text:
What is the lewis structure of NO
I am a bit confused about the lewis structure of $\ce{NO}$ The answer from my worksheet is $\ce{N=O}$ with nitrogen having a free radical. But I drew the structure as N triple bonded to $\ce{O (:N#O:)}$ and oxygen has the free radical. I am confused as to why my answer is not the case

You'd get  more then 8 electrons on N.

It's always a good idea to check formal charges. In N=O, the formal charge is zero on each atom. In N triple bonded to O with a single electron on N, the formal charge is 1 and -1 on the N and O respectively. Thus, N double bonded to O is correct.

related: Where does the 9th electron go in a N=O bond?, Is Bond Formation “Strictly” Exothermic?

As an extension of Mithoron's comment, if you drew the structure as N triple bonded to O, you would fulfil both the octets without placing the free radical electron.
Now, you'd have to extend the octet of either nitrogen or oxygen, but you have no free orbitals to place the extra electron.
Hence, the first one is the correct Lewis structure.


text:
What old chemistry textbooks are good until present day?
I have found old chemistry textbooks like Walter Moore's Physical Chemistry and Morrison&Boyd's Organic Chemistry to be really good source of information, as good as actual reference textbooks.
What other books do you recommend, regardless the area, that still useful nowadays despite its age?
The reason I ask this is because the language used by these authors seems to be different from today, and I find it to be a pleasant reading.

If the books were written in the 1960's then they should be ok; the basic science hasn't  changed so these books should still be good and topics such as chemical bonding in transition metals were well developed by then.  Buy clearly topics such as lasers,  atomic force/scanning tunnelling microscopes and lots about DNA etc. had not been developed or discovered, and probably also several synthetic methods. NMR/FTIR & optical spectroscopy/mass spectrometry  methods have greatly developed since then also.

Charles E. Mortimer's Chemistry book was very good for our General Chemistry I & II courses. It has detailed Molecular Orbital diagrams, coordination chemistry information, sufficient examples of electrochemical cells and also a good text of nuclear chemistry. It does not have too many colorful images and pages so it is easier to focus on the knowledge.
In general, old books are only for general chemistry and history of chemistry research etc. When one starts to delve into things, it is dangerous not to use the most up-to-date textbook.

General Chemistry by Linus Pauling, there are several additions. It is better than a couple modern college texts that I have seen.

Additionally, many other Dover Books are still useful, at least for certain things. Modern Quantum Chemistry is seminal. Symmetry and Spectroscopy is also quite good.

source: Stack Exchange

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

sort: votes


text:
Is there a mathematical way to determine the bond angles in a molecule without using the standard VSEPR model?
I was wondering if anyone knew of a way to mathematically determine the bond angles in a molecule without using the standard VSEPR model. I was curious if such a thing exists.

It doesn't.$;$

You would have to do a complicated molecular orbital calculation using software such as Gaussian. That is of course, 'mathematical' solving the Schroedinger equation but probably not in the sense you are thinking of. For small molecules the Walsh rules give an indication of what bond angles should be.

Alright thanks a lot, I will definitely check that out, I'm a chem/math major and would rather use something more calculation based than Lewis Structures.

That may not be right. Being good at chemistry (as with other sciences) frequently involves knowing when a very simple model will give you essentially the answer you need without the using a more complicated model.

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


text:
Could a strong acid become a strong base?
Is it technically possible for a strong acid to change into a strong base, through a chemical reaction?
Example: if an acid with pH 1 became a base with pH 12.

Take nitric acid for instance. Reduce the nitrate until ammonia forms. Then using a stronger base than amide ion, you can generate amide base.
Therefore, it is required to make a number of reactions to get a strong base from a strong acid. If you meant that could a strong acid be changed to a strong base by one step reaction, it might be possible with complex big organic or organoboron molecules by replacement of acid functional group with a basic one, but I am unsure about this part.


text:
Obtaining values for electrical mobility from ion transport numbers in a Hittorf cell
I am following the experiments detailed in the links below in order to find the electrical mobility of a variety of ionic species:

Wikipedia article on ion transport numbers and Hittorf method
PHYWE series of publications - "Transport numbers" PDF
"Determination of Transference Number by Hittorf’s Method" - PDF

I know that the ion transport number of an ionic species is the ratio of the current carried by that particular species over the total current conducted through the Hittorf cell. The change in concentration of the electrolytes, the mass of the electrodes, the voltage and current of the cell can be used to calculate this number. But, using these results, how do I calculate the drift velocity of those ions and hence values for their electrical mobility?
I tried looking at the Einstein-Smoluchowski relation but the concepts are a bit over my head right now. Any other tips, notes and comments on this experiment would also be appreciated.

You can't directly derive electrical mobilities from ions from their transport numbers; transport numbers are a relative measurement of charge transfer that, precisely, reduces absolute mobilities for different ions to contributing fractions. Absolute mobilities are lost when we use transport numbers, so if all we know about an electrochemical system is transport numbers, we cannot recover ionic mobilities.
However, we can recover drift velocities if we have additional information on the system - specifically, total current $I_\mathrm{total}$ and distance between plates in the cell $d$. Alternatively, if we're interested in ion mobilities, we will also need to know voltage $V$.
Let's look at a simple system with only one electrolyte involved in the electrochemical reaction. Total current will be the sum of the contributions of the anion and the cation:
$$ I_\mathrm{total} = I_+ + I_- =  \cfrac{n_+ z_+ \nu_+ e}{d} +  \cfrac{n_- z_- \nu_- e}{d} $$
where $n_i$ is the number of ions $i$, $z_i$ is the charge of ion $i$, $\nu_i$ is the drift velocity of ion $i$, $e$ is the electron charge and $d$ is the distance between the plates. Per the requirement of electroneutrality,
$$ n_+ z_+ = n_- z_- $$
so
$$ I_\mathrm{total} = \cfrac{ n_+ z_+ e \left( \nu_+ + \nu_- \right) }{d} = \cfrac{ n_- z_- e \left( \nu_+ + \nu_- \right) }{d} \tag{1}$$
We can calculate transport numbers as
$$ t_+ = \cfrac{I_+}{I_\mathrm{total}} = \cfrac{n_+ z_+ \nu_+ e /d}{n_+ z_+ \left( \nu_+ + \nu_- \right) e / d} = \cfrac{\nu_+}{\nu_+ + \nu_-} $$
$$ t_- = \cfrac{I_-}{I_\mathrm{total}} = \cfrac{n_+ z_+ \nu_- e /d}{n_+ z_+ \left( \nu_+ + \nu_- \right) e / d} = \cfrac{\nu_-}{\nu_+ + \nu_-} $$
Note that we could use the definition of transport numbers to obtain a specific drift velocity, $\nu_i$, from the total drift velocity, $\nu_+ + \nu_-$:
$$ \nu_+ = t_+ \left( \nu_+ + \nu_- \right) $$
$$ \nu_- = t_- \left( \nu_+ + \nu_- \right) $$
Let's go back to $\mathrm{(1)}$ to derive an expression for total drift velocity, $\nu_+ + \nu_-$:
$$ \nu_+ + \nu_- = \cfrac{d}{n_+ z_+ e} I_\mathrm{total} = \cfrac{d}{n_- z_- e} I_\mathrm{total} $$
We can therefore derive drift velocity for an ion, $\nu_i$, from its transport number, $t_i$:
$$ \nu_+ = \cfrac{ t_+ d }{ n_+ z_+ e } I_\mathrm{total} $$
$$ \nu_- = \cfrac{ t_- d }{ n_- z_- e } I_\mathrm{total} $$
Note that it is necessary to know current $I_\mathrm{total}$ and cell size $d$ since precisely these values introduce the time and distance dependence necessary to recover drift velocity $\nu_i$ from transport number $t_i$.
Assuming the system can be approximated by parallel plates, i.e. assuming electric field is constant, ion mobility, $\mu_i$ can be derived from drift velocity through voltage, $V$, and distance between plates, $d$:
$$ \mu_i = \cfrac{\nu_i}{E} = \cfrac{\nu_i d}{V} $$
So the expressions for cation and anion mobility would be
$$ \mu_+ = \cfrac{t_+ d^2}{n_+ z_+ e V} I_\mathrm{total} $$
$$ \mu_- = \cfrac{t_- d^2}{n_- z_- e V} I_\mathrm{total} $$
For a more complex system with $N$ ions contributing to current,
$$ I_\mathrm{total} = \cfrac{e \sum\limits^{N} n_i z_i \nu_i }{d} $$
$$ t_i = \cfrac{ n_i z_i \nu_i }{\sum^{N} n_j z_j \nu_j } $$
and drift velocity and mobility can be computed as
$$ \nu_i = \cfrac{ t_i d }{ n_i z_i e} I_\mathrm{total} $$
$$ \mu_i = \cfrac{ t_i d^2 }{ n_i z_i e V} I_\mathrm{total} $$


text:
Are the best buffers those that are in a 1:1 ratio with its conjugate acid/base?
I wanted to comment this, but I don't have enough reputation, so I just made this a question. I know that buffers and their conjugate base/acid must exist in similar amounts for the buffer to be effective, so would the buffer be most effective when the ratio is close to 1:1 (of the buffer and its conjugate acid/base)?
Also I'm confused about how much of a weak base/acid buffers should be. If you have a very weak acid, then wouldn't you have a very strong base. In this case will the buffer be ineffective? So the upshot of this paragraph is asking will buffers still work if you have a very weak acid?

Yes, the best buffers are those that are in a 1:1 ratio with its conjugate acid/base, but in a somewhat indirect way; it's because buffer capacity is maximum when $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$, and for that value of $\mathrm{pH}$, the proportion of the acid/base conjugates is 1:1.
The contribution of an acid/base conjugate pair to buffer capacity is
$\beta_{\ce{HA/A-}} = C_\mathrm{A} \cfrac{10^{-\mathrm{pH}-\mathrm{p}K_\mathrm{a}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{a}} \right)^2} \ln{10}$
(For derivation, check this response to another question)
The maximum of buffer capacity is achieved when $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$, and is
$\beta_\mathrm{max}= \cfrac{1}{4} C_\mathrm{A} \ln{10}$
As you deviate from $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$ in either direction, concentrations also deviate from 1:1 (as per Henderson-Hasselbalch) and buffer capacity will fall.
Note that the maximum buffer capacity does not depend on the strength of the acid or base, only on the total concentration of the conjugate system $C_\mathrm{A}=\ce{[HA]}+\ce{[A-]}$.
What $\mathrm{p}K_\mathrm{a}$ will determine is at which $\mathrm{pH}$ value will this maximum buffer capacity be achieved, so you will choose an acid/base pair with a $\mathrm{p}K_\mathrm{a}$ that will allow the system to act as a buffer around the $\mathrm{pH}$ that you want to preserve; ideally by choosing a pair with a $\mathrm{p}K_\mathrm{a}$ equal to the target $\mathrm{pH}$, which is when you will obtain the maximum buffer capacity.

