III), 6 water molecules coordinating the iron ion, with the oxygen atoms being at the vertices of an octahedron.

Hexaaqua iron(III) has a high spin d5 electron configuration.

All the d-d transitions are spin forbidden.

Hexaaqua iron(III) is nearly colorless.

This answer mentions about colours of aqueous iron solutions and how to test them (only)
$$\ce{10 FeSO4 + 2 KMnO4 + 8 H2SO4 -> 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8 H2O}$$

Iron(II) salts are usually very pale green in aqueous solution and iron(III) salts are yellow in aqueous solution. You can test what form of iron is present by adding thiocyanate, which gives red precipitate exclusively with iron(III): Place a drop of test solution on a spot plate place and add one drop of 1 per cent ammonium thiocyanate solution. A deep-red colouration may appear. Nitrites should be avoided in acid solution as they form nitrosyl thiocyanate $\ce{NOSCN}$, which yields a red colour, disappearing on heating similar to that with iron(III)
$$\ce{Fe^3+ + 3 SCN- ->Fe(SCN)3}$$

And iron(II) can be detected most reliably with 2,2'-bipyridine reagent where deep-red complex bivalent cation $\ce{[Fe(C10H8N2)3]^2+}$ is formed in mineral acid solution. Test a drop of the faintly acidified test solution with one drop of the reagent on a spot plate; a red colouration may be obtained.


C-N bond lengths in 2-bromo-1,3,5-trinitrobenzene Are all the $\ce{C-N}$ single bonds in 2-bromo-1,3,5-trinitrobenzene the same or different? I am of the view that the $\ce{C_1-N}$ and $\ce{C_3-N}$ bonds are of the same length and they are longer than the $\ce{C_5-N}$ bond, due to steric hindrance. Am I right or am I missing out something?

(source: nist.gov)

By rotating the molecule 180° about the $\mathrm{C_2}$ symmetry axis that passes through the bromine and the nitro group in the 5-position, you can interconvert $\ce{C_1-NO2}$ and $\ce{C_3-NO2}$.  Groups that can be interconverted by a rotational axis must be equivalent.  There is no symmetry element that interconcerts $\ce{C_1-NO2}$ (or $\ce{C_3-NO2}$) and $\ce{C_5-NO2}$, therefore these groups cannot be equivalent. So symmetry tells us that the  $\ce{C_1-NO2}$ and $\ce{C_3-NO2}$ bond lengths are exactly the same, and both are different from the $\ce{C_5-NO2}$ bond length.

Symmetry can only tell us if things are the same or different, it can't tell us the magnitude of any difference.  So I don't know for sure which set of carbon-nitro bonds is longer or shorter, but your argument based on steric strain seems reasonable.  It is also possible that the bulky bromine causes the adjacent nitro groups to twist out of the plane of the aromatic ring, reducing resonance overlap between nitrogen and carbon and also lengthening those adjacent nitro bonds.

The bond lengths of carbon 1 & 3 are same and they are longer than that of carbon 5 and its attached nitrogen. This is due to resonance and in resonance we see 1 structure at a time.


Solute and Solvent classification in a Solution When I look up the definition of solute and solvent, the definition by the Oxford English Dictionary is
solute

/ˈsɒljuːt,sɒˈljuːt/

noun

the ***minor*** component in a solution, dissolved in the solvent.

What does minor imply here? Does it classify them as solute and solvent as per mass/volume comparison? Or does it refer to their concentration?

By major/minor they mean whichever component has the greater/lesser presence in mass or moles.


What is the pKa of the second ionizable group? One compound contains an amino group with $\mathrm{p}K_\mathrm{a}$ of $8.8$, and one other ionizable group with a $\mathrm{p}K_\mathrm{a}$ of $5$ to $7$. To $100\ \mathrm{mL}$ of a $0.2\ \mathrm{M}$ solution of this compound at $\mathrm{pH}\ 8.2$ was added $40\ \mathrm{mL}$ of a solution of $0.2\ \mathrm{M}\ \ce{HCl}$ and the $\mathrm{pH}$ changed to $6.2$. What is the $\mathrm{p}K_\mathrm{a}$ of the second ionizable group?

I am a little confused because I am not sure what to do when the $\mathrm{pH}$ does not equal the $\mathrm{p}K_\mathrm{a}$ of the amino group. I'm thinking since the known $\mathrm{p}K_\mathrm{a}$ of $8.8$ is highest at that point there should be no free $\ce{H+}$.

Since you're given that the maximum $\mathrm{p}K_\mathrm{a}$ of the second group is $7$, you can assume that the amino group will fully protonate before the second group starts accepting electrons.  (Perhaps this is what you meant by "at that point there should be no free $\ce{H+}$"?)
Therefore, from Henderson-Hasselbalch$^\ddagger$, you can calculate the fraction of the amino group that is protonated based on the starting $\mathrm{pH}$ and the amino group $\mathrm{p}K_\mathrm{a}$. (Hint: Yes, the amino group should be mostly, but not entirely, protonated.)
It's straightforward to then calculate:

The total number of moles of acid added
The number of moles of not-yet-protonated amino groups before adding the acid

which is equal to

The number of moles of acid consumed to protonate the rest of the amino groups

and, therefore, you can find

The number of moles of acid left over to protonate a portion of the second ionizable group in the molecules.

Having calculated #4, you can then calculate what fraction of the second ionizable group is protonated by the leftover acid.  This feeds back into Henderson-Hasselbalch, from which you can calculate the second $\mathrm{p}K_\mathrm{a}$ value since you are told the value of the final $\mathrm{pH}$.
$^\ddagger\,\mathrm{pH} = \mathrm p K_\mathrm a +\log\left(\frac{\ce{[A-]}}{[\ce{HA}]}\right) = \mathrm pK_\mathrm a + \log \left(\frac{[\mathrm{A_{total}} ] - [\ce{HA}]}{\ce{[HA]}}\right) = \mathrm p K_\mathrm a + \log\left(\frac{1}{f_\mathrm{protonated}}-1\right)$

pH=pKa + log ([A]/[HA])
pH goes to 6.2
8.2=8.8 + log([A]/[HA])
8.2-8.8 = log ([A]/[HA])
the difference is negative but since it represents just "the difference" we can use positive.
.06= log ([A]/[HA])
so using the new pH,
6.2= pKa + .06
6.2-.06= pKa
pKa= 5.6


What is the transparent film that covers water-soluble dishwasher tablets? These dishwasher tablets (Finish Powerball Shine) seem to be covered in plastic!

But that can't possibly be true since there's no unwrapping, and they're completely soluble in water.

