text:
Mechanism for reaction of NOCl (Tilden reagent) with primary amines
According to The Pearson Guide To Organic Chemistry For The IIT JEE by Atul Singhal, the reaction between $\ce{NOCl}$ and $\ce{CH3CH2-NH2}$ is as follows:

$$\ce{CH3CH2-NH2 + NOCl -> CH3CH2Cl + N2 + H2O}$$

But how exactly does the reaction proceed?
The first step (what I think) could be that $\ce{NOCl}$ dissociates into $\ce{NO+}$ and $\ce{Cl-}$ ions and then the $\ce{Cl-}$ ion attacks the amine group and replaces it by $\mathrm{S_N2}$ mechanism.
But according to paperplane.io (via the Internet Archive), a diazonium product is produced. Even though a diazo product is formed, how does it produce an alkyl halide according to my textbook?

If you know how to form a diazonium salt with NOCl (or HONO/HCl) then you are an SN2 reaction away from the answer.

Please cite the textbook (as in 'according to my textbook'), or leave out that statement if it is not relevant. Always cite the sources you are using.

Why would the chloride ion attack the electron rich amine? The amine is the better nucleophile in the system, and will attack the $\ce{NO+}$. This gives a $\ce{RNH-NO}$ nitroso species which is then protonated on oxygen, loses water, and gives the diazonium cation. The resultant diazonium cation loses nitrogen rapidly and get attacked by chloride to give the primary alkyl chloride.
More can be read about this process here (which is where I sourced the reaction scheme) or by searching the term "nitrosation of primary amine"

I don't understand how amine is a better nucleophile than chloride ion.

@praig_ Well, what you don't understand? You have some serious misunderstandings there. Probably wanted to get too much while not knowing necessary facts. This is such fact. You can get amine from halide and ammonia, not otherwise (unless it's silyl amine)

@user85683: You may need to read this to start.


text:
How does a vitamin differ from a mineral
At the molecular level, I would like to know fundamentally how does a vitamin differs from a mineral. I understand both have, to say, Fe (Iron), then what exactly makes it a mineral Iron or a vitamin Iron.  Does the properties of Iron changes from Vitamins to minerals?

Iron is an element, and minerals and vitamins are classes of chemical compounds. You can ask the same question about oxygen in quartz and oxygen in vitamin A. Same element, different compounds with different properties, not much to say, really. There is no such thing as "mineral element" or a "vitamin element".

Since you are asking about vitamins vs. minerals, I assume what you mean by mineral is a "dietary mineral" (as opposed to the more typical meaning of mineral, which is a naturally-occurring inorganic compound).  So:
A dietary mineral is an element that is essential for life (e.g., K, Ca, Fe, Se, Cr, P, I, Cl, Mo, etc.), and is required in small quantities.  This rules out C, H, O, and N which are essential for life in large quantities as part of carbohydrates, fats, and proteins.[For more info., see: https://en.wikipedia.org/wiki/Mineral_(nutrient)]
For vitamin, I'll just give you the definition from the dictionary on my Apple computer: A vitamin is "any of a group of organic compounds which are essential for normal growth and nutrition and are required in small quantities in the diet because they cannot be synthesized by the body."
Thus while fats, carbohydrates, and proteins are all organic compounds essential for life, they are not considered to be vitamins, because they are used in large quantities as either energy sources or building blocks.
So, in sum, in their dietary usage, a mineral is an essential element needed in small amounts, and a vitamin is an essential organic compound needed in small amounts.

source: Stack Exchange

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

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text:
Can anyone suggest a mechanism for hemiaminal metal complex to imine complex?

Can anyone suggest a probable reason for this conversion? This seems like a simple conversion from hemiaminal to imine.
For secondary amine this kind of conversion might be possible by
but in the case of tertiary amine how can something like this form? Is there a chance of removal of RCH2OH?
The metal here is copper(II). Is it possible that the metal has some role to play?

I think you have to consider the influence of the metal. This cannot be compared to a simple hemiaminal to imine because of the electrons donated to the metal-N complex

The metal used here is copper.