The buffer capacity is usually used as indication of the 'strength' of a buffer. It is defined as the gradient of total concentration with pH, and is in effect the number of moles of acid/ base needed to change the pH by $\pm 1$. i.e. $\displaystyle \beta = \frac{dC}{dpH}$ and has the form (using concentration instead of number of moles just for simplicity);
$$\displaystyle \beta=2.303\left[\frac{K_w}{[\text{H}^+]}+[\text{H}^+]+\frac{CK_a[\text{H}^+]}{ ([\text{H}^+]+K_a)^2 }  \right]$$
where $K_w=\ce{[H^+][OH^-]}=10^{-14}$ and the equilibrium constant is $K_a$.
Usually the first two terms in $\beta$ are small compared to the last one so the buffer capacity is proportional to the total concentration of the species present which is what is expected as the acid and base act as reservoirs as the pH is changed.
The buffer capacity has the form of a peak near to the $\mathrm{p}K_a$.

The figure is that for benzoic acid/sodium benzoate (each at 0.01 M) with a $\mathrm{p}K_a=4.2$. The maximum is virtually at the $\mathrm{p}K_a$.
The Henderson - Hasselbalch equation, although it is an approximation, can be used to work out the pH of a buffer solution and in its general form is
$$\mathrm{[H^+]} = K_Ac_a/c_b,  \qquad \mathrm{pH}=\mathrm{p}K_A - \log_{10}(c_a/c_b)  $$
so that when the concentrations are not equal the $\mathrm{p}K_a$ and pH differ slightly.
The answer in this post How to set up equation for buffer reaction? gives the derivation of the last equation.


text:
obabel doesn't split molecules
I have a .xyz input file
8
Energy =
C     0.0354972    0.7111560   -1.7500000
C    -0.0354972   -0.7111560   -1.7500000
C     1.1823040   -1.4277968   -1.7500000
C     2.3594716   -0.7301614   -1.7500000
C     0.0354972    0.7111560    1.7500000
C    -0.0354972   -0.7111560    1.7500000
C     1.1823040   -1.4277968    1.7500000
C     2.3594716   -0.7301614    1.7500000

(this is shortened, problem stays the same) which I would like to split with obabel. However, with
obabel coord.xyz -o split.xyz --seperate -m

I only get one outputfile split1.xyz containing the whole supermolecule. This is also the case if I enlargen the z-distance between the two molecules.
Does anybody either:

Know why this doesn't work
Has a different program which I can use

I don't have the time to manually split 500 molecule files.

Not knowing which version you are running, it is difficult to actually help you.
I am using Open Babel 2.4.1, and had absolutely no problem separating the two ethylene fragments from the following file (super.xyz):
12
symmetry c1
C       -0.498388809     -1.662363855     -2.140955132
C       -0.498388809     -1.662363855     -0.810797132
H       -0.498388809     -0.739022855     -2.713905132
H       -0.498388809     -2.585704855     -2.713905132
H       -0.498388809     -0.739022855     -0.237847132
H       -0.498388809     -2.585704855     -0.237847132
C        1.730729269     -0.175065928     -2.629761741
C        1.730729269     -0.175065928     -1.299603741
H        1.730729269      0.748275072     -3.202711741
H        1.730729269     -1.098406928     -3.202711741
H        1.730729269      0.748275072     -0.726653741
H        1.730729269     -1.098406928     -0.726653741

Using the command
$ obabel super.xyz -Ofrag.xyz --separate -m

results in frag1.xyz and frag2.xyz containing one ethylene each.

I'm using 2.3.2. Your input file with your command works. This helps me debug the problem at least, thank you.

@Fl.pf. your input worked for me, too (after fixing the atom count). Maybe it's the lower case o which should indicate the file type, not the output location.

yeah that, and the fact that I'm an idiot (it's --separate, not --seperate). Thanks for being my rubber ducky though :D


text:
Why is the difference between the melting point and boiling point large for some compounds, and small for others?
I was looking at a list of melting and boiling points of various compounds and I realised some had very large differences while some have very less differences between their melting and boiling points.
Example:

for cyanogen, boiling point is -21.2°C while melting point: -27.9°C
for propane, boiling point: -42 °C while the melting point:  -144°C

Why does this happen?

Boiling points and melting points are based on the stability of different phases. There's no reason to assume that the factors that govern the stability of solids governs the stability of liquids in the exact same way and to the exact same degree.

well thinking that the same factors influence the melting and boiling point of materials is possibly wrong.
high melting point is due to strong packing(highly symmetric molecules) while  high boiling point is due to large surface area
so if a compound has good packing and less surface area will end up to have very less difference in melting and boiling point and sublime


text:
Why is the initial carbocation not formed adjacent to the phenyl moiety?

I know that after rearrangement the carbocation is more stable, but we should protonate the compound first where the carbocation it is more stable, i.e.  adjacent to the phenyl group. Why do they suggest a different pathway? Is the shown correct?

That seems reasonable.

Steric interference maybe?

I agree with your sentiment, so I run a very quick calculation with a semi-empirical method (GFN-xTB) to find the protomers. This has been done using water as solvent. Note that I did not choose to include any deuterium.
First I calculated the conformational space of the plain molecule (NIST: 1,1'-(3-methyl-1-propene-1,3-diyl)bis-benzene, InChI=1S/C16H16/c1-14(16-10-6-3-7-11-16)12-13-15-8-4-2-5-9-15/h2-14H,1H3/b13-12+, Google search InChIKey: GNQWHYWLSGTMSL-OUKQBFOZSA-N) to obtain the lowest lying conformation and show you the HOMO:

As you can see the double bond is a good protonation cite, with both carbon atoms having about equal contribution; one would therefore expect that the most stable intermediate will form. Undoubtedly this should be the one where the charge can be delocalised into the aromatic π-system.
The automatic generation of the protomers agrees here, and the lowest lying conformation (with partial π-stacking) is shown below, the protonated carbon indicated in bright white:

There is a tentative pathway via a four-membered (green) transition state to rearrange to the product shown in the exam question, but I do not have the time to investigate. Therefore here is my proposed mechanistic scheme:


text:
What's the relation between molarity and molality?
I am trying to derive the relation between the molarity and molality of a solution

Mole concept $\displaystyle n = \frac{M'}{M''}$

Here, ($n$) stands for number of moles
$\ce{(M^{'})}$ stands for mass of substance in gram
$\ce{(M^{''})}$ stands for molecular/atomic mass of substance

Let $V_{\text{sol}}= $ volume of solution
Molarity (M) $= n/V_{\text{sol}}$

The above equation of molarity ($M$) shows the concentration of solution of substance, i.e. ($M$) is the ratio of number of moles ($n$) with respect to it's volume ($V_{\text{sol}}$)

let $W$ = mass of solvent in kilogram
molality $(m) = n/W$

The above equation of molality ($m$) shows the concentration of solution of substance
i.e. ($m$) is the ratio of number of moles ($n$) with respect to mass of it's solvent ($W$)

What is the relation between molarity ($M$) and molality ($m$)?

Molarity ($M$) = $n$/$V_{\text{sol}}$
Molality ($m$) = $n$/$W$

I'm confused on how to relate the $W$ with $V_{\text{sol}}$.

$W$ and $V_{\text{sol}}$ can be related via the density of solution. We have, mass of solution = mass of solute + mass of solvent.
Hence,
$$ M_{\text{sol}} = W + M’ $$
Let us assume the density of the solution to be $\rho$. Since, density is equal to mass over volume, we have,
 $$ \rho=  \frac{M_{\text{sol}}}{V_{\text{sol}}} = \frac{W + M’}{V_{\text{sol}}} $$
Thus, you have relation between $W$ and $V_{\text{sol}}$.


text:
Can alkyl fluorides be formed by the Hunsdiecker reaction?
We know that the Hunsdiecker reaction is
$$\ce{RCOOAg + X2 ->[CCl4] RX + AgX + CO2}.$$
This reaction is successful with chlorine and bromine. Also, we know that if $\ce{X}$ is iodine, we will get an ester.
Can we form alkyl fluorides using this reaction? If not, why?

Are you trying to use fluorine gas? :/

Yes @Zhe. Generally we do that only...

That was a rhetorical question........

Fluorine shows anomalous behaviour among the halogens. While chlorine, bromine and iodine produce alkyl halides with alkanes in presence of light, fluorine produces carbon and HF in a violent reaction.

The classic reaction of silver salt of fatty acid and halogen to produce alkyl halide was first studied by Alexander Boronin in 1861. But the reaction was generalized by Heinz Hunsdiecker due to which the reaction is named Hunsdiecker reaction. Hunsdiecker used bromine to make methyl bromide. Chlorine was then used to make alkyl chloride. When iodine was used, then ester was formed. This reaction was called Simmonini reaction. Fluorine was not yet discovered then and hence alkyl fluoride was not made through this reaction (fluorine was discovered in 1886 by Henry Moissan).
But there are some speculation discussed among the chemist community whether Boronin actually used some fluorine containing reagent to make alkyl fluoride. It is discussed in this forum that Boronin might have used "acidic potassium fluoride" or $\ce{KHF2}$/potassium fluoride($\ce{KF}$) to make benzoyl fluoride in 1862 but there is no report or scientific paper on this reaction making it a speculated information. (Note: fluorine was discovered by electrolyzing $\ce{KHF2}$).
Fluorine containing compounds were made through Hunsdiecker reaction though proper alkyl fluoride synthesis through this reaction is yet to be reported:

Perfluoroalkyl iodides were made through this reaction:
$$\ce{R_fCO2Ag + I2 ->[\Delta] R_fI + CO2 + AgI}$$
This compound is used to make numerous fuctionalized fluorocarbons and
first Grignard reaction of this class.

Perfluoroalkyl diiodides has also been prepared by heating perfluorocarboxylic acid in presence of iodine. It has been used to make diacids having ether linkage.

Fluorinated ester of type $\ce{RFCO2CF2RF′}$ has been was obtained from reaction of silver 3,6,9-trioxa-F-undecanoate and iodine with a diluent at 130 °C. Substitution products, $\ce{RFCOY}$ (Y = nucleophile) and pyrolysis products, $\ce{RFCOF}$ and $\ce{RF′COF}$ were isolated and characterized. It appears probable that the acyl hypoiodite and the iodo-F-alkane reacted by an $\ce{S_Ei -}$ type process to give the completely fluorinated ester.

Fluoro-functionalized graphene oxide  is made via the Hunsdiecker reaction. It is made by treating of carboxylated graphene oxide with selectfluor at 90 °C for 10 h under an atmosphere of nitrogen, using silver nitrate as a catalyst.

References:

Organofluorine Chemistry: Principles and Commercial Applications
edited by R.E. Banks, B.E. Smart, J.C. Tatlow - page 18
Well-Architectured Fluoropolymers: Synthesis, Properties and Applications
By Bruno Ameduri, Bernard Boutevin - page 108
Brace, N. O. Some approaches to the synthesis of fluorinated alcohols and esters. I. Completely fluorinated esters from the hunsdiecker reaction of silver F-alkanoates with iodine. J. Fluorine Chem. 1981, 18 (4), 515–524. DOI: 10.1016/S0022-1139(00)82668-8.
Xing, R.; Li, Y.; Yu, H. Preparation of fluoro-functionalized graphene oxide via the Hunsdiecker reaction. Chem. Commun. 2016, 52 (2), 390–393. DOI: 10.1039/c5cc08252h.