What is the film made of?

It is made using Polyvinyl Alcohol (Mowiflex material).


Sodium azide to Nitrogen ratio in airbags I've been given this question and I don't know how to solve it.

Airbags are designed to protect the occupants of a car during a collision. They use the decomposition of sodium azide, $\ce{NaN3}$. Three chemical reactions occur rapidly after impact and produce nitrogen gas ($\ce{N2}$) to fill the airbag.
The equations summarise the overall process that takes place.

$\ce{2 NaN3 -> 2 Na + 3 N_2}$
$\ce{10 Na + 2 KNO3 -> K2O + 5 Na2O + N2}$
$\ce{K2O + Na2O + 2 SiO2 -> K2O3Si + Na2O3Si}$

What is the ratio between the number of sodium azide ($\ce{NaN3}$) molecules used and the number of nitrogen ($\ce{N2}$) molecules produced in the overall process?

I initially just used the first equation to get a ratio of 2:3, but obviously that's not right. I have no clue where to start. How do you do this? (How this can be considered a Year 10 question baffles me)

Assuming you have an excess of $\ce{KNO3}$, the ratio of $\ce{NaN3}$ consumed to $\ce{N2}$ produced would be 5:8.

Let's renormalize the first two equations, dividing the first through by 2, and the second through by 10.  By doing this we can see, directly, that in the overall reaction, 3/2 + 1/10 = 8/5 = 1.6 molecules of $\ce{N_2}$ are produced for every molecule of  $\ce{NaN3}$ consumed.  If you prefer an integer ratio, it's 1:(8/5) = 5:8.

  $\ce{NaN3 -> Na + 3/2 N_2}$
  $\ce{Na + 1/5 KNO3 -> 1/10K2O + 1/2 Na2O + 1/10 N2}$

The initial ratio 2:3 is correct. $\ce{2NaN3 = 2 Na}$ molecules as well as $\ce{6 N}$ molecules. The products of the reaction, $\ce{2Na}$ is equal to two $\ce{Na}$ molecules and $\ce{3N2}$ is equal to $\ce{6 N}$ molecules. So as you can see the equation is perfectly balanced. Hope this helps.


Entropy change in an adiabatic expansion Question
  One mole of ideal gas initially at a pressure of 1 atmosphere and $T = 298\ \mathrm K$, is
  expanded into a volume 50 % larger adiabatically.

In this adiabatic process, no work is done on the surroundings and $\mathrm dQ=0$. And entropy is defined as $\mathrm dS=\frac{\mathrm dQ_\text{rev}}{\mathrm dT}$. However, there must be a change in entropy surely?

Is the definition $\mathrm dS=\frac{\mathrm dQ_\text{rev}}{\mathrm dT}$ only at a constant volume? Because I have also seen entropy defined as $\mathrm dS=\frac{C_V}{T}\,\mathrm dT+\frac{R}{V}\,\mathrm dV$ and there would be a change in both temperature and volume. However, how do I find what these changes are? Would it make a difference if the expansion was reversible or irreversible (the question didn't make it explicitly clear)

"One mole of ideal gas initially at a pressure of 1 atmosphere and T = 298 K, is expanded into a volume 50% larger adiabatically."

The question doesn't have sufficient information for a solution, because we don't know if the adiabatic expansion is reversible or irreversible.
If reversible:
If the expansion is done reversibly, then we know entropy of the universe can't be changing (because reversible), and that heat flow from the gas to the surroundings is zero (because adiabatic).  Without heat flow, the only way the entropy of the universe can stay constant is if both the entropy of the universe and of the gas stay constant.  In this case you can apply both of your equations.  But keep in mind that your equation for $dS$ is only true for ideal gases.
$dS = \frac{\delta q_{rev}}{dT} = 0$
$dS = 0 = \frac{C_V}{T}dT + \frac{R}{V}dV$
$\frac{dT}{T}=(-\frac{R}{C_V})\frac{dV}{V}$
This is now a differential equation you can solve for temperature if you know volume.
As the other answers have indicated, a reversible expansion will involve the gas doing work on the surroundings.  One way to think about why this is so is that for the process to be reversible, the potential of the gas to do work must not be lost, and so if this potential for doing work doesn't remain in the gas, it must go to the surroundings.  A loss in the potential to do work (of the universe) is an increase in entropy.  And in a reversible process that can't happen.
Think of gas in a piston, where the piston is held down by a pile of sand.  The reversible expansion involves slowly removing sand, grain by grain, from the piston.  As the grains are removed, the piston rises and the gas lifts the remaining sand, doing work against gravity.
If irreversible:
...then we still don't have enough information because there are many ways the process can be irreversible.  But one particular, extreme case of irreversibility is if the adiabatic expansion occurs without transfer of any energy, then we can solve the problem.  In this extreme scenario, neither heat nor work are not transferred.  You can thus infer that $dU=0$ for the gas, so
$dU=-TdS + p dV=0$
$TdS = p dV$
Using your definition for $dS$ for an ideal gas,
$T \frac{C_V}{T}dT + T \frac{R}{V}dV = p dV$
Keep in mind that $\frac{RT}{V}$ for an ideal gas is $p$, so you should get that
$C_V dT = 0$
Which is also a differential equation that you can integrate to get the temperature change.  (The obvious solution is that the temperature does not change in this case.)
In this second case, now that we know that $T$ is constant, we know $dT$ is zero, and thus using your equation for $dS$, we get that $dS = \frac{R}{V}dV > 0$, i.e., the entropy of the gas increases, which makes sense because the process was irreversible.  The potential to do work has been lost.
Think of gas confined to one half of a cylinder, and the other half of the cylinder is empty (i.e. vacuum).  If the divider between the filled and unfilled halves is suddenly removed, that would be a case of adiabatic, isoergic expansion.

An important thing to know is that on the adiabatic processes, the system is isolated, due to that no heat can be given or taken, but that doesn't mean that the volume can't change. Because we can't take heat, we can lower/rise the temperature from T1 to T2 . We don't have any work done only when the volume is constant. Entropy will be at a constant volume or pressure only when you see the index v ** or **p in the formula. If the process would be reversible ( which i think it is) only the change of temperature should be in infinitezimal amounts. Hope I understood the question right


Is there a commercial liquid with refractive index greater than n=2? For an application I need a commercial, transparent liquid with refractive index $n>2$.

Is this kind of liquid available?
What's the chemical composition of this?

On the web, only $n=1.81$ is routine.

Definition of refractive index:

$n=\frac {c}{v}$

If n>2 then this means that the velocity of light in the material is less than half than in vacuum so there must be a strong interaction with the material.