I think the electron donation to copper would weaken the N-H bond and make a proton shift from N to O easier. Once that occurs the loss of water is highly favoured because of the driving force of conjugation of the C=N with the aromatic ring

what about the first case? the nitrogen is tertiary there? this might be the case for secondary amine where there is a presence of N-H bond.

I think a similar weakening of the C-N bond is reasonable

Should that result in the removal of an alcohol? I still can't think of a proper mechanism.

A fairly similar copper complex is discussed in http://dx.doi.org/10.1002/asia.201500889, take a look, probably there is something relevant (or not, I didn't dive deep).

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


text:
Increasing temperature in system of dynamic equilibrium
According to the article Summary of Le Chatelier's Principle on dynamicscience.com.au, for the process

$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}\quad ΔH = \pu{-92 kJ/mol}$$

the forward reaction is exothermic, while the reverse reaction is endothermic.
If the forward reaction is exothermic, increasing temperature decreases $K,$ because it drives the equilibrium backwards, in order to absorb the heat energy.
However, in the graph, there is an abrupt change in the reaction rate of forward and reverse reactions. Why does the forward (exothermic) reaction rate rise up so abruptly, while the reverse (endothermic) reaction does not go up so high?
Shouldn't it clearly be the opposite, as increasing temperature favors the reverse reaction?

Dont try to learn chemistry off dodgy websites ;)

Shouldn't it clearly be the opposite, as increasing temperature favors the reverse reaction?

Yes, they made a mistake. For all the other scenarios, they paired a figure of the rate changes with a figure of the matching concentration changes. For this scenario (increase in temperature) they matched it with a correct figure of concentration changes when the temperature is decreased:

If you change the label in the figure posted by the OP, it would reflect what happens when increasing the temperature: Both rates increase, but the reverse rate increases by a larger factor, resulting in a net reverse reaction until equilibrium is reached again.
Here are sketches for the rate changes both for increase in temperature (low to high, upper panel) and decrease in temperature (back to low temperature, lower panel). As you can see, the reverse rate is more sensitive to temperature (higher activation energy because the product is lower in energy than the reactant for this exothermic reaction):


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Reaction of phenol with concentrated sulfuric acid
What is the product when phenol reacts with conc. $\ce{H2SO4}$ in excess?
I have been told that the major product will be ortho at low temperature and para at high temperature. Why cannot it form both ortho- and para-substituted product as in the case of picric acid?

https://chemistry.stackexchange.com/questions/117608/difference-in-major-product-of-sulphonation-of-phenols

Also: https://chemistry.stackexchange.com/questions/44090/preparation-of-picric-acid

It's a reversible reaction, so at higher temperature the thermodynamic para product is formed from the kinetic ortho product.
Further reading here.

Why does it only form a monosulphonated product and why not a trisulphonated product

Because the mono-product is strongly electron-withdrawn and the electrophile is not as good as NO2+

And because the forcing conditions needed to further sulfonate cause a range of other oxidative reactions leading to a complex mixture.

Why is the ortho product the kinetically controlled product? I would've assumed that the attack at the para position is faster since the ortho position suffers the negative inductive effect of the oxygen.


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Substitute Sr with Mg in MgO
Could someone please help me with the following question.
Due to the perspective of voids in the MgO structure, what is the maximal amount of Sr that can substitute with Mg.
And why is it unlikely that it will happen?
Since the quota between the radiis is 0.84 and the size of the octahedral void is 0.414 the atom should substitute all of the Mg atoms. But is it unlikely because of the size of the Sr ion or is it some other factors involved?


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How to increase the conversion for the reaction between 2-acetylpyridine and diamine?

I tried refluxing in toluene using activated molecular sieves to absorb any water produced (didn't use Dean-Stark trap as it's a very small scale reaction).
This reaction isn't going by changing the time or temperature conditions. So, I added some (0.5 equivalent) triflic acid to increase the electrophilicity of the carbonyl. The reaction is proceeding but conversion isn't high. Is triflic acid a good choice for this reaction? Should I try using equivalent amount of triflic acid to increase the conversion?