Borodin did synthesize benzoyl fluoride. But he made it by reacting benzoyl chloride with KHF2. Even then, benzoyl fluoride is not an alkyl fluoride. See http://www.chtf.stuba.sk/~szolcsanyi/education/files/Organicka%20chemia%20II/Prednaska%205_Aldehydy%20a%20ketony_Reakcie%20II/Doplnkove%20studijne%20materialy/Aldol%20reactions/Alexander%20Borodin%20-%20Composer%20and%20Chemist.pdf.

See https://encrypted.google.com/patents/US4704484. This mentions that Hunsdiecker reaction is limited to halogens except fluorine.

@ShoubhikRajMaiti Yes, that's the point. Borodin tried to make a fluorine compound from Hunsdiecker reaction but was unable to synthesize an alkyl fluoride.


text:
How can a weak base act as a strong nucleophile (and vice versa)?

Carboxylate ion ($\ce{RCOO-}$) is a weak base as the negative charge on oxygen is delocalised due to resonance. So, it has less tendency to donate lone pair of electrons to H+ ion. Similarly, it has less tendency to donate lone pair of electrons to a carbocations. But in my textbook, it is given that, it acts as a weak base but strong nucleophile. Why does it act as strong nucleophile?
LDA (lithiumdiisopropylamide) is a strong bulky base. Why it is considered as a weak nucleophile?

Nucleophilicity and basicity aren't the same concept.

Nucleophilicity is tendency of a nucleophile to donate electron pair to carbocation,isn't it ?

Nucleophilicity is a kinetic concept. Basicity could be kinetic or thermodynamic.

Sorry sir. I didn't understand you.

Right. And I think that's your problem. You're confusing two separate concepts that have some overlap. One does not imply the other or vice versa.

LDA is a strong base, but it is bulky. A nucleophile must have the capacity to attack a electron deficient centre and form a bond. But if it is bulky, it cannot form a strong bond, as the nearby atoms would hinder it. So, LDA becomes a bad nucleophile.


text:
Reaction between phenol and nitrous acid
My question is about the products formed when phenol reacts with nitrous acid. I googled it and reached this PubChem paper (DOI: 10.1016/S0045-6535(02)00857-3), which states:

It was found that phenol reacts with nitrous acid to produce cyanide ions. Cyanide ion generation is attributed to the conversion of phenol to nitrosophenol through the well-known nitrosation reaction, and decomposition of benzoquinonoxim to form cyanide and aliphatic compound.

Is this correct? If not, then what is the actual reaction?

The reference is behind a paywall, I assume you bought it or otherwise gained access.  Do tell what they did (procedure, conditions) to find said cyanide ions, then we can make an objective evaluation.

@Nilay that gets us to the quinoxime,but then what? Assuming the quinoxime forms, it could ultimately release cyanide if either a water molecule or the oxime hydroxyl group attacks adjacent to the oxime carbon.
If this occurs, it breaks both the carbocyclic ring and the N-O bond from the oxime.  But, is such a reaction plausible?  We need access to the paper to know, maybe?

@OscarLanzi, I simply typed phenol with nitrous acid on Google and tht was the first link shown by Google..

I found out that this mechanism is only valid in wastewater treatment from all the research papers I gathered and not discussed anything about cyanide formation in general analytic chemistry. It only says that $\ce{NO+}$ is the electrophile in this reaction and nothing about cyanide.

What the reference says
The reference gives only following figure to support their theory. A weakness of this mechanism is requiring a high energy intermediate like sp2-carbocation.

An alternative proposal
This alternative avoids high-energy intermediates and thus appears more likely:


text:
Shifting of groups in carbocation rearrangement

Phenyl has greater migratory power. However, why did methyl migrate in step 2 of the reaction?

What makes you think the reaction depicted is right?

^ In other words, what is the source for this mechanism? Note that your thought may possibly be correct as well.

One might compare the stabilities of the cations formed from phenyl vs. methyl migration.

In general, migratory aptitude of different groups follow following order:
$\ce{-Ph} > \ce{-C(CH3)3} > \ce{-CH(CH3)2} > \ce{-CH2CH3} > \ce{-CH3} > \ce{-H}$
For example, 2,3-diphenylbutane-2,3-diol (1) was allowed to react under homogeneous (25 mol% of p-toluenesulfonic acid) or heterogenious (20 mol% of graphene oxide) catalytic conditions using toluene as a solvent at $\pu{100 ^\circ C}$. Under both conditions, only pinacol rearrangement with phenyl migration was observed affording 3,3-diphenylbutan-2-one (2) in a 90 and 95% conversion. Methyl migration has not been observed [Reference].

Reference:
Pinacol Rearrangement and Direct Nucleophilic Substitution of Allylic Alcohols Promoted by Graphene Oxide and Graphene Oxide $\ce{CO2H}$: M. Gomez-Martinez, A. Baeza, D. A. Alonso, ChemCatChem, 2017, 9, 1032–1039 (https://doi.org/10.1002/cctc.201601362)


text:
Is ground-state structure of a molecule entirely determined by the positions of the nuclei?
I'm a computer science with only a weak undergraduate chemistry background, but have recently become interested in simulation methods for computational chemistry. Many simulation methods, such as DFT, make the Born-Oppenheimer approximation (nuceli are static) and solve for some properties of the electrons (in the case of DFT, the density). In fact, when using Gaussian-16, I only need to specify the locations of the nuclei in space.
What then, is the role of actual bonds? Are bonds, bond order, etc. just a convenient abstraction for what the electrons are doing? If I tell you the location of some nuclei in 3D, can you then (theoretically) uniquely determine the bond structure?

Yes, bonds are just a convenient abstraction. Nature does not know about them.

Born-Oppenheimer approximation is only this - approximation. For example ground state of $\ce{CH5+}$ cation is fluxional and position of it's hydrogen atoms is changing constantly no matter how low is the temperature.

Yes, these methods (both wavefunction-based and electron density-based) typically solve a purely electronic system defined by an electrostatic potential which is produced by the nuclei; the Born-Oppenheimer approximation can be understood as treating the electronic system as a function of nuclear geometry, which implies that (1) nuclear dynamics are not strongly affected by electrons and (2) electrons will almost immediately adapt to any new position of the nuclei. Born-Oppenheimer can also be understood in terms of electronic and nuclear motion (and the corresponding energies) being separable; that is, electronic structure takes shape after (and as a consequence of) nuclear dynamics, but does not directly affect nuclear dynamics themselves. Since chemistry is usually interested in stationary solutions as snapshots, these approximations are fitting for pretty much all chemical questions.
In practice, nuclei are not even particles in electronic structure calculations - in both Schrödinger equation methods (HF and post-HF) and Hohenberg-Kohn methods (DFT), the only role of nuclei is to generate an external electrostatic potential, $V_{ext}$, that will be part of the Hamiltonian/energy functional of the system. The specification of nuclear positions that you mention is just a more convenient, and chemically intuitive, way of describing that external electrostatic potential, but there is no mathematical or conceptual difference if we consider that this external potential is arbitrarily imposed from outside; it's just that it's easier to define a number of point charges (with their charge defined by their atomic number) than to provide a complex 3D potential function.
Once we assume that Born-Oppenheimer is a valid approximation and that nuclei merely exert an external electrostatic potential on electrons, the fact that electronic structure for the ground state is entirely determined by the positions of the nuclei is axiomatic: it is a consequence of the postulates of quantum mechanics (and the basis of all variational methods) for wavefunction-based calculations, and one of the Hohenberg-Kohn theorems for density functional calculations.
Geometry optimisation is also performed under the same assumptions - electronic structure is calculated under a given geometry, and then resulting forces over nuclei are estimated; a small motion is performed along these tensions, and the process is repeated iteratively until forces experienced by nuclei are small enough to fall below a certain tolerance level. Note that there is no explicit treatment of the interaction between electrons and nuclei; only two process that interact iteratively through electrostatics: electronic structure is minimised for each geometry, and then geometry is relaxed, but not minimised, according to that electronic structure. We obtain a quantum mechanics depiction of the electronic system for each geometry, but we do not perform a quantum mechanics treatment of the nuclei. Ab initio molecular dynamics is a similar approach, in which nuclei move through classical dynamics but electronic structure is calculated through QM at each step.
In all these cases, chemical bonds are only emergent interpretations of the energetics of electronic structure, but their identification is largely arbitrary, as energetic effects are not localised. While we can calculate the energy of two molecular fragments as they approach from infinity, and we'll observe interactions that we identify with bonding, the limits of what is a bond are not clear, and at a quantum chemical level, it makes more sense to talk about energies associated with different geometries than to think in terms of bond "objects".
It is in principle possible to perform QM computations which treat both nuclei and electrons as quantum particles, thus obtaining a single wavefunction describing both, but the complexity of this treatment (especially given the large difference in scale for electronic and nuclear motion) restricts calculations to a very small number of particles - the highest number I remember reading about is 8, although I might have missed recent studies. In addition, the problem of describing interactions accurately (which is behind the variety of quantum chemistry methods used just to describe electrons) grows considerably worse, to the point that quantum Monte Carlo may be a better option than quantum chemistry methods, as happens with bosons.

As far as Quantum mechanics is concerned, there is no such thing as bonds or bond order. Any QM method is just dependent on finding the wavefunction that satisfies the Schrodinger (or Dirac if concerned with relativity) equation (or some approximation of it). Within the Born-Oppenheimer approximation for a particular method, the only user inputs needed are the location and type of all the nuclei, and the basis set(s) you are using to construct your MOs. Its important to note that basis sets are empirically optimized to describe particular atoms, so if you were making your own electronic structure program from scratch, it wouldn't be enough to just have descriptions of nuclei; you would need to incorporate preexisting basis sets or determine your own. So I would say no, the ground state structure depends not only on the positions of the nuclei, but also on the basis functions and method you use to evaluate the energy/wavefunction.


text:
Why is it that all strong bases contain at least one hydroxide bond?
Why is it that all strong bases contain at least one hydroxide bond? What is going on here at the chemical level? For example, we know that $\ce{NaOH}$ is a strong base. I know it is strong because it completely dissociates in water. And under the Bronsted-Lowry definition of bases, I know that is is considered a base because it accepts a proton. So what is it about having a hydroxide bond specifically that lets it dissociate?

Not clear what you are asking here. NH3 is a base, but doesn't have any hydroxide, Fe(OH)3 has plenty of hydroxide but is not basic.

NH3 is a weak base though. I am specifically talking about all strong bases. And I don't mean to say that all compounds with hydroxide bonds are strong bases, just the other way around. (Or at least the vast majority of strong bases)

There’s no hard and fast rule that a strong base should have a -OH group. Refer Which is the strongest base?

Sodium hydride and sodamide are strong bases with no hydroxide bond present

There is not such thing as a "hydroxide bond" afaik.