 One of the highest refractory index in liquids is 1.79(20°C) possessed by a solution of sulfur in methylene iodide($CH_2I_2$). Liquids are less dense than solids and this explains why it's difficult to have a refractory index greater than 2.
I found a nature article that says:

  Selenium monobromide, Se2Br2, has a higher
  refractive index than that of any pure liquid hitherto
  recorded. Prepared by direct combination, the value
  for $\nu$ is 1.96±0.01 rising to 2.02 on exposure to the
  atmosphere, owing to decomposition of the bromide,
  with separation and reabsorption of selenium.

https://www.nature.com/articles/133066b0

EDIT: From a more recent paper(2016), selenium monobromide is confirmed as one of the liquid with the highest refraction index(>2). However, diiodomethane is more easily accessible commercially.

  A promising candidate solvent is the organic liquid diiodomethane (CH2I2), which is one of the liquids with the highest
  known refractive index values (n = 1.74). While other high
  refractive index liquids exist (phenyldi-iodoarsine (C6H5AsI2)
  with n = 1.85 and selenium monobromide (Se2Br2) with n=
  2.1), diiodomethane has the key advantage of being commercially available. In addition, diiodomethane is an excellent
  solvent, and many liquid formulations using salts dissolved in
  diiodomethane are reported to increase the refractive index
  and are even available commercially.

https://www.osapublishing.org/ao/abstract.cfm?uri=ao-55-12-3165

Sucrose or glucose in H2O = approx 1.4-1.5 depending on composition and concentration

Glycerol = 1.473
Toluene = 1.5
Chlorobenzene = 1.524

Getting higher than that without very high flammability/toxicity/reactivity is hard


Reactivity of chlorobenzene and benzene in electrophilic substitutions As we know, chlorobenzene is a deactivating substituent and benzene has a stable structure. Now I'm confused.

My question is — considering electrophilic aromatic substitution is chlorobenzene or benzene more reactive?

considering electrophilic aromatic substitution is chlorobenzene or benzene more reactive?

Let's expand the question to include all of the halobenzenes.

All of the halogens ($\ce{F, Cl, Br, I}$) are more electronegative than hydrogen.  Therefore they will inductively remove electron density from a benzene ring. However, you can also draw resonance structures for the various halobenzenes where the halogen substituent is donating electron density to the ortho and para positions of the benzene ring through resonance.

Therefore, in the case of the halobenzenes these two effects are working in opposite directions.  In order to determine which effect controls the reaction, we need some data.

The following table compares the relative rates of electrophilic aromatic nitration of the halobenzenes to the rate for benzene itself. If we assign benzene a relative rate of 1, we see that all of the halobenzenes react slower than benzene itself. This tells us that the inductive effect is stronger than the resonance effect in the halobenzene series.
\begin{array} \hline
 \text{Relative rates of aromatic electrophilic nitration} \\ \hline
\end{array}
\begin{array}{|c|c|c|c|} \hline
\ce{Ar-X} & \text{Relative rate}  \\
\hline
\ce{Ar-H} & 1.0 \\ \hline
\ce{Ar-F} & 0.11 \\ \hline
\ce{Ar-Cl} & 0.02 \\ \hline
\ce{Ar-Br} & 0.06  \\ \hline
\ce{Ar-I} & 0.13 \\ \hline
\end{array}

(data source: see p. 26 here)

However, when the halobenzenes do undergo electrophilic aromatic substitution, they react preferentially at the ortho and para positions, because, as explained above, the resonance effect donates electron density to the ortho and para positions.  Therefore, these two positions are the least deactivated and react preferentially.

Trying to make it easy: electrophile means electron loving so in haloarenes the halogen group pulls electron from benzene. But electrophiles need electrons so benzene is more active.


Regioselectivity in Diels–Alder reaction between 2-nitrobuta-1,3-diene and acrylonitrile Since $\ce{-CN}$ shows strong −I effect, there will be δ+ charge on the carbon adjacent to it. Similarly, $\ce{-NO2}$ shows strong −I and –M effects. So, I proposed that π-electrons are being donated to the electron-deficient parts:

But the answer given in my book has $\ce{-CN}$ group at the third position to $\ce{-NO2}$ group. Can someone please tell me where I have got my concepts wrong?

You pointed out correctly that $-CN$ has a great $-I$ effect but it won't bring a $+\delta$ on adjacent carbon atom, reason being it's instability. Apply electromeric effects on propanenitrile.

If you need help then just comment below :-)


Synthesis Golf X: TAK-457 I shall try to revive this!
A full FAQ post has been written on chemistry.meta.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

TAK-457 is a drug (more accurately, an injectable prodrug of TAK-456) that displays antifungal activity:[1,2,3]

Some rules, as usual:

At most one chiral centre may be bought in your starting material(s). [Chiral catalysts, ligands, auxiliaries, etc. do not fall under this rule.]
Your synthesis must feature a method for construction of either the triazole or tetrazole ring (i.e. do not buy both).
The counterion doesn't matter.
Bonus points for synthesis that can be adapted to produce the other three stereoisomers of the drug (preferably via late stage diversification as opposed to, say, using a different enantiomer of starting material). These stereoisomers were of interest.[4]

References

Ichikawa, T.; Kitazaki, T.; Matsushita, Y.; Yamada, M.; Hayashi, R.; Yamaguchi, M.; Kiyota, Y.; Okonogi, K.; Itoh, K. Optically Active Antifungal Azoles. XII. Synthesis and Antifungal Activity of the Water-Soluble Prodrugs of 1-[(1R,2R)-2-(2,4-Difluorophenyl)-2-hydroxy-1-methyl-3-(1H-1,2,4-triazol-1-yl)propyl]-3-[4-(1H-1-tetrazolyl)phenyl]-2-imidazolidinone. Chem. Pharm. Bull. 2001, 49 (9), 1102–1109. DOI: 10.1248/cpb.49.1102.
Hayashi, R.; Kitamoto, N.; Iizawa, Y.; Ichikawa, T.; Itoh, K.; Kitazaki, T.; Okonogi, K. Efficacy of TAK-457, a novel intravenous triazole, against invasive pulmonary aspergillosis in neutropenic mice. Antimicrob. Agents Chemother. 2002, 46 (2), 283–287. DOI: 10.1128/AAC.46.2.283-287.2002.
Tsuchimori, N.; Hayashi, R.; Kitamoto, N.; Asai, K.; Kitazaki, T.; Iizawa, Y.; Itoh, K.; Okonogi, K. In Vitro and In Vivo Antifungal Activities of TAK-456, a Novel Oral Triazole with a Broad Antifungal Spectrum. Antimicrob. Agents Chemother. 2002, 46 (5), 1388-1393. DOI: 10.1128/AAC.46.5.1388-1393.2002.
Ichikawa, T.; Yamada, M.; Yamaguchi, M.; Kitazaki, T.; Matsushita, Y.; Higashikawa, K.; Itoh, K. Optically Active Antifungal Azoles. XIII. Synthesis of Stereoisomers and Metabolites of 1-[(1R,2R)-2-(2,4-difluorophenyl)-2-hydroxy-1-methyl-3-(1H-1,2,4-triazol-1-yl)propyl]-3-[4-(1H-1-tetrazolyl)phenyl]-2-imidazolidinone (TAK-456). Chem. Pharm. Bull. 2001, 49 (9), 1110–1119. DOI: 10.1248/cpb.49.1110.