Triflic acid is just going to protonate your diamine. Something like Ti(OiPr)4 might be a better option

https://doi.org/10.1016/j.poly.2008.04.028.
This might be helpful. The starting material is slightly different (phenyl group instead of methyl) but I hope following similar procedure might help.

@SoumikDas I have read it already but as I said increasing temperature isn't working for me. here they have used reflux conditions only.

@SuchiGhosh This may be inappropriate to the question you asked. But I just wanted to know do you really need this particular reaction if it's not happening. Is there a possible way to bypass this step if you are doing a multi-step synthesis ? What might be occuring is that due to lot of methyl groups in the product, the conversion rate is not so high due to steric factors, so I was suggesting whether there can be any alternate way to reach the final goal.

Actually I do need to synthesize the imidazolidine ring where there is a methyl substitution at the imidazolidine-pyridine junction. The steric effect is  one of the reasons that this isn't going as expected. But in presence of acid it is converting to the desired product as the carbonyl will be more electrophilic. But I am not sure about the choice of acid.

@SuchiGhosh Hmmm... I see. May be you can try with glacial acetic acid if that works.

You haven't said how you are conducting the reaction. Try Dean-Stark trap and toluene solvent at reflux. No acid.

@user55119 tried refluxing in toluene using activated molecular sieves to absorb any water produced. Didn't use dean-stark as it's a very small scale reaction


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Can we use HPLC to purify an organic reaction product?
I know HPLC is a form of column chromatography that uses high pressure to pump a sample mixture/eluent through a column. However, I'm not sure if this technique is being used to purify and isolate organic products? After doing some research, I found that automated flash chromatography systems use LPLC technique rather than HPLC. I can't understand why high pressure can't be used in these automated systems?

Yes you can. Reverse phase is usually more useful than normal phase, MeCN/water elution. Prep HPLC has saved more of my syntheses than I care to admit to.

You have analytical HPLC, semi-prep (more to do with scale than anything),  preparative and then process scale chromatography.  For commercial purposes,  easy to remove solvents are used, which usually isn't an aqueous system. Glaxowellcome (now GSK), used this on high value products and Novasep use simulated moving bed (SMB) to deliver tonnes of material per year.

Next to HPLC the MPLC may provide advantages of scale and resolution, too (https://www.bio-rad.com/en-us/applications-technologies/introduction-medium-pressure-chromatography?ID=MWHBD3HYP, an application of natural product isolation https://link.springer.com/article/10.1007/BF02258984).

For semiprep applications, you need larger columns, because you cannot just increase the concentration (at least not a lot) with a normal one. You can get along with an analytical column using an automated system running over the weekend, but that too fails if you want to scale up by another order of magnitude.
A HPLC pump that can run at 100x the standard flow rate is likely twenty times as expensive. The column will be much more expensive. The larger volume needs totally different safety considerations. And so on ...
As Waylander commented, you can do preparative HPLC at a pinch and save your day (or your dissertation), but to use it as a standard lab procedure is most often un..sustainable. ;)

Thanks for the explanation. This clears things up!


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How do I scale up to semi-prep from an analytical HPLC
I am looking for an equation that would allow me to change the variables injection volume, column size, and flow rate scaling up from an analytical HPLC to a semi prep instrument.

Best practice is to try and maintain the same column length between scales.  Then the scale up factor is simply the ratio of the square of the radii.
If the lengths are different,  try and keep the aspect ratio in the same ballpark e.g. if the length to diameter ratio is not the same but between 4 to 5 for both, you're in a better place. Then simply scale by the ratio of the two column volumes, using the radii calculated from the internal diameters.
Using the ratio of column volumes should work in other circumstances but scaling from a long, thin column to a pancake may cause some issues.
Make sure to keep your particle size of your stationary phase the same too so you have similar packing.
All things being scaled appropriately,  the back pressure should be the same too.


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Acidity order of o-nitrophenol vs. o-chlorophenol

It is given that the above statement is true. Why does o-nitrophenol, in spite of intramolecular hydrogen bonding, have a lower $\mathrm{p}K_\mathrm{a}$ than o-chlorophenol?