All strong bases you know contain at least one hydroxyl group because we live in a world dominated by water, almost like in that old movie with Kevin Costner. Water chemistry is of paramount importance to us, and acid/base reactions in water pretty often boil down to $\ce{H+ + OH- ->H2O}$.
There is nothing special about the hydroxyl group that lets it dissociate. Come to think of it, the group itself does not even dissociate in bases - I mean, $\ce{O}$ and $\ce{H}$ remain together, the bond between them does not break. (It does in acids, but that's another story.) If you mean the dissociation of metal hydroxides, there is nothing special here either. $\ce{NaOH}$ dissociates like any water-soluble ionic compound. On the other hand, $\ce{Fe(OH)3}$ refuses to dissociate, like any water-insoluble ionic compound, of which there are plenty.
In a wider chemical world outside, there are bases which are many orders of magnitude stronger than that, and they don't contain any hydroxyl groups whatsoever. $\ce{NaNH2}$, $\ce{NaH}$, $\ce{BuLi}$ come to mind. You don't see them much in everyday life, though, because they don't survive in our world. Water act as an acid on them, and instantly transforms them to mere $\ce{NaOH}$, etc.
So it goes.

Not all strong bases are hydroxides.  Guanidine  has no oxygen at all and is not ionic in its free base form, yet it competes with alkalies.  The highly stabilized, delocalized electron structure of the cation formed upon protonation gets credit in one of the strongest non-ionic bases.


text:
Why or why not can hydroquinone and 2H-pyrrole act as ligands?
I want to understand why hydroquinone and 3H-pyrrole can or cannot act as ligands in transition metal complexes. If they are ligands, then what is their denticity?

I'm trying to understand the overall co-ordination behaviour of these two molecules, so please consider all atoms and rings as possible electron donors.
I think the denticity for the second molecule should be 1, since there's only 1 donor atom N, but would it be a weak ligand as the nitrogen is double bonded to a C?

Your second molecule is actually pyrrole in disguise.

This is an under-specified question because you didn't specify the binding mode of each ligand. For example, in the hydroquinone, do you want an oxygen atom to be the electron pair donor or do you want the benzene ring? And if the benzene ring, do you want to coordinate 2, 4, or 6 carbon atoms? Same types of questions with the second structure.

Would pyrrole even donate via the N-Atom? I think in pyridine the lone-pair on the nitrogen is orthogonal to the aromatic pi-system but in pyrrole it is parallel to the aromatic pi-system. Therefore I'd expect it to rather coordinate like in metallocenes.

@Zhe I'd appreciate if you could explain donation from both the O and the benzene ring (and the N atom and the ring respectively). I'm just trying to understand the co-ordination behaviour, sorry for not making that clear.

Sorry, if my explanation doesn't make sense, I'm not sure trying to explain further will clear things up for you. You should look at a good reference text on coordination chemistry or organometallics.

Zaworotko, M. J.; Stamps, R. J.; Ledet, M. T.; Zhang, H.; Atwood, J. L. Heterocyclophane complexes of transition metals. 1. Synthesis and crystal structure of both the (.eta.5-2.2pyrroloparacyclophane)tricarbonylchromium and the (.eta.6-2.2pyrroloparacyclophane)tricarbonylchromium. Organometallics 1985, 4 (10), 1697-1700. DOI: 10.1021/om00129a001

A chromium tricarbonyl complex of a pyrrole is reported, here the pyrrole coordinates with all four carbon atoms and the nitrogen atom. I think that the pyrrole is a six electron donor to the chromium.

Bentivegna, B.; Mariani, C. I.; Smith, J. R.; Ma, S.; Rheingold, A. L.; Brunker, T. J. Formation, Stability, and Structures of Borenium and Boronium Cations Derived from Pentamethylazaferrocene–Boranes by Hydride or Chloride Abstraction Reactions. Organometallics 2014, 33 (11), 2820-2830. DOI: 10.1021/om500348u

A pyrrole binds with all five non hydrogen atoms to an iron which is also bonding to a pentamethyl cyclopentadienyl anion. This is forming a protonated version of CpFe(C4N) which looks very much like ferrocene. In this case the bonding is the same as it was in the chromium compound.

Yamamoto, K.; Kimura, S.; Murahashi, T. σ-π Continuum in Indole-Palladium(II) Complexes. Angew. Chem. 2016, 128 (17), 5408-5412. DOI: 10.1002/ange.201601992

A pyrrole binds with two carbons as a pi ligand to palladium. These selected results suggest to me that pyrrole can bind as an alkene or as if it was a Cp ligand to transition metals.

Son, S. U.; Reingold, J. A.; Carpenter, G. B.; Czech, P. T.; Sweigart, D. A. Charge-Assisted Hydrogen Bonding and Other Noncovalent Interactions in the Self-Assembly of the Organometallic Building Block [(η6-hydroquinone)Rh(P(OPh)3)2]+with a Range of Counteranions. Organometallics 2006, 25 (22), 5276-5285. DOI: 10.1021/om0604425

It reports the binding of a rhodium to all six carbon atoms of hydroquinone. Here it acts as a six electron donor.

So there is nothing to stop the molecules in the question acting as pi ligands to transition metals.


text:
How many protons does phenolphthalein lose in basic solutions?
So, I've looked it up and Wikipedia claims that two protons are released by Phenolphthalein when pH is over 8.2, but I've seen other sources that claim only one proton is lost. So, is it really just one or two? Which form is responsible for  its colour? (I would expect a fully deprotonated PhPh in more basic solutions, so I guess the PhPh that released two protons is the one responsible for the solution's colour)

Can you please cite the "other sources"? Their context may be different, for example, they may lose only one proton in stronger acidic conditions.

ChemSpider phenolphthalein's $\mathrm{p}K_\mathrm{a}$ value: 9.7 (at room temperature). Here they've is only given one $\mathrm{p}K_\mathrm{a}$ value, but it is still reasonable to assume that there are other forms in other pH conditions, like in this image:

(source: Wikipedia)
In other sources like academic ones, it is always shown with 2 forms, and these are associated with the supposedly (especially back titration from pink to colourless is very hard to catch) sharp color change in pH 8-10 range. If one is to use an indicator for that pH region, phenolphthalein is chosen but not for the other regions so no one is interested in possible other forms.
It is also understandable if people actually don't even know the other forms, or try to use them also since pH meter(also possibly UV-Vis spectrometry) is used for a long time to track the pH, I have only seen these indicators in student laboratories for teaching purposes, not in any of research laboratory.


text:
Nucleophilic Aromatic Substitution in ring with 4 Substituents

Does the chlorine still get substituted by OH in this case?
I was also wondering, if it does get substituted, does anything happen between the ethanol substituent and the phenol -OH? I'm suspecting that there might be some kind of ring formation...just a strange feeling.
Thanks!

I think you are right. The tert-butoxide ion being a bulky base will not do an ipso substitution and rather abstract an H+ from -OH and then this -O- will attack the carbon with chlorine to form a ring.


text:
Are ions actually charged particles?
For instance, if an atom loses an electron and becomes a cation, such as sodium, how does that make the sodium atom positively charged? The sodium's atomic radii decreases due to less electron-electron repulsions which in effect makes the sodium atom more electron dense.
The common response I get from TA's is that positive and negative charges cancel each other out so essentially one proton isn't getting canceled out which results in a net positive charge. However, protons and electrons don't cancel each other out in an atom (effective nuclear charge), so I know that response is incorrect since the electrons closest to nucleus of an atom feel a greater positive charge than the valance electrons.

Comparatively, the atomic radius of Sodium in its neutral form is greater than the atomic radius of Sodium in its cationic form so there's a higher probability of Sodiums electrons being closer to the nucleus relative to before which is what I meant by electron density.

Yes ions are charged, and you cannot hide the fact that if there are more protons than electrons the overall change will be positive  and vice versa, no matter about screening or any other fancy thing you can invent.

Yes, ions are actually charged particles and neutral atoms are not. The TA's version is correct: the positive and negative charges cancel each other out. In a positive ion, we have one extra proton that gives the whole thing a net positive charge. In a neutral atom, electrons and protons cancel each other out exactly. How can they not to? After all, their electric charges are precisely identical, just with opposite signs.
Now when we speak about cancelling out like that, we only mean the net charge. Take a mole of neutral atoms and a mole of positive ions (just theoretically, of course, otherwise the consequences might be catastrophic), and even a crude electroscope will tell you the difference between the two right away.
If you specifically intend to ask about one ion, let's look at it from a different perspective. A neutral atom or molecule is, well, neutral, but that doesn't mean it would not participate in electrostatic interactions: there are those small positive and negative charges inside it. They seem to be moving, or at least not being all in one place, and you can feel tiny electric fields resulting from that. The corresponding electric potential decays with distance as $\frac1{r^2}$ for a dipole, and even faster than that for a quadrupole or other configurations of charges. The same applies to an ion, but it also has a net charge, and its potential behaves like $\frac1r$. When looking from a far enough distance, you will only feel the pull of the net charge. Look at a neutral atom from the same distance, and you will feel nothing.
Sure enough, if you look very closely, or even inside an atom (or an ion), you will see a lot of things going on around you, and the net charge will not be all that important.
So it goes.

The atom would not become more electron dense in your case. Electrons repel each other and that effect + Z number is the main explanation for size in (small, non relativity affected) atoms. The attraction between core and every electron stays the same, it is just the repulsive forces between the screen of electrons and a single electron that is less for each electron lost.
Your TA is right. The net charge of the atom is now positive. Only for very large atoms with a correspondingly large electron "screen" will the screen effectively shield the surroundings from the charge at the core.

Yes, of course an ion is a charged species. When you say that the charges on electrons and protons can not cancel out each other then, I guess, you are thinking of them far apart from each other. But within an atom, the two charged things (electrons and protons) are sufficiently close to each other to cancel their charges. Even if you go to molecules, it may have different charges on different atoms in it but the overall charge on the molecule will be equal to the sum of charges on each atom. You can think of for example, CO2. In CO2, the C-atom is positively charged and O-atoms are negatively charged but CO2 molecule is neutral. So, charges really cancel out each other and hence after losing an electron an atom becomes a positively charged ion.


text:
Hydroboration oxidation reaction on trans,trans,trans-1,5,9-cyclododecatriene with carbon monoxide

The given solution:

I got this question in an entrance exam coaching test paper in India. Product P is easily covered by the mechanism of simple hydroboration-oxidation reaction, and I've understood it well. However, product Q is what intrigues me.
Looking at the product and the reactants, I can guess that the carbon in $\ce{CO}$ functions as the source for the bridging carbon in the product Q. However, I could not figure out the mechanism for the same. It is not given in my textbook, and wasn't found via Google searches either.
I'v well understood how the intermediate product (bridged by boron) had formed. However, how is the carbon monoxide supposed to displace boron from there? I can guess that once a carbonyl group bridge is formed in the middle, the alkaline hydrogen peroxide will reduce the ketone to an alcohol (to get product Q), but I am not too sure.

What's the stereochemistry of the final product? (One containing 3 -OH groups)

@schrodinger_16 You may look at the hydroboration-oxidation mechanism for that - which I've since forgotten. The stereochemistry was not mentioned in the question.

My instructor told me that only two OH groups are on the same side in the final product, and I've been wondering why ever since then.

@schrodinger_16 I'm afraid I actually don't know the stereochemistry.