Fragment #1 Preparation

1) An exact preparation of an aryl imidazolidinone (54% Yield)1
2) A modified preparation of a tetrazole from an aniline derivative (90% yield with pure aniline)2
3) A modified acetal formation (protocol modifications based on urea) (no reported yield)3

Fragment #2 Preparation

1) An exact preparation of an alpha-bromophenone (Hell-Volhard-Zelinski Reaction) (97% yield --likely prior to purification)4
2) Modified coupling of 1,2,4-triazole and alpha-bromophenone (100% yield with chloroacetone -- likely prior to purification)5
3) Wittig reaction modification (73% yield with phenyl methyl ketone)6
4) A modified triazolium formation (82% yield with methyl iodide)7
5) Stereoselective Jacobsen epoxidation (92% yield with 1-phenylcyclohexene)8

Final Fragment Coupling

1) Deprotonation of the amine with LDA
2) Epoxide ring opening followed by the protonation of the alkoxide and hydrolysis of the acetal to yield the final product

Stereochemical and General Considerations

The use of the Jacobsen epoxide formation allows for the use of the (L, L) Jacobsen catalyst in the final steps of the synthesis to yield the enantiomer of TAK-457. Synthesis of the other stereoisomer would require modification of earlier steps in the reaction.

Longest Linear Sequence: 5 steps
Total Number of Steps: 9 steps (I consider the final two steps as one since they are performed without subsequent purification)

Comment on 1,2,4-Triazole Ring: Yes, I did not prepare the triazole ring directly; however, this seems to be the more efficient route.

Comment on Wittig Reaction: This is the step I am probably the most worried about. The steric bulk from the phenyl group may be enough to slightly favor the desired enantiomer; however, this cannot be determined until further experimentation is performed.

References

Stabile, P.; Lamonica, A.; Ribecai, A.; Castoldi, D.; Guercio, G.; Curcuruto, O. Mild, Convenient and Versatile Cu-mediated synthesis of N-Aryl-2-imidazolidinones. Tetrahedron Lett. (2010) 51 (24). 3232–3235. DOI:10.1016/j.tetlet.2010.04.064
Kundu, D.; Majee, A.; Hajra, A. Indium Triflate-Catalyzed One-Pot Synthesis of 1-Substituted-1H-1,2,3,4-Tetrazoles Under Solvent-Free Conditions. Tetrahedron Lett. (2009) 50 (22). 2668–2670. DOI:10.1016/j.tetlet.2009.03.131
Shuai, F.; Wang, X.; Zhang, J. Preparation of Dipyridamole Active Pharmaceutical Ingredient. CNIPA Patent 108069972, Nov 16, 2016
Alonso-Alija, Cristina; et al. Preparation of Amino(monocyclic aroyl)pyridinones that Inhibit p38 Map Kinase for use as Antiinflammatory Agents. WIPO Patent 2003076405, Sep. 18, 2003
Sun, Y.; Huang, N.-Y.; Ding, M.-W. Efficient Synthesis of 6-(1H-1,2,4-Triazol-1-yl)-thieno[2,3-d]pyrimidin-4(3H)-ones via an Iminophosphorane. Synth. Commun. (2010) 40 (13). 1985–1991. DOI: 10.1080/00397910903219351
Pratsch, G.; Overman, L. E. Synthesis of 2,5-Diaryl-1,5-dienes from Allylic Bromides Using Visible-Light Photoredox Catalysis. J. Org. Chem. (2015) 80 (22). 11388–11397. DOI:10.1021/acs.joc.5b01962
Myles, L.; Gathergood, N.; Connon, S. J. The Catalytic Versatility of Low Toxicity Dialkyltriazolium Salts: in Situ Modification Facilitates Diametrically Opposed Catalysis Modes in One Pot. Chem. Commun. (2013) 49 (46). 5316. DOI:10.1039/c3cc41588k
Brandes, B. D.; Jacobsen, E. N. Highly Enantioselective, Catalytic Epoxidation of Trisubstituted Olefins. J. Org. Chem. (1994) 59 (16). 4378–4380. DOI:10.1021/jo00095a009


Why is the rate of reaction of polar EAS favored by inductive effects and the rate of less polar EAS reactions through hyperconjugation? 1. Reactions involving strong electrophiles (nitration) are
       said to be governed
       by
       Inductive effects thus the order of nitration is
                                       D>C>B>A
  2.Chlorination is a less polar reaction thus chlorination said the be governed by hyper conjugation.
                                        A>B>C>D

My question is,  why?

The factors controlling EAS in both the cases will be  the same for both the substitutes.
Hyperconjugation and inductive effects will both operate without distinction in both the cases.

So why is it that
The rate of reaction of polar EAS favored by inductive effects and the rate of less polar EAS reactions through hyperconjugation.

SOURCE   A guidebook to mechanism Peter Sykes

EDIT
I have been asking around for an answer to this question.

No one has been able to answer it conclusively , the primary school of thought   is that

  In Highly polar reactions , inductive effects dominate as more electron density is available , hyperconjugation on the other hand is just redistribution of electrons.

First of all is this correct ?   Even if it's correct but how does it explain the opposing rate orders for nitration and bromination?

I may be oversimplifying things.

A look at electron density

Let's pick up the extreme cases of the substituted benzenes . $$\ce{A -> ArCH3}$$
$$\ce{D -> Ar(CH3)_3}$$

Case A shows the maximum hyperconjugation. Case D on the other has the maximum +I effect.

For a moment let's imagine that the electrons are coins.
Case D , has more coins, more methyl groups (let's say 3).
Case A, on the other hand has 1 coin , hyperconjugation would simply mean that I am  simply moving that coin in circles. It's simply redistribution of electrons.it doesn't actually increase the density.