The powerful inductive effect of the nitro group if far stronger than any H-bonding effect

@AniruddhaDeb Btw the above statement is true. o-chlorophenol has greater pKa (i.e. less acidic) than o-nitrophenol.

@Waylander Could you give a more rigorous answer? Are there any more examples of compounds where the inductive effect dominates over H-bonding? Or is o-nitrophenol an exception?

@SoumikDas edited accordingly.

The thing you have to realise is that H-bonding can lower the ease with which the acidic $\ce{H}$ is removed, but it does not render it completely 'locked' within the molecule. What you have to understand is, the hydrogen bond shown here is not that strong as compared to the other kinds of H-bonds which have been observed in chemistry
Let us consider some values from Wikipedia:

Hydrogen bonds can vary in strength from weak ($\pu{1–2 kJ mol−1}$) to strong ($\pu{161.5 kJ mol−1}$ in the ion $\ce{HF2−}$. Typical enthalpies in vapor include:
$\ce{F−H···:F}$ ($\pu{161.5 kJ/mol}$ or $\pu{38.6 kcal/mol}$), illustrated uniquely by $\ce{HF2−}$, bifluoride
$\ce{O−H···:N}$ ($\pu{29 kJ/mol}$ or $\pu{6.9 kcal/mol}$), illustrated water-ammonia
$\ce{O−H···:O}$ ($\pu{21 kJ/mol}$ or $\pu{5.0 kcal/mol}$), illustrated water-water, alcohol-alcohol
$\ce{N−H···:N}$ ($\pu{13 kJ/mol}$ or $\pu{3.1 kcal/mol}$), illustrated by ammonia-ammonia
$\ce{N−H···:O}$ ($\pu{8 kJ/mol}$ or $\pu{1.9 kcal/mol}$), illustrated water-amide
$\ce{HO−H···:OH+3}$ ($\pu{18 kJ/mol}$ or $\pu{4.3 kcal/mol}$)

And from here, value of bond dissociation energy of phenolic $\ce{O-H}$ bond is approximately $\pu{83.3 kcal/mol}$.
Now, look at the value of the $\ce{O−H··:O:}$ bond and think, that while an extra 5 kcal/mol is needed to release the acidic proton, it's not that different as compared to a relatively quite large amount of $\pu{83.3 kcal/mol}$, which you have to anyhow supply as a base amount to break the phenolic $\ce{O-H}$ bond.
To cause appreciably high hindrance in releasing the acidic proton, there are far higher values of H-bonds available, up to $\pu{38.6 kcal/mol}$, which is actually becoming comparable to the half of $\pu{83.3 kcal/mol}$ (i.e. $\pu{41.65 kcal/mol}$).
Hence, the proton can still be released fairly easily in o-nitrophenol, and if you compare the stability of conjugate base with radically less stabilising substituents like that in o-chlorophenol (as Waylander said in the comments), you can see a vast difference in which the reaction will be pushed towards conjugate base in the former case, so after a little bit of initial difficulty (relatively) in proton releasing, Le Chatlier principle will start pushing o-nitrophenol to it's conjugate base by itself.
However, it would become difficult to compare in neck-to-neck cases with another substituent of the caliber of -$\ce{NO2}$ (for example, -$\ce{F}$)

Remember Bond Dissociation Energies are a measure of homolysis, not heterolysis, as your link mentions.

@user55119 So,the values I have are not for the hydrogen bonds and phenolic O-H bond?

In your discussion below the chart the value of 83.3 kcal/mol, which is taken from your link, is the BDE(O-H) reported for phenol. It is the value for homolysis as the the authors explain. I have no comment on your arguments about acidity.

@user55119 Oh, okay, thanks for the info! I am also adding a link in relation to this which can help people looking at this answer

$\mathrm{p}K_\mathrm{a}$ is inversely proportional to acidity of a compound similar to $\mathrm{p}K_\mathrm{b}$ which is inversely proportional to basicity as it is $-\log K_\mathrm{a}$ and $-\log K_\mathrm{b}.$
In this case nitro group is a electron-withdrawing and deactivates the phenyl group the most, whereas $\ce{Cl}$ is a moderate deactivating group but ortho, para directing group. Hence more of the tendency to polarise the $\ce{OH}$ bond and increasing acidity is with the nitro substituted phenol.