You guessed it right: a ketone will be formed.
This reaction involves migration of alkyl groups from boron to the carbon atom of CO. There will be three migrations after which you will get your alcohol:

Image taken from National Programme on Technology Enhanced Learning (NPTEL). Lecture 26 : Boron Containing Compounds. Image link: Scheme 9
See also Handouts for Organic Chemistry Lectures given at Imperial College London, Chemistry, by Henry Rzepa.


text:
Cyclic elimination reaction

Which of the following will give cyclic elimination by heating?

phenol
propanoic acid
trimethyl isopropyl ammonium hydroxide
cyclohexyl propanoate

Cyclic elimination by heating is Syn Pyrolytic elimination reaction. This is given by ester, quaternary ammonium salt, and amine oxide. The reaction is according to Hoffman. This cancels out option (1) and (2).
Option (4) is ester and hence the answer to the question.
But why does the ammonium salt not give the reaction? What am I getting wrong?

Example of a similar tertiary ammonium reaction with mechanism
The thing is, it does react but there will be no cyclic product, in the end, I think that is the reason it is not the answer here.


text:
Why is (tri-)ammonium phosphate ((NH4)3PO4) so unstable?
Why is triammonium phosphate, $\ce{(NH4)3PO4}$, so much more unstable than diammonium phosphate, $\ce{(NH4)2HPO4}$, or monoammonium phosphate, $\ce{NH4H2PO4})$?

Ammonium phosphates are solids between 0 and 75°C. Because ammonium
  hydroxide $(\ce{NH4OH})$ is a much weaker base than the common metal
  hydroxides, ammonium phosphates are comparatively unstable.  Both
  triammonium phosphate ($\ce{(NH4)3PO4}$) and the double salt
  ($\ce{(NH4)3PO4.2(NH4)2HPO4}$) are unstable at room temperature and
  evolve ammonia to form diammonium phosphate. Even the commercial mono-
  and diammonium phosphates exhibit an ammonium vapor pressure in the
  solid form and in solution. Although MAP is more stable than DAP, it
  decomposes at high temperature to give ammonia and polyphosphoric
  acid. Diammonium phosphate decomposes to give ammonia and monoammonium
  phosphate at around 70°C.

Source: (2005) Ammonium Phosphates and Ammonium Sulfate. In: Synthetic Nitrogen Products. Springer, Boston, MA. DOI: 10.1007/0-306-48639-3_12.

I honestly don't understand why $\ce{NH4OH}$ being a weaker base makes TAP and DAP be unstable.

Looking at vapor pressures of NH3 for those solids would be a good idea...


text:
How to attach soft tubing to glass
I plan to do some amateur chemistry with some basic equipment. I've watched hours of youtube videos for how to bend and polish glass tubing and push it through rubber stoppers, and I've seen diagrams of such tubing attached to rubber or silicone tubing, but nothing showing how to do that. I've even searched through catalogs of equipment and seen lots of specialized gadgets with hose barbs and such, but nothing basic. I just want to do a simple thing: have a rubber stopper with a glass tube in it, and attach that directly to a flexible tube. Nothing fancy, just one tube connected to another. I see it in diagrams like http://www.mctcteach.org/chemistry/C2224/labman/vacfilt/filter7.htm , but it gives no explanation. What am I missing?

Why the downvote? Is this not a sincere question about lab procedure that's not answered by a Google search?

Heat up the soft tubing to make it easier to fit over the connection. Most labs have a heat gun that is used for this (amongst other uses). A hairdryer will serve, or put the end of the tubing in hot water. Many glass connection have what is called an olive which you have to push the tubing over while hot, when it cools it contracts giving a good seal (particularly important for the water lines of condensers which may be left overnight)

What "connection"? That's the whole of my question--plain glass tubing has no hose barb or anything like that. Am I supposed to just assume that the ID of the hose is a bit smaller than the OD of the glass? Do I attach a tube with an olive to the plain tube by melting them together?

For plain glass with no olive I would try to choose soft tubing with a smaller internal diameter and soften it so that in can be pushed over (carefully) the glass tube. This is not that satisfactory but should hold if there's not too much pressure in the system.

That should work. It just seems odd that there are hundreds of different fittings for special applications, but not a simple one just for this.

You do this with great care, putting rubber or plastic tubing onto glass tubing or taking it off is a major cause of injuries in a chemistry lab.
I always saw that the most dangerous object in my lab is a traditional liebig condensor with short bits of glass tube which you push the plastic water hoses onto.
The best way to do it with rubber hoses (such as semipressure / vacuum hose) is to dip the end into dichloromethane or chloroform to make the rubber more soft, then to push the hose gently onto the glass tubing. To remove it, it if often best to cut the tubing off rather than pull it off if the tube will not come off with ease.
For plastic hoses such as PVC it is best to heat the plastic tube up with a hot air gun to make it more soft before trying to put it onto the glass tubing.


text:
Is instant coffee an impurity when you add it to water?
I have this question for chemistry, but can't know whether it is an impurity or not, please help.

Is your water still pure after you add coffee to it?

Isn't this more of a linguistics question?

An impurity is pretty much by definition something that is present that is not desired. If you have deliberately added something, in this case instant coffee, then it is not an impurity but a component, something you intend to have present.


text:
Reaction of Seretonin with aqueous Bromine
I have a question that I am stuck with. Serotonin (figure below) reacts with $\ce{Br2 (aq)}$ to give a compound. The marking scheme I am following says that Bromine only reacts in the ortho position of the phenol group. But I think it should also react with the double bond. Shouldn't the bromine react with the double bonded carbon?
Additionally when I react serotonin with $\ce{CH3COCl}$, shouldn't phenol and both the $\ce{NH}$ react? My marking scheme says it won't react with the double substituted $\ce{NH}$ group but does with the $\ce{NH2}$ group? Who is wrong here?

If the phenol gets acetylated that product acts as an acetylating agent as phenol is a fairly good leaving group. This is why you only see acetylation of the NH2

Well, looks you didn't hear about pyrrole...

@Mithoron Thank you. Did some searches on pyrrole. Pyrrole seems to fit my criteria. I would appreciate if you could answer my other query as well.

Talking about your double substituted -NH
The p orbital of nitrogen containing the lone pair overlaps with π bond of C=C, resulting in two pairs of electrons delocalized (spread) over these two centres (N & C), making the lone pair on N less available for donation to acylium ion (acyl chloride with chloride ion removed). Similarly, the electron density of C=C is also decreased which means it no longer shows addition reaction with Br2 (aq).
The delocalization of electrons is responsible for the unusual stability of some compounds and their reluctancy to react, as in the case of benzene ring.
While solving chemistry questions keep in mind the basic chemistry & try to solve things with reason and logic first.

For further understanding, a delicious information is available here: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/07%3A_Acid-base_Reactions/7.06%3A_Acid-base_properties_of_nitrogen-containing_functional_groups#:~:text=The%20aromatic%20compound%20pyridine%2C%20with,for%20bonding%20with%20a%20proton.

You may want to render the structure in this manner:

With this rendering, not only is the six-atom ring conjugated with six pi electrons, but so is the five-membered ring including the apparently nonbonded electron pair on the nitrogen atom. Thus the five-membered ring is also aromatic, specifically derived from pyrrole.
So the proposed double bond and nitrogen nonbonding pair in the five-membered ring do not effectively exist; they are delocalized/smeared into an aromatic ring. Accordingly the bromine and acetyl chloride do not have these functioms to react with and thus exclusively react elsewhere (bromine with the activated benzenoid ring, acetyl chloride with the unhindered amino group on the side chain).


text:
Why do intramolecular forces weaken due to heat?
In my most recent research into thermoplastics, I was unable to find any information regarding what actually happens to intramolecular forces when heat is applied.

They don't weaken. It is just that the molecules start to move faster, so you can't hold them in place using the same force as before.

You can think of intermolecular forces as having an anharmonic and dissociating potential , such as  Morse or Lennard -Jones. The potential will have vibrational energy levels and as heat is applied more of these levels become excited till eventually the dissociation limit is reached and the molecules separate. So the intermolecular 'bond' is weaker as energy is suppled as levels close to the dissociation limit are populated and this is also where the separation of molecules is larger.


text:
Is one syn aldol product preferentially formed over the other?
In Carruthers' Modern Methods of Organic Synthesis, the following aldol reactions of two (Z)-enolates with benzaldehyde are shown:

The two transition states shown seem to be different, and lead to different stereochemistry in the product (the first product has both Me and OH pointing away from the plane, whereas the second has both pointing out of the plane).
Is this significant, and if it is, how can we predict which TS is used for which problem specifically?

Ignoring the difference between the SPh and Et groups, the two products shown are enantiomeric, not diastereomeric. Both of these reactions are not enantioselective in any manner, so both enantiomers will be formed in equal amounts, via both transition states. Presumably, in the book, only one enantiomer is shown to save space and to avoid redundancy.
The only thing that needs to be illustrated here is that the phenyl group is equatorial in the lower-energy transition state. That leads to the diastereoselective formation of the syn aldol product (but no enantioselectivity between the two syn products).
So, there is no difference at all between the two models you have used. Both are the same thing and are equally applicable. If you had a chiral auxiliary on boron or something of that sort, the story might be different. But that isn't the case here.


text:
Why isn't orthoboric acid soluble in cold water even though it has polar B-OH bonds?
This is the information I found regarding orthoboric acid's solubility:

Textbook (NCERT India): Sparingly soluble in water (didn't say what temperature - presumably cold) and highly soluble in hot water.
Wikipedia: "Boric acid is soluble in boiling water."

My question: why is it only sparingly soluble in cold water even though it has polar $\ce{B-OH}$ bonds? Also, what happens on increasing the temperature that increases the solubility?

It's mostly because orthoboric acid doesn't react with water the way other acids do, while other acids dissociate into protons and conjugate bases, orthoboric acid actually absorbs the hydroxyl from the water, thus releasing protons from the water itself. I'm sure a better answer is on the way though...

The pKa of all the hydrogens are 9.2, 12.4, and 13.3, so it doesn't dissociate in neutral water (https://en.wikipedia.org/wiki/Boric_acid). This also suggests the molecule is less polar, so solubility will be lower.

see section 2.1 of this: https://digital.library.unt.edu/ark:/67531/metadc843233/m2/1/high_res_d/1025802.pdf


text:
Is there always a limiting reagent in every equation?
Is there always a limiting reagent in every equation? For instance, is there a limiting reagent in $\ce{CaCl2 + Na2CO3 -> 2NaCl  + CaCO3}$ where the same number of moles are reacting with each other?

Considering that there are $6.022\times10^{23}$ molecules per mole, it seems unlikely that an exactly 1:1 ratio could ever really be achieved.

@MaxW It is possible using e.g. AFM tips to nudge individual atoms and molecules and cause them to react, which I suppose counts as exact in a very technical manner.

An interesting philosophical question. We are given a chemical equation: $$\ce{CaCl2 + Na2CO3 -> 2NaCl  + CaCO3}$$ and that the same number of moles of the reactants are used. Now the OP poses the question "Is there a limiting reagent?"
As chemists we can say that 1.00 moles of $\ce{CaCl2}$ were reacted with 1.00 moles of $\ce{Na2CO3}$. We'd pragmatically say that the moles were equal and that there was no limiting regent.
However this is a significant figure argument, not reality. Avogadro's number is about $6.022\times10^{23}$. There is no lab in the world that can weight out precisely one mole of $\ce{CaCl2}$ accurate to 23 decimal places. We don't even know Avogadro's number accurate to 23 decimal places! The best value is $6.022140857(74)\times10^{23}$.
So in reality, despite our best efforts as chemists, one reagent or the other will be in some small excess which we'd consider to be insignificant given the overall precision of the experiment.