So the first point which I'm going to use to answer the question is that -

Inductive effects increases  the electron density, hyperconjugation simply redistributes the electron density.

A look at Coulomb's law

$$\ce{ F \propto Q_1*Q_2}$$

More polar reactants will have a greater charge on them as compared with less polar reactants.
Since inductive effects concentrate  electron density at one area the negative charge concentration will be greater.

HOMO/LUMO

Unoccupied orbitals are high in energy and occupied orbitals are low in energy. For the best overlap the energy gap between the HOMO and LUMO must be small.

Rate Order for more polar reactions

In more polar reactions the electrostatic forces of attractions play the primary role as the polar reactants is highly charged. As established earlier ,inductive effects lead to a electron density thus a greater pull. Homo/Lumo interactions are secondary as the overlapping happens after a collision occurs.

The electrostatic forces are pulling the benzene and reactant. That would by itself increases the number of collisions.
A greater electron density leads to the better stabilization of the +tive charge on the reactant.

inductive effects support more polar reactions

Rate order for less polar reactions

The electrostatic forces of attraction are lower in the case as as the reactant is less charged.
So the control of the reaction is primarily through Homo/Lumo interactions.

Unoccupied orbitals of the reactant are high in energy.( They are Unoccupied).
To bridge the HOMO/LUMO gap the Lumo of benzene carbon must be high in energy. Non bonded electrons or the electrons which are delocalized through hyperconjugation are high in energy as they are held by only one atom.
So the HOMO/Lumo gap reduces in the case of hyperconjugation thus hyperconjugation supports less polar reactions.

Chlorine is a weak electrophile, while the nitronium ion is a strong electrophile.
Methyl groups are +I groups and they are electron releasing groups.

The NO2 group will attack a negatively charged carbon attack readily.
The +I effect is permanent, and its magnitude decreases with increase in distance, and therefore, the nitronium ion will attack the carbon nearest from the methyl groups.

The Chlorine will not get attracted to to a negatively charged carbon atom as strongly as NO2 will. In hyperconjugation. the charge is delocalized i.e., it is distributed throughout the structure, in various resonance forms. Hence, the effect of hyperconjugation is weak. The Chlorine atom. therefore, will get attracted to to a weakly charged carbon atom, determined by hyperconjugation.

The hyperconjugation won't attract NO2, as the charge is delocalized, and not one specific carbon atom, or there is no carbon atom that is more charged than the others.


Which is a better nucleophile: hydroxide anion or amide anion? I know that nucleophilicity order depends on what type of solvent we take, polar protic or polar aprotic and also on the basicity of nucleophile.

  In DMF, which is a better nucleophile $\ce{NH2-}$ or $\ce{OH^-?}$

Since the solvent is aprotic polar, it doesn't have any donor hydrogen to form  hydrogen bonds with nucleophile. So, the nucleophilicity should depend on which among them is more basic. Now, since $\ce{N}$ is less electronegative than $\ce{O}$, it's lone pair is more readily available than that of $\ce{OH-}$.

So, according to me, $\ce{NH2-}$ should be more nucleophilic than $\ce{OH-}$, but the answer given is the opposite, that is, $\ce{OH-}$ is more nucleophilic than $\ce{NH2-}$. How is my reasoning incorrect?

And also what are the factors we should keep in mind while deciding nucleophilicity?

I was not able to find data on this nucleophilicity of $\ce {NH2^-}$ and $\ce {OH^-}$ in dimethylformamide, $\ce {Me2NCHO}$. However, I have a plausible explanation as to why the answer may not be what you expected.

Firstly, it is important to remind ourselves that nucleophilicity is always relative to the electrophile that we are concerned with. However, it is also possible to discuss nucleophilicity generally, i.e. not with respect to any particular electrophile. This is the case now.

The fact that the solvent is polar and aprotic suggest that hydrogen bonds cannot be made with the $\ce {N}$ and the $\ce {O}$ of the nucleophiles. However, it does not restrict us to form hydrogen bonds between the lone pair on the carbonyl oxygen with the partially positively-charged $\ce {H}$ atoms on the ions. This is not at all unlikely considering that the strong delocalisation of the lone pair on the $\ce {N}$ in amides, gives a significant resonance structure with this carbonyl oxygen bearing a formal negative charge.

$\ce {NH2^-}$ has two such $\ce {H}$s and $\ce {OH^-}$ has only one such hydrogen. Thus, perhaps, we may still expect greater extent of solvation for the former. Relative to $\ce {OH^-}$, the greater amount of solvation of $\ce {NH2^-}$ increases the activation energy required to initiate the subsitution reaction since solvation can be said to bring down the energy of the reactant in the reaction energy profile. Thus, the rate of substitution with $\ce {OH^-}$ would be faster than $\ce {NH2^-}$ in this polar aprotic solvent.

Alternatively, the answer provided could very well be wrong. That is a possibility that cannot be discounted...

I think that We can not consider the strength of nucleophile based on the electronegativity in this case, but rather than the Hydrogen bonded to Oxygen and Nitrogen.
Positive charge of Hydrogen diminishes the nucleophile strength of Nitrogen or Oxygen via sigma bonding. The more hydrogen bonded with Oxygen or Nitrogen, the more nucleophile decreases.


Is PCl5 really trigonal bipyramidal in shape? We know that the structure of $\ce{CH4}$ is tetrahedral, and not planar tetragonal, as a tetrahedron enables maximum distribution of $\ce{H}$-atoms in space, as they repulse from each other. This is proved by the fact that each $\ce{H-C-H}$ bond angle is equal in magnitude.

In the case of $\ce{PCl5}$, why should the shape be trigonal bipyramidal? Obviously the top and bottom $\ce{Cl}$-atoms, above and below the plane subtend $90^\circ$ with each of the $3$ planar $\ce{Cl}$-atoms; whilst planar ones subtend $120^\circ$ in between each other. This means all bond angles aren't equal. Why so? Shouldn't there be regular distribution of $\ce{Cl}$-atoms in space?

To my opinion, the XY-plane distribution of $\ce{Cl}$-atoms should be such that each $\ce{Cl-P}$ is $72^\circ$ apart from the other $\ce{Cl-P}$ bond (atoms would not be in the same plane at all. But we should see that the orthogonal projection of each on XY-plane subtends equal angles, i.e. $72^\circ$). Same for YZ-plane. Hence there is regular distribution of atoms along all planes. Now it is to be calculated what the ultimate structure results in.

Sorry, I don't have the necessary tools/knowledge to represent my hypothesis. Else could've just modelled this figure and shown a representative model.

The trigonal bipyramidal shape of $\ce{PCl5}$ is indeed the lowest energy conformation.

I have had a very hard time figuring out what your description of the structure meant and I am still not convinced I have actually understood correctly. However I was trying to model it in my head I either arrived at a variety of different angles, or at least two of them were equal to $72^\circ$, or both of these cases.

The only structure $\ce{ML5}$, where all $\ce{L-M-L}$ bond angles are equal is a regular pentagon, i.e. all ligands are in one plane, with $\angle(\ce{L-M-L})=72^\circ$, e.g. $\ce{[XeF5]-}$.

Another possibility for (almost all) equal bond angles is a square pyramid, where $\angle(\ce{L-M-L})=90^\circ$, e.g. $\ce{ClF5}$. Note however, that there are also other angles, so not all bond angles are equal, $\angle(\ce{L-M-L})=180^\circ$.

There is no better solution to arrange 5 point 'uniformly' on a sphere than the trigonal bipyramid. This is also known as the Thompson Problem, which is addressed in the linked question on Mathematics.se, provided by orthocresol: 5 Points uniformly placed on a sphere.
Another point of view would be to ask the question whether there is a way to put 5 points on the surface of the sphere so that they are indistinguishable, which is also answered there. Spoiler warning: there isn't.

You can dive deeper into the matter. For example for $\mathbb{R}^n$ there can only be $n+1$ points equally equidistant to each other. In our perceived reality $n=3$, which makes the tetrahedron the most complex molecule with this property. See on mathematics.se: Proof maximum number of equidistant points for a given dimension (duplicate), Prove that $n+2$ points in $\mathbb{R}^n$ cannot all be at a unit distance from each other (duplicate), $n$ points can be equidistant from each other only in dimensions $\ge n-1$? (Which are all duplicates, but still have answers attached to them.)

And then there are only five Platonic solids, none of which has five vertices.

Furthermore, I would like to add to all of that another word of caution: Molecules are flexible. The trigonal bipyramidal arrangement of ligands just happens to be the low energy conformation for a variety of reasons (considering the necessary and useful approximations, e.g. Born-Oppenheimer approximation).
The molecule (and other analogues) are special though, as the undergo pseudorotation according to the Berry Mechanism (see Wikipedia).

Side note:

  We know that the structure of $\ce{CH4}$ is tetrahedral, and not planar tetragonal, as a tetrahedron enables maximum distribution of $\ce{H}$-atoms in space, as they repulse from each other. This is proved by the fact that each $\ce{H-C-H}$ bond angle is equal in magnitude.

This is only half the truth. It is certainly true that protons repel each other, more importantly though is that electrons repel each other, too. The most important part of which is also built into the very simple VSEPR theory. While this theory has its uses, there is a lot of criticism attached to it and one should be aware of that. I shall stress again at this point that VSEPR theory is unrelated to orbital hybridisation; there are quite a few textbooks which do make this post-hoc rationalisation, which can be misleading and dangerous.

The shape of any molecule may be obtained by applying the VSEPR model (Valence Shell Electron Pair Repulsion). In this theory the angle between the bonds must be as large as possible. With your idea, the angles between P-Cl bonds are equal to 72°. It is not very much. In the bipyramidal model, the angles are 90° and 120°. This is much better than with your geometry.


What is oil smoke point? I understand that the smoke point is the temperature at which oil starts to produce smoke. But how exactly is that smoke produced? I've read that it is related to hydrolysis and oxidation, but I'd like to know a more precise link.

Typically smoke point has to do with the composition and amount of free fatty acids in the fat/oil. Most fats and oils that you see on the market are mostly composed of triacylglycerols, with only small amounts of free fatty acids. The more free fatty acids you have the lower the smoke point. Similarly unrefined oils have a low smoke point. If you have fats/oils that contain high levels of unsaturated fatty acids, then that product will also have a lower smoke point than say a fat that contains significant amounts of saturated fat. In regards to the chemical mechanism, upon heating the triacylglcyerol structure gets broken down into free fatty acids and glycerol via a hydrolysis reaction. The free fatty acids volatilize and cause the visible smoke. Unsaturated fatty acid are more sensitive to heat and will be hydrolyzed quicker than saturated fatty acids. For more info read this article: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4424769/

The smoke point also known as
burning point of an oil or fat is the
temperature at which, under specific
and defined conditions, it begins to
produce a continuous bluish smoke
that becomes clearly visible.


What is the major product of the reaction of 2-methylbuta-1,3-diene with HBr? What is the product of the reaction of 2-methylbuta-1,3-diene (isoprene) with HBr?

What I understand is that carbon (2) will get a positive charge after protonation.
But after this, which should be the major product?

3-bromo-3-methylbut-1-ene – because in this case the carbocation would be stable because of being tertiary, but the double bond would have less number of hyperconjugated structures and will be less stable

1-bromo-3-methylbut-2-ene –  as the carbocation would be less stable because it would be primary but the double bond would be much more stable because of 8 $\alpha$ hydrogen available

The regioselectivity of the overall reaction depends on the temperature.  The carbocation intermediate can pass through the two different transition states  that lead to the 1,2- and 1,4- products, respectively. If we choose a temperature low enough, then the product distribution will reflect the difference in energy between the two activation energies Ea (1,2) and Ea (1,4). So long as the reaction is not reversible, the product with the lower energy transition state will dominate which is decided by the stability of the intermediate, so in that case 3-bromo-3-methylbut-1-ene (product-1) will be more stable product.
At higher temperatures, the reaction has the potential to be reversible, this means that both the 1,2- and 1,4- products can ionize This sets up an equilibrium. The product ratio will now reflect the relative stabilities of the 1,2- and 1,4- products, not the transition states leading to their formation. In the case of 2-methylbuta-1,3-diene, the product 1-bromo-3-methylbut-2-ene (product-2) will be more stable and dominant at higher temperatures.


Ozone resonance structure: differing opinions I've always thought that the two-form resonance structure of ozone was the scientific concensus (from Wikipedia]:

Then today I found two other opinions. A three-form structure:

And a four-form structure:

Is there a consenus? Are the two and three-form representations just simplifications of the four-form structure? Please include references.
EDIT: The second resonance structure on the four-form image above is drawn incorrectly, as pointed out by @ron. The upper oxygen atom should have a lone pair and a positive charge, just like the second structure of the two-form.

Those are all valid resonance structures for ozone (except the second from the left in the "four-form structure" row, that one has two negative charges, maybe it's a typo).  In fact you can draw even more structures like this one, where this is no single bond between the middle and terminal-right oxygens.