Note that, since 20 May 2019, the Avogadro number is (by definition) exactly $6.022,140,76\times10^{23}$.

There can't be any limiting reagents in the equations. Equations are purely theoretical expressions and are always balanced in terms of moles.
"Limiting reagents" arise in real world chemical reactions.

The questions on limiting reagents are usually formulated in a form of "we have a reaction according to this equation, and we have X grams of this reagent and Y grams of that reagent, which one is limiting?"
A question "which reagent in this equation is limiting?" without additional info makes no sense.

The OP has clearly stated "same number of moles are reacting with each other", so the amount of the reagents are given. Using the molar masses you can get to the actual weight. What additional info is required?

I'd like to explain this problem with an example. The chemical equation in question is:$$\ce{CaCl2 + Na2CO3 -> 2NaCl  + CaCO3}$$
Since $K_\mathrm{sp}$ of $\ce{CaCO3}$ is $\pu{3.36\times10^{-9} mol^2\cdot L^{-2}}$, it is safe to assume that this reaction would have proceeded to completion, if $\pu{1 mol}$ of $\ce{CaCl2}$ is reacted with $\pu{1 mol}$ of $\ce{Na2CO3}$ to give $\pu{1 mol}$ of $\ce{CaCO3}$ solid and $\pu{2 mol}$ of aqueous $\ce{NaCl}$. Therefore, no limiting reagent in that calculation because we used exact stoichiometric amounts of starting materials.
What if you have mixed $\pu{100 mL}$ of $\pu{2 M}$ $\ce{CaCl2}$ with $\pu{100 mL}$ of of $\pu{1 M}$ $\ce{Na2CO3}$? You can calculate how many mols in each solution:
$$
\begin{align}
\text{Number of moles of}~ \ce{CaCl2}~ \text{in solution} &= M_1V_1=2~\pu{mol L-1}\cdot \pu{0.100L}=\pu{0.2mol}\\
\text{Number of moles of}~ \ce{Na2CO3}~ \text{in solution} &= M_2V_2=1~\pu{mol L-1}\cdot \pu{0.100L}=\pu{0.1mol}
\end{align}$$
Now, according to the chemical equation, only $\pu{0.1 mol}$ of $\ce{CaCl2}$ would react with available $\pu{0.1 mol}$ of $\ce{Na2CO3}$ to give $\pu{0.1 mol}$ of $\ce{CaCO3}$ solid and $\pu{0.2 mol}$ of aqueous $\ce{NaCl}$. The rest of $\pu{0.1 mol}$ of $\ce{CaCl2}$ would remain unreacted in the solution. In this situation, we called $\ce{Na2CO3}$ is the limiting reagent because all of them would be consumed by the reaction. In other word, $\ce{CaCl2}$ would be in excess.

Okay, your entire solution revolves around a given example, but is missing a conclusion. So, are you implying that whether there would be a limiting reagent or not depends on the amount of the reagents used, correct?

@ Gaurang Tandon: correct. There is always a limiting reagent, if at least one of the starting materials is in excess in the reaction mixture.


text:
What is number of 1H NMR signals for toluene?
I am searching the web for the number of 1H NMR signals of toluene and I find two different results. Some say there are two signals while others say there are four (According to me, it should be 4). For the two signals, it suggests that all five aromatic hydrogens are taken as equivalent. A similar problem arises for compounds like ethylbenzene, benzyl alcohol, and anisole. How can I reconcile these different results?

In general, the aromatic region of any $\ce{^1H-NMR}$ spectrum would display a complex coupling pattern even if for mono substituted nucleus such as toluene. However, your $\ce{^1H-NMR}$ spectrum is taken by using a low field NMR instruments such as 60 MHz, you'd not able to get one with good resolution to show these patterns (other factors involve in having broad multiplets instead of resolved peaks are low quality calibration, RF instability, etc.). The following spectrum is taken using a high resolution NMR instrument. The $\delta~7.70-7.70$ region was expanded to show the coupling pattern of 5 aromatic hydrogens. In a low-field instrument, those peaks would overlap to give you a broad multiplet. Read this article to learn about basic principles of coupling patterns.

Thanks for answering my query, but then what should I answer for a question when asked me about number of signal should I consider it as a high resolution ?

The phrase, "number of signals"  has a broad meaning, and it is based on the person who asking the question in this case. For example, there are signals for three magnetically different hydrogens in this spectrum. But, your teacher may consider a doublet as 'two signals.'

A lot of times in answering questions like this it is best to state your assumptions. So when answering the question -- "Assuming that the number of signals means the number of peaks, then..."

@ MaxW: Thank you for your valuable instructions.

Question should ask for the number of non equivalent protons rather than for a variable depending on experimentals. Or both the asker and whom is asked should be open for circumstances


text:
Calculating surface area from BET results
I have been trying to work out the workings of the BET method and I have come to a standstill.
The surface area of a sample is given by
$$
S_{\text{BET}} = \frac{\nu_{\text{m}}Ns}{V}
$$
where $\nu_{\text{m}}$ is the amount in the monolayer (given in $\mathrm{cm^3/g}$), $N$ is Avogadro's constant, $s$ is the adsorption cross-section of the adsorbate ($\pu {0.162 nm^3}$ for $\ce{N2}$) and $V$ is the molar volume of the gas ($\pu {22.4 L/mol}$).
If I run this sample through a commercial software that comes with the adsorption equipment, for a given mesoporous sample, it gives me $\nu_{\text{m}} = 209.6160~\mathrm{cm^3/g}$, and a final result of $S_{\text{BET}} = 912.5 ~\mathrm{m^2/g}$.
If I try to calculate this myself (converting everything into SI units), however, I get that same result, but divided by a factor of $1\times 10^9$.
I think I'm making some stupid unit conversion mistake, but I can't see it.

Your problem is not the unit conversion, but the units you have used. The adsorption cross-section of $\ce{N2}$ ($s$) has been found using BET method is $\pu{0.162 nm^2}$ (Langmuir). Yet, you have used $\pu{0.162 nm^3}$, more than a factor of $1 \times 10^9~\mathrm{nm/m}$.

It is always a good idea to use dimensional anaylsis to check an equation if something seems to be "going wrong". Also you can use it to check equations if you are not sure if the equation is right.

Thanks, I knew it was something stupid that I failed to check.
Thanks also for the edits, I'm not very familiar with S-X's LaTeX markup so I wasn't sure how to enter units.


text:
What is the future of water treatment, and why we don't use enough purified sea and freshwater?
I understand thermal and membrane distillation are very expensive and that slowed down their usage, but now we can bring 100% clean energy to power them. Except the set-up, the price of materials and the maintenance, I understand these other costs are not that heavy compare to energy from the grid ?
Many global companies have been building global purification centers but I'm thinking if local initiatives can be cheaper and interesting even in developed countries ?


text:
How to tell whether lone pairs in heterocycles contribute to aromaticity?
How can I decide whether lone pair electrons take part in the continuous overlap of electron or not?
For example, the lone pair in pyridine is not a part of the conjugated system, whereas the lone pair in pyrrole is.
I had read that this was due to the lone pair in pyridine being localised in an sp2 orbital, while the lone pair in pyrrole is delocalised.
How do we identify this?

What would it look like if the pyridine N would put its pair to the p-orbital and the remaining electron to the sp2 that points outward? A wannabe-aromatic ring with 7 electrons and a lone unpaired electron on sp2.

Related: Why is only one lone pair in imidazole delocalised?; Why is the lone pair of pyridine's nitrogen atom not part of the aromatic ring?

For the most common situations use this rule: a conjugated ring atom needs either a ligand or a localized electron pair to replace a missing ligand.  Any remaining electron pairs are then part of the conjugated pi system.
Thus pyridine needs a localized lone pair on nitrogen to replace a missing ligand, and that leaves only three bonding pairs for the pi system.  Pyrrole has a ligand on nitrogen so the lone pair is conjugated into the pi system along with two carbon-carbon bonds.  Furan has two lone pairs on oxygen but a missing ligand, therefore oxygen contributes one pair to the pi system.

Okay so whenever the contribution of lone pair may result in aromaticity, it will always participate in the conjugated system ? Also thanks!

The ability to form an aromatic ring is favorable, but in most cases the sigma orbital requirements come first.  Check that, then count what's left.


text:
Reaction of dihalides with excess of NaNH2
Gem-dihalides react with $\ce{NaNH2}$ to give terminal alkynes. Why is a terminal alkyne prefered?
For example,

I had thought that 2-pentyne would be preferred over 1-pentyne as it would be more stable due to hyperconjugation, but it says the final product is terminal alkyne.
Why?

It's all about how the reaction is run. 2-Pentyne can form under milder conditions. More vigorous conditions induce the "zipper reaction" producing the anion of 1-pentyne. 1-Alkyne (pKa ~ 25; NH3, pKa ~ 35.

this once was my doubt...terminal alkynes are preferred because the acidic hydrogen at the terminal carbon is acidic(sp hybrid of Carbon) and which when taken up by the base gives a species with a negative charge on sp hybridized carbon which is indeed stable...so this provides the incentive for forming a terminal alkyne as the final product.
Trusted source: Jerry March Organic Chemistry 6th edition.

When the base is NaNH2 1-alkynes predominate (where possible), because this base is strong enough to form the salt of the alkyne, shifting any equilibrium between 1- and 2-alkynes. When the base is $\ce{OH-}$ or $\ce{OR-}$, the equilibrium tends to be shifted to the internal alkyne, which is thermodynamically more stable.


text:
Melting point of trans-4-methylhex-4-enoic acid and cis-4-methylhex-4-enoic acid
My book said that the cis isomer has a higher melting point. But I dont understand isn't that the trans isomer can pack more efficiently, because the trans isomer less bulky/ create less 'kink' than the cis isomer?

Between Trans and Cis isomer Cis has the boiling point as it's µ ≠ 0 ,Both of the isomers have exactly the same atoms joined up in exactly the same order. That means that the Van der Waals dispersion forces (London Forces) between the molecules will be identical in both cases. The difference between the two is that the Cis isomer is a polar molecule whereas the trans isomer is non-polar.And since Trans form is symmetric along CH₃ molecule, the main thing to check is polarity hence boiling point is compared.

It's probably fair to say that the VDW forces are comparable, but it's certainly not identical.

the dipole moment of the trans isomer is not zero as well.

If I understand correctly, part of the issue stems from your question referring to the cis/trans isomers differently from what's shown in the image. In the answer to 17(a,b) the book identifies X as the trans isomer, but your question appears to describe X as being the cis isomer with higher boiling point.
I took a look in Scifinder but it doesn't seem like there is isomer-specific experimental thermal data for the two compounds. I would argue that a "packability" judgement is pretty difficult to make with this pair of isomers because of their low symmetry and the minimal difference between isomers.

so, which one should be cis and which one is trans?