All of these structures contribute to the "real" structure of ozone - however, some resonance structures contribute more and some resonance structures contribute less.  Here is a link to a nice set of rules (see page 2, the "Resonance Contributor Preference Rule" section) that describes how to weight the contribution of various resonance structures.  The most important rule is that each atom in the resonance structure should have an octet of electrons.  Only the structures labeled I and II in your "three-form structure" row meet this requirement, so they are the two most important resonance structures for ozone - but all of the others do contribute, just to a lesser extent.

Covalent bonds suppose to help an atom achieve filled valence shell by sharing electrons. Resonance structures need to have maximum atoms adhering to the octet rule. Look at III, the oxygen atoms on the right and left have only 7 electrons in their valence shall even AFTER forming covalent bonds. That is why it is not one of the resonance structures.


Mechanism for Amino-Imidazole Ring Fission A reaction I've tried several times in the past is the hydrogenation of nitroimidazole derivative shown below. The amino product is extremely unstable and rapidly decomposes.

This instability is noted in may pieces of literature and the explanation tends to only go far as saying 'ring fission' occurs.

Question: What is the mechanism of this ring fission? To be clear, I am not interested in getting this reaction to work.... It is no longer a target of mine.

Perhaps the entire mechanism is not determinable, but I presume the amine of one molecule attacks another molecule's imidazole ring? Where would this be? How is fission occuring?

EDIT: Possibly something to do with the apparently favoured tautomerism between these two forms:

http://imgur.com/tZMLWkG

I'm afraid the answer to how this ring fission occurs is... it doesn't, you just made some mistake.

Here and here the authors used the same reaction as you and got the product in good yields. I think you might not realized that the product is a very hydrophilic amine (or salt, depending on workup). If you did an extraction of your reaction mixture then the product surely ended up in the aqueous phase.

One explanation could be that the imadazole is hydrogenated as well, leading to a decomposition of the starting material to phenylamine, ethylamine and ethane:


Nucleophilic addition of grignard reagent Here why isn't the phenyl group attacking the carbonyl carbon? Isn't the first step in addition of grignard reagent the attack by the carbanion on the carbonyl carbon ?

The attacking of the Grignard reagent in these kinds of $\alpha$-$\beta$ unsaturated carbonyl compounds depends on two factors mainly. $1$) Hardness of the acid centre $2$) Steric repulsion between the surrounding groups and incoming nucleophile.When the attacking nucleophiles are Grignard reagents,as per your equation, they are generally hard bases.According to Pearson's HSAB Theory, Hard bases prefer to bind to the Hard acid centre and soft bases prefer to bind to the soft acid centre.  In your compound i.e in $\alpha$-$\beta$ unsaturated carbonyl compounds, the carbonyl carbon centre is the Harder acid centre, whereas the $\ce{C=C}$ double bond centre is the softer acid centre.So, Grignard reagents try to first bind with the carbonyl carbons unless there is no other destabilising factor.The second governing factor is the steric crowding, which can sometimes make the attacking nucleophile attach to the least sterically crowded region even though it is violating HSAB theory. So, the addition compound actually depends on these two competititive factors.  Now, in your given compound the attacking nucleophile is $\ce{PhMgBr}$.It is undoubtedly Hard Base. But if it were to bind to the carbonyl carbon i.e. $\ce{C=O}$ double bond, there will be huge repulsion between the lone pair of oxygen and $\pi$-electron cloud of Phenyl ring which actually makes the previously stated second factor to dominate. So, due to this huge steric repulsion, The attacking nucleophile has to attack the less sterically crowded(Note: The repulsion is also there at the double bonded carbon centre , but repulsion between methyl groups and phenyl rings are only due to inefficient spatial orientation which is definitely smaller than inter electronic strong columbic repulsion forces in close vicinity in the case of $\ce{C=O}$) softer basic $\ce{C=C}$ centre which infact gives the product.


Why does this compound have two benzene rings? If we look at carbons 1 and 2, they have 5 electrons. 3 are directly involved in bonding, and 2 are involved in delocalisation to create the 2 benzene rings. This would mean that they had 5 valence electrons, which isn't possible, and thus there should be only one benzene ring.

Chemists are already accustomed to this type of formula, even though it is not absolutely correct. More exact would be drawing all bonds with double solid/dashed bonds to express that the bond order is between 1 and 2, approximately 1.5, but is almost never done

Rarely can be seen formula with the two circles merged (and maybe dashed also), like

But the electrons are in fact delocalized across all bonds of the polycycle structure in "electron cloud" that looks somewhat like the infinity sign ∞ in this case (naphthalene) above and below the planar molecule. What matters is that the count of π (those from the "second bond" of the double bond) electrons is 4n+2 for n is some natural number (here it is n=2). It is a special feature called aromaticity, which has something to do with advanced topic of quantum chemistry; and it greatly stabilizes the molecule. Any of these formulas is unable to describe reality.

I find this representation to be kind of sloppy for the reason you mentioned. The $\pi$ system is delocalized throughout the molecule, but that doesn't mean it's two aromatic systems.

I find this representation much better:


The effect of hyperconjugation on the stability of alkenes with MO theory Hyperconjugation stabilizes carbocations and that makes sense because electrons are given to the empty p orbital.

But how does it stabilize alkenes? Can molecular orbital theory be used to explain it?

Since you are familiar with how carbocations are stabilized via hyperconjugation I will keep this brief: With alkenes, it really is the same principle at work. Instead of the empty p orbital, you have to consider the empty $\pi^{*}$ orbitals of the $\ce{C=C}$ double bond. They interact with the filled high-lying neighboring $\ce{C-H}$ or $\ce{C-C}$ $\ce{\sigma}$ bonds in $\pi$-bond-like manner

and so from those two fragment orbitals you will get two molecular orbitals: one from the in-phase combination of $\pi^{*} (\ce{C=C})$ with $\sigma (\ce{C-X})$, which will be lower in energy then the fragment orbitals, and one from the out-of-phase combination of $\pi^{*} (\ce{C=C})$ with $\sigma (\ce{C-X})$, which will be higher in energy then the fragment orbitals. Since in this interaction there are only two electrons involved, i.e. the ones in $\sigma (\ce{C-X})$, you get a stabilization from the interaction since those two electrons will now occupy the in-phase MO and have thus lowered their energy. The out-of-phase MO is empty and doesn't contribute to the energy of the molecule.

When the sp2 hybridized Carbon atom directly attached to sp3 hybridized Carbon atom having at least 1 hydrogen atom(called α hydrogen), this condition is known as hyperconjugation.
Stability of any hyperconjugative hydrocarbon depends upon no.  of α hydrogen present in compound..

More the no. of α hydrogen, more is the stability of compound


Work done in expanding a gas reversibly and irreversibly So, my chemistry teacher gave the class following $P_{external}$ versus $Volume$ diagrams for reversible and irreversible expansion of a gas which are as follows.

(Reversible expansion)

(Irreversible expansion)
Then he gave us following results to memorize:
$$\mid \left(W_{reversible}\right)_{expansion}\mid \gt\mid \left(W_{irreversible}\right)_{expansion}\mid$$
Where W is for work.
According to him, these results are always true.
But consider the following graph

If I'm not wrong, this can also be considered an irreversible expansion.
But as we can see clearly,
$$\mid \left(W_{reversible}\right)_{expansion}\mid \lt\mid \left(W_{irreversible}\right)_{expansion}\mid$$
This thought is really bothering me. Where did I go wrong?

I will call your processes (1), (2), and (3). I assume (1) is reversible isothermal expansion, (2) is irreversible isothermal expansion, and (3) is unspecified.
The key idea is that the inequality $w_\text{rev} \geq w_\text{irrev}$ can only be applied to reversible and irreversible analogues of the same process. (1) and (2) satisfy this requirement because (1) is the reversible analogue of (2), but (1) is not the reversible analogue of (3).
Physically, (1) is like having a pile of sand on top of a container (containing an ideal gas) whose lid is a piston, and then slowly blowing away the sand. (2) is the same scenario, except you sweep off the sand in one grand gesture. But (3) is a different scenario entirely.
If we refer to the first law, $\Delta U = q + w$, it's clear that, given an initial and a final state, we can make $w$ as large as we want as long as we have a commensurate value of $q$. Just because processes (1), (2) and (3) all have the same initial and final states does not mean that it makes sense to apply the inequality $w_\text{rev} \geq w_\text{irrev}$ to any two of them.

You are misunderstanding how pressure is changing.
Work is given by $$\delta W = -p_{ext}dV$$
Work is performed when the external pressure on the system is smaller than the internal pressure of your gas.
$p(A)$ and $p(B)$ are the final and initial external pressures of your system. If you are expanding isobarically (as in your second image), $$W = -p(B) \Delta V$$
in the reversible and non-reversible case. Therefore, work done in an isobaric expansion is the same for reversible and non-reversible process. It may happen than once the pressure drops, the system might expand again, but we are not interested in that regime right now. No work will be done in an infinitesimal amount of time after a sudden instantaneous drop in pressure.
If the pressure change is adiabatic from $p(B)$ to $p(A)$, you will have the curve in image 1.
If the pressure change is irreversible i.e. you have a sudden change in pressure (in image 2, the pressure drops suddenly, not in a smooth, differentiable manner), you can apply
$$W_{irrev} = -\int _{V_A} ^{V_B} p_{ext} dV = p(A)\Delta V$$
If it were reversible and adiabatic,
$$W_{rev} = -\int p_{ext}dV = -nRT\ln \left( \frac{V(B)}{V(A)}\right)$$
assuming an ideal gas. You can check that
$$W_{rev} \geq W_{irrev}$$
Note the $\geq$ rather than $>$ in the above expression. That was a slight error on the part of your teacher.


Metallic and magnetic nature of transition metal oxides Which of the following oxides is both metallic and ferromagnetic:
  $\ce{VO2}$, $\ce{CrO2}$, $\ce{MnO2}$, $\ce{TiO2}$?

This is how I went about solving it:

Since ferromagnetic materials also show paramagnetism, so the condition for paramagnetism must be checked.

In $\ce{TiO2}$, $\ce{Ti}$ has no unpaired electrons, it cannot be paramagnetic and as a consequence it cannot be ferromagnetic.

$\ce{CrO2}$ is amphoteric. A metallic oxide would show basic properties in most cases. So $\ce{CrO2}$ may not be the right answer.

$\ce{VO2}$ is basic. All three except $\ce{TiO2}$ are paramagnetic.

Then I googled, I didn't get an exact answer. But I read that $\ce{CrO2}$ is termed as a "half metal" which is related to its magnetic properties, most part of which I didn't understand.

Then I also found articles about half metallic ferromagnetism of $\ce{TiO2}$. But that goes against my assumption.

So my questions are:

  Explain where I'm wrong.
  Why $\ce{CrO2}$ is the answer? (Found that when I finally referred to the solutions of the question paper. However the solutions given
  can have mistakes. So it is possible that $\ce{CrO2}$ may not be the
  true answer).
  If you think $\ce{CrO2}$ is not the answer, then why?

This question should be cross-posted to Physics.SE, it is not a simple chemistry question. I'll give short rationalization of the fact, but beware: it is a rationalisation, not a deduction. It is also given in layman's terms.

We know the crystal structure of $\ce{TiO2}$, it is stereotypical and quite common. It contains octahedrons $\ce{TiO6}$ with oxygens shared between three octahedrons. Since the four compounds have similar composition and the cations have similar size, we may expect them to have similar structures. This is, however, a mere guess, but lets' go with it.

For a compound to have metallic conductivity, it must have mobile electrons, i.e. partially occupied group of orbitals. Thus, $\ce{TiO2}$ is obviously out.

It is known, that $d$-orbitals of a metal in octahedral neighbourhood split: three are lower in energy then two other.

In $\ce{MnO2}$ there are three electrons, so all three lower orbitals are occupied. This makes electrons immobile: for an electron to move to a neighboring atom, it has to pair with one of the already present electrons, which requires energy. I do not expect, however, that $\ce{MnO2}$ shall be a great insulator: the pairing energy is quite low, and furthermore, any impurity shall create charge carriers. So, $\ce{MnO2}$ is probably a semiconductor.

To choose from remaining two, we should recall that ferromagnetism is common among compounds with a lot of half-filled d-orbitals. This suggests that $\ce{CrO2}$ has higher chances, than $\ce{VO2}$ Still, the latter should be paramagnetic and conducting under right conditions.

Again, you should cross-post the question on Physics.SE and probably read a good solid state physiscs book.

$\ce{TiO2}$ is diamagnetic. $\ce{VO2}$ has only 1 unpaired electron, definitely not ferromagnetic. But I am still confused between $\ce{MnO2}$ and $\ce{CrO2}$. Both are ferromagnetic. Metallic character is decided by the strength of the Metal-Oxygen bond. In case the metal is in a very high O.S. (e.g. $\ce{MnO4-}$, $\ce{Cr2O7^2-}$) the M-O bond is more covalent and cannot be ruptured by water to give a base.

In this case however, both $\ce{MnO2}$ and $\ce{CrO2}$ seem appropriate.