Some of this ambiguity may be due to the use of cis and trans, rather than the preferred E/Z. Some people may refer to C/C=C(C)/CCC(=O)O (a SMILES string) as the cis isomer because the two methyl groups are cis, whereas others refer to C/C=C(C)\CCC(=O)O as the cis isomer because the two highest (and two lowest) priority groups via the CIP system are cis. Can this be clarified.


text:
Calculating the EMF of cell

Find the EMF of the following cell :
  $$\ce{Pb(s)}, \ce{PbSO_4}|\ce{SO_4^{2-}}(\pu{0.100M})||\ce{Pb^{2+}}(\pu{0.004M})|\ce{Pb(s)}$$
  Given:  $E^0_{\ce{PbSO_4|Pb,SO_4^{2-}}}=\pu{-0.359V}$
  and $E^0_{\ce{Pb^{2+}|Pb}}=\pu{-0.126V}$

I first found the $\ce{Pb^{2+}}$ concentration in the oxidation half cell using the sulphate ion concentration and the solubility product of lead sulpahte ($2.53\times10^{-8}$) and found the concentration to be $(2.53\times10^{-7})$
And accordingly $E^0_{\text{cell}}=0.359-0.126=0.233$
Then using the Nernst equation,
$$E_{\text{cell}}=E^0_{\text{cell}}-\frac{RT}{nF}\ln\frac{[\text{Products}]}{[\text{Reactants}]}$$
And substituting,
$n=2$,
$[\text{Products}]=2.53\times10^{-7}$,
$[\text{Reactants}]=0.004$,
$T=\pu{298K}$,
$R=\pu{8.314JK^{-1}mol^{-1}}$, and $F=\pu{96500C}$. Hence, I got $E_{\text{cell}}=\pu{0.357V}$
Whereas, the answer given in my book is $E_{\text{cell}}=\pu{0.133V}$.
So, is my answer correct or have i  misunderstood something?

Sorry for being rusty, but can you clarify the setup more? What reactions are happening, are the cells compartmentalized and what exactly are the values given in brackets in the question?

Use RT/F = 0.059 or 0.06
It will simplify your calculations.
Note:- I have converted ln into log.
So formula becomes:-
$$E_{\text{cell}}=E^0_{\text{cell}}-\frac{0.059}{n}\log\frac{[\text{Products}]}{[\text{Reactants}]}$$
Just try this in problems. Calculations would become easier.

But the answer i am getting by doing these calculations is 0.357 whereas the answer given in the book is 0.133.
So my question is where exactly am i making a mistake..


text:
On the addition of HCl to a conjugated diene

Is this allylic chloride formed by an ionic or radical mechanism? Is it the more stable isomer?

Don't base product distributions on thermodynamics unless there is an equilibrium that can be established. You should be looking at kinetic considerations instead.

I wonder if this actually is a free radical reaction. Afaik, the peroxide effect operates only in the case of HBr...

Yes you are right.

But the question still remains unsolved.

Which is the best site for the carbocation?

Peroxides affect the regioselectivity with HBr. It will certainly help promote the radical pathway here.

Hmm, well, maybe not: https://pubs.acs.org/doi/abs/10.1021/ja01650a041?journalCode=jacsat

Benzylic carbocation is formed as it is stable due to resonance.
So Cl attaches to benzylic position.
More substituted alkenes are more stable so the terminal double bond changes its position.
(Also this isn’t free radical reaction)


text:
calculate SPE using dummy/ghost atoms in NWChem
I have been making some failed attempts to calculate interaction energy of 2 monomers using a dummy/ghost atom in NWChem6.5.
The code I was using to calculate the interaction energy is:
title 'bond_energy'
start dimer
geometry units angstrom
  N 5.286 28.124 48.085
  H 5.966 28.410 48.528
  O 7.110 29.056 50.252
  C 7.758 28.682 51.230
end
basis "ao basis"
  * library 6-311+g*
  bqN library N 6-311+g*
  bqH library H 6-311+g*
  bqO library O 6-311+g*
  bqC library C 6-311+g*
end
dft
  xc m06-2x
  direct
  grid fine
  iterations 10000
end
bsse
  mon first 1 2
  mon second 3 4
end
task dft energy

However, I'm not being able to replace H atom with a dummy/ghost atom (with no charge) in the calculation.
If I use bqH 5.966 28.410 48.528 in the geometry section and calculate odft then also the calculation an error.
Any suggestions or modification needed in the code to achieve this target will be highly appreciated.
Thank you.

You should start by just copy-and-pasting the example using their BSSE correction example here.  If not, you can just do the BSSE correction by hand by specifying the ghost atoms in the script and giving a task directive below each geometry.

I don't want to do that. I want to calculate the bsse energy of NHOC bond, in absence of H atom. That's why I'm trying to replace it with dummy atom.

Using the notation @ for a ghost atom, does this mean you want the BSSE of $\ce{N@H\bond{...}OC}$? That is, keep the hydrogen as a ghost atom during all 5 parts of the BSSE calculation? If so, you may have to do the whole thing "by hand" and piece the BSSE expression together manually, otherwise the program will try and replace a ghost atom with a ghost atom, which is an edge case.

Something else that's missing: what do you expect the charge and spin multiplicity to be on each of your fragments?

Sorry, 3 calculations, not 5. (AB in the basis of AB) - [(A in the basis of AB) + (B in the basis of AB)], all with ghost H.


text:
Effect of halogen atom on Markovnikov's rule
I know that the Markovnikov's rule is because of the stability of carbocations. If a halogen atom (electron-withdrawing group) is directly attached to the carbon atom carrying a positive charge, would it destabilize the carbocation (for example dichloroethene reacts with HBr)and 'break' the Markovnikov's rule?

After the halogen attaches, there's no longer a carbocation. It's just like any other alkyl halide.

Markovnikov's rule is just a simple rule for predicting the structures of  alkenes in some reactions.
It is not advisable to apply this rule to all cases,it is often helpful to check the relative stability of carbocation intermediates.
In your question the carbocation will be stabilised by +M
effect of halogen as halogens have lone pair of electrons.


text:
How to tell the difference between octane-1-ol and oct-1-ene?
How to tell the difference between octane-1-ol and oct-1-ene? The things I can do are:

check their solubility in water
use bromine water
use $\ce{KMnO4}$
use sulfuric acid

I assumed that checking the solubility would be a good idea because of the presence of an alcohol but will I notice a difference since the alcohol has 8 carbon atoms. And they all react with bromine water, $\ce{KMnO4}$ and $\ce{H2SO4}$. I think I'm missing something.

Do you have access to IR or NMR?

Alkene is best characterized, then, by its property
of decolorizing both a solution of bromine in carbon tetrachloride  and
a cold, dilute, neutral permanganate solution (the Baeyer test). Both
tests are easily carried out; in one, a red color disappears, and in the other, a purple color disappears and is replaced by brown manganese dioxide.

See whether in addition of Br₂ in CCl4 it gives red color or not ,It is a test to see the presence of the double bond and you can try Lucas test on the alcohol

In a decent university chemistry lab you could try treatment with ozone, followed by triphenyl phosphine or reduction with zinc / acetic acid and then steam distillation. The octene will form formaldehyde and heptanal while the octanol should do nothing.


text:
Activation of alcohols by forming sulphonate ester
In activation of alcohol by tosyl chloride, why does the chloride ion not displace the poor leaving group?
I think that's because chlorine is a poorer base than that thing which will be formed. But that should not be true considering delocalisation.
Additional Information:
I know this question is unclear because I don’t know the name of the things formed when alkyl group has left the sulphonate ester so please edit this.

Chloride can displace the tosylate group if one is not careful in the preparation of the tosylate. Usually, the tosylation is done in the cold.

Yeah, it can happen.

This is particularly observed with tbe std conditions of Et3N/MsCl/DCM for mesylate formation where Cl displacement of the mesylate can lead to significant amounts of the chloride

@user55119 Two questions: Even if tosylate gets displaced we get alkyl hallide so that is not bad right? And if the preparation is done in cold due to activation energy barrier rate of displacement will be very slow is that true?

If you are forming the sulfonate to then displace with a nucleophile (or to eliminate) then the presence of chloride is of no significance as it will follow the same pathway, particularly if you add catalytic iodide along with the nucleophile. Cooling should reduce the amount of displacement, but it will also slow the formation of the sulfonate.


text:
No phenolphthalein mono-anion?
This is a follow-up to the question here.  There is no mono-anionic stage shown, does it "disproportionate" to the neutral molecule and the di-anion?

It just means second pKa is not significantly higher then first.

So maybe when we form the mono-anion the conjugation of the two phenolic rings favors driving the next proton off to form a highly stabilized "quinone-aromatic" structure, thus lowering that pKa?

This is an answer to a question of mine, which I posted a few days ago, Deprotonated form of phenolphthalein.
I found this paper (it's in Japanese, but the relevant things are legible) https://www.jstage.jst.go.jp/article/yakushi1947/117/10-11/117_10-11_764/_pdf, which says that phenolphthalein has 2 pKas, pKa1 = 9.05 and pKa2 = 9.50. It also says that the pink form is the twice deprotonated; the monoanionic form exists, but is colourless.


text:
Sign of work done by the system in first law of thermodynamics equation
In the equation, $\Delta E = q - W$, according to my book, $W$ is positive when it is done by the system and negative when it is done on the system.
But in this problem:

$\pu{3600 kJ}$ heat is entered in a gas cylinder, calculate the value of $∆E$ if the volume is not constant and work of $\pu{-800 kJ}$ was performed by the gas.

the sign is different. How should I make the equation?

Who says the sign is different?

Here the work was done by the system but it is stated as negative.

All that means is that the surroundings in this problem are doing work on the system.  Have you never worked with negative numbers?

@ChesterMiller is "Have you never worked with negative numbers?" really necessary? Please be nice to users. Thanks!

I always got lost in remembering the rules for the signs, so I just ignore them and think about the two parts from the standpoint of the system (cylinder):
If heat enters the system, then it gets hotter so that contribution is positive. This part's fairly intuitive.
If I let the cylinder expand, it's relaxing to a less compressed state and so loses energy, which means the work contribution is negative. In order to compress the cylinder I have to push on it, adding some of my own energy to the system, so the work contribution is positive.
In your case, the change in energy from the heating (positive), and loss of energy through expansion (negative) is:
$$ \Delta E = 3600kJ - 800kJ $$
Edit: The problem itself is ambiguous regarding whether it's expansion of contraction. My answer assumes expansion since that's what would naturally happen upon heating a gas cylinder. Gaurang's answer is correct if you read it as contraction.

I am not sure if I understood the main part of your answer. That said though, our final values are contradictory. Please have a look at my answer below.

We're both saying the same thing, but we're interpreting the question differently: expansion vs contraction. I assumed expansion because if you added heat to a cylinder the gas would naturally expand.

Since in chemistry, work $=-P\Delta V$, that means there definitely is an expansion in the given question (since work given is negative). But, you seem to imply that I assumed contraction of volume. Though, I don't know why. I never said in my answer that I'm assuming contraction. In fact, I didn't even mention volume changes at all! ;)

Oh, so we agree it's expansion then? A compressed piston has a higher potential energy than an expanded one, so the when the cylinder expands it has to lose energy. The only way this works is if the $w_{PV}$ winds up being subtracted from the heat.

No, sorry. With that last comment I ended up contradicting myself. You're right, though, I was assuming contraction. This problem is indeed ambiguous.

You can interpret the process of work in the reverse direction. A work of $\pu{-800kJ}$ is done by the gas, that implies $\pu{800kJ}$ work is done on the gas.
Now, since positive work is done on the gas, positive energy would be supplied to the gas. Recall that work is essentially the transfer of energy. So, here the internal energy of the gas would increase.
Thus, final $\Delta E$ would be $(\pu{+3600kJ}) + (\pu{+800kJ})=\pu{+4400kJ}$.


text:
Different values of concentration of product at equilibrium

$\pu{60g}$ of acetic acid and $\pu{46g}$ of ethyl alcohol are mixed with each other at a constant temperature and allowed to attain equilibrium. At equilibrium $\pu{58.2g}$ of ethylacetate and $\pu{12g}$ of water were present: find the concentration equilibrium constant ($K_\mathrm{c}$)

Here is my solution:

The correct answer is 4 but I'm getting 3.7 something. When I calculate x with the help of made of water it is slightly different than when I do it with the mass of ester. Shouldn't they both give the same answer? If not, which substance should I use?

In the given image, your $K_\mathrm{c}$ is indeed exactly equal to 4. (last step -> $K_\mathrm{c}=\frac{0.66^2}{0.33^2}=4$). Is there any other calculation page that you forgot to show us here where you're getting 3.7 as the answer?

No, the calculation is wrong. 1 - 0.66 = 0.34. Then Kc would be  3.7. The question is: can two products with the same molar ratio can have different concentrations at equilibrium?

0.666 is actually the approximated value for 2/3. So, 1-0.66 is more accurately 1-2/3 which equals 1/3 which equals approximately 0.33 as expected.

The concentration of ethyl acetate at eq. is 0.661363636 and the concentration of water at eq. is 0.666666666. That's a slight difference but they give different values of Kc: 3.7 and 4 respectively. Is that an error in the given data that the two products with the same molar ratio have different concentrations at equilibrium or is there some rule by which I should prefer one product over the other for determining x which in turn helps calculates Kc?


text:
How to synthesize tetrabutyl titanate from titanium dioxide powder?
How do you synthesize tetrabutyl titanate $(\ce{[CH3(CH2)3O]4Ti})$ from titanium dioxide ($\ce{TiO2}$) powder? As I know, we can get $\ce{TiO2}$ from tetrabutyl titanate, but how can we do the opposite?

Welcome to Chemistry.SE. We’d like you to take the Tour to get familiarized with the site. Meantime, let us know your thoughts about it as well. It’d help us to agree or disagree and give our feedback.

there is not practically viable way to do at at lab. It thousands time cheaper to go with titanium tetrachloride.

Treat with a chlorination reagent the titania to form titanium tetrachloride. You could do this with coke and chlorine in an electric arc furnace. Then distill the titanium tetrachloride. Next react it with a mixture of sodium butoxide and butanol at a low temperture.
Next vacuum distill the product to obtain pure titanium butoxide


text:
Bridged ring system reduction through Wolf Kisher: Is it possible?

Is it possible for  a ketone like this to undergo the Wolf Kishner mechanism? If so, what would be the product? It seems unlikely to me as disrupting the stable equitorial carbonyl seems unfavorable...

If you are thinking of reducing the ketone with hydrazine under alkaline conditions then I see a problem. The hydrazine will also act as a nucelophile on the alkyl bromide in the molecule. I think that you will have a SN2 reaction on the alkyl bromide to form an alkyl hydrazine at the same time as the hydrazine is reacting with the ketone. If you then heat it up with base you will remove the ketone but assuming that you use a large excess of hydrazine you will end up with an alkyl hydrazine in the top left of the molecule rather than an alkyl bromide.

Why not react the original compound with benzyl alcohol using a mild base to form a benzyl ether at the start. Then do the reduction with hydrazine, then hydrogenate with hydrogen gas and palladium on carbon. This hydrogenation will remove the benzyl group and give you a alcohol. Then you can react the alcohol with tosyl chloride/pyridine and use the tosylate as an electrophile. If you very much want an alkyl bromide then react the tosylate with sodium bromide in DMF.


text:
What bond exists in "red oxygen", and can something similar happen with hydrogen?
"Red oxygen" is the nickname for O$_8$. It only exists at extremely high pressures.
What kind of bond does it have? Can hydrogen theoretically have something similar at very high pressures too?

https://chemistry.stackexchange.com/questions/37898/how-are-the-4-molecules-of-o2-bonded-in-the-o8-molecule-at-low-temperature-and-h

That is what I found on wikipedia. It has a cubic structure and it was not predicted theoretically.
here the pressure is not very high but is high(10 GPa)
High-pressure studies of hydrogen typically employ a diamond anvil cell (DAC) made of two diamonds holding a few-micrometer-thick gasket containing a tiny sample of hydrogen.
These are some more images I could found of hydrogen at high pressures, but I do not think any of them are like red oxygen.

Sources

Bonding, structures, and band gap closure of hydrogen at high pressures

This is a must read: Solid Hydrogen at High Pressure

Hydrogen forms similar clusters as cations (here).  Some do not even need high pressure.  The trihydrogen cation, with a triangular structure resembling an aromatic cyclopropenyl cation ring, is found in interstellar space and some planetary atmospheres, and the six-atom cation with its peculiar "turnstile" strcture (two triangles joined vertex to vertex and twisted 90 degrees from each other) is observed in solid hydrogen.


text:
How to calculate the percentage yield for enantiomers in free radical halogenation?

How many mono-chlorinated products of 2-methyl butane will be formed upon chlorination in the presence of sunlight?
Find the percentage of major product formed.

Six products were formed: two pairs of enantiomers and two achiral chloroalkanes.

Until here my answer was correct, confirmed by my teacher.
For the second part that is the percentage yield I had an issue with the enantiomers. I calculated the relative yield ratios by multiplying probability factor and rate factor.
For cases A and B, where enantiomers are being formed, I further multiplied $p_\mathrm{f} \times r_\mathrm{f}$ by half since there's an equal likelihood of each of the enantiomers to form. My teacher said this is wrong, as we do not multiply it by half.
So, my question is why?

I'll say that this is just a matter of convention, and not really worth worrying about. Your underlying method and concept is 100% correct. Your final question boils down to:

Are enantiomers separate chemical compounds?

To elaborate, if the question was:

Find the percentage of major product formed, counting the enantiomers separately.

Your approach would have without doubt been correct.
But since the question only says:

Find the percentage of major product formed.

It is difficult to ascertain what the question wants. Enantiomers themselves have different chemical reactions - so they may be interpreted as different chemical products. But, your teacher has interpreted them as a single compound, presuming that optical isomers are just one compound.
Again, I guess this is just convention, because I haven't been yet able to find references to favor either definition.

Finally, to satisfy both the parties involved, and write a foolproof answer that guarantees a complete score, you can frame your results like so:

Percentage of (R)-2-chloro-3-methylbutane = $x$
Percentage of (S)-2-chloro-3-methylbutane = $x$
Total percentage of 2-chloro-3-methylbutane = $2x$

where $x$ is your calculated value for a single enantiomer.


text:
How to determine which ions will pair to form a compound in an aqueous solution?
Let us assume we have an aqueous solution with the ions $\ce{A+}$, $\ce{B+}$, $\ce{C-}$ and $\ce{D-}$. Which ions will form a bond and why? Will the compound formed be $\ce{AC}$ and $\ce{BD}$ or $\ce{AD}$ and $\ce{BC}$?
Is there any general rule to identify the compounds we have at the end?

Your question arises because you believe that ions in an aqueous solution remain bonded with each other. That isn't true though! In fact, ions in an aqueous solution freely roam around throughout the solution.
Note that most ionic compounds have ions bonded to each other in solid state only. Once you dissolve that ionic compound in water, the ions quickly dissociate (the ionic solid dissolves) and become solvated (surrounded by water molecules).
So, to answer your current question: no, there won't be any compound formed, and all the ions would be separate from each other.
Though, another possibility is of precipitation. In that case, the ions involved ($\ce{AD}$ for example) would separate out of the solution and settle down as a solid. Indeed, there may be more than one possible precipitates. So, to decide between any given number of compounds, as to which precipitates first, we'll need to check if their ionic product exceeds their $K_\mathrm{sp}$. The compound for which this happens earlier, due to lower $K_\mathrm{sp}$, would precipitate earlier.

Bonus content: Your question would make more sense if the ions were given in gas phase instead. Now, when the oppositely charged ions would come together, they would stabilize each other via the large amount of electrostatic energy released, and thus forming a stable compound.
Whether $\ce{A+}$ would combine with $\ce{C-}$ or $\ce{D-}$, would depend on their energy of interaction $\left(=\displaystyle -\frac{Kq_1q_2}{r}\right)$. More the energy released for a particular bond, more would be the probability of formation of that bond.


text:
What is the difference between nuclear charge and ionic charge?
There was a question asking why magnesium metal has a higher melting point than sodium.
In the answer, they said:

Magnesium has more delocalised electrons.
Magnesium ions have a greater positive charge.
Magnesium has a greater electrostatic attraction between ions and delocalised electrons.

However, in the mark scheme in the additional guidance, it also said the mark couldn't be given if you said:

Magnesium ions have a great nuclear charge rather than greater positive charge.

I was confused as to what the difference is.

What is your weight? Now what is the weight of your skeleton? Is there any difference?

Comments are not for extended discussion; this conversation has been moved to chat.

Nuclear charge is the charge present inside a nucleus, due to the protons. For example, a carbon atom has $+6e$ nuclear charge.
Ionic charge is the charge due to the gain or loss of electrons in the valence shell. So, the sodium ion has an ionic charge of $+1e$ (due to loss of one valence electron). That ion's nuclear charge is still $+11e$ (there are still 11 protons only).
The reason because nuclear charge as an answer to this question is unaccetable is because nuclear charge - being inside the nucleus - has largely no affect on the intermolecular forces between the magnesium atoms. Those magnesium atoms have intermolecular forces governed only by the valence electrons only.


text:
What products come from the electrolysis of sodium bicarbonate?
I did an experiment finding the electrolysis of various solutions. One of them was sodium bicarbonate and tap water. I want to be able to scientifically explain my results, using chemical equations and explanations of those equations. Can somebody help?
-Thanks


text:
Which are the major truths of paradigms on which the whole body of chemistry is built?
For example, as evolutionary theory is fo biology or the existence of numbers for math, the existence of an independent reality for physics, etc.

The periodic table

Well the most important driving force I see in chemistry is for atoms and molecules to be most stable when the atoms have a noble gas electronic configeration. Many reactions will occur to try to reduce the energy of the atoms and molecules by gaining / losing or rearranging electrons to bring as many atoms as close as possible to a noble gas configeration.
Another important driving force is to reduce or avoid the mechanical strain in a molecule (Steric effects)
There are effects in pi systems such as aromaticness (4n + 2 pi electrons in rings)
There are entropy effects, by increasing the entropy of a system then an endothermic reaction can be made to occur. For example the dissolution of ammonium nitrate in water occurs with ease but this is an endothermic process.