Phenols have higher acidities than aliphatic alcohols due to the aromatic ring's tight coupling with the oxygen and a relatively loose bond between the oxygen and hydrogen of the hydroxyl group directly attached to the aromatic carbocyclic nucleus. For this reason, electron withdrawing groups (EWG) and/or electron donating groups (EDG) attached to aromatic nucleus play a significant role in acidity of phenols.
Phenols are more acidic when the ring is substituted with a EWG regardless of the position. For example, unsubstituted phenol has a $\mathrm {p}K_\mathrm{a}$ of approximately $10$. However, the acidity of each of o-, m-, and p-chlorophenol shows higher acidity than that of phenol ($\mathrm {p}K_\mathrm{a}$s are $8.56, 8.97$, and $9.14$, respectively) due to the negative inductive effect ($\mathrm{−I}$) of chlorine atom (the closer the $\ce{Cl}$ atom to oxygen of phenol, the bigger the inductive effect). Some GWGs has higher effect than that induced by inductive effect alone. The best example of this kind is nitrophenols. For instance, o-nitrophenol, m-nitrophenol, and p-nitrophenol have $\mathrm {p}K_\mathrm{a}$ values of $7.2, 8.36,$ and $7.15$, respectively. This higher acidity is due to additional negative mesomeric (resonance) effect ($\mathrm{-M}$) of nitro group compared to that of chloro group (which is actually shows some $(\mathrm{+M})$-effect). The extra resonance stabilization of conjugate base (phenoxide ion) is possible only when nitro group is at 2- or 4-position of the phenol. $\mathrm{M}$-effect is not possible when nitro group is at 3-position of the phenol (it is only $(\mathrm{-I})$-effect that make 3-nitrophenol more acidic than phenol).
Keep in mind that $\pm$-mesomeric effect is essentially same for ortho- and para-cases while inductive effect magnitude is distance dependent. For example, it would be dominant when nitro/chloro group is present in ortho-position rather than para-position. Based on those facts, we can conclude followings:

Magnitude of $(\mathrm{-I})$-effect of nitro group ($(\mathrm{-I})_\ce{NO2}$) is larger than that of chloro group ($(\mathrm{-I})_\ce{Cl}$), based on lower $\mathrm {p}K_\mathrm{a}$ of m-nitrophenol ($8.36$) compared to $\mathrm {p}K_\mathrm{a}$ of m-chlorophenol ($8.97$): The acidity of meta-substituted phenols are solely based on $(\mathrm{I})$-effect.
Magnitude of $(\mathrm{+M})$-effect of chloro group ($(\mathrm{+M})_\ce{Cl}$) is smaller than magnitude of $(\mathrm{-I})$-effect of chloro group ($(\mathrm{-I})_\ce{Cl}$) when it is at para-position: This conclusion is based on the comparison of $\mathrm {p}K_\mathrm{a}$ of p-chlorophenol ($9.14$) and that of phenol ($10$).
Hence, magnitude of $(\mathrm{-M})$-effect of nitro group ($(\mathrm{-M})_\ce{NO2}$) is much larger than magnitude of $(\mathrm{+M})$-effect of chloro group ($(\mathrm{+M})_\ce{Cl}$): This conclusion is based on above two conclusions and the comparison of $\mathrm {p}K_\mathrm{a}$ of o-nitrophenol ($7.2$) and that of o-chlorophenol ($8.56$).

Thus, when both groups are at ortho-position, acidity of o-nitrophenol (with $(\mathrm{-M-I})_\ce{NO2}$) should be higher than that of o-chlorophenol (with $(\mathrm{+M-I})_\ce{Cl}$).


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what causes the smell of a room ionizer? If ozone, what are we actually smelling when ozone interacts with nasal mucous?
This touches on how smell works, but my intent is ask about the mechanism for why do ionised smell like they do?  I would like to get one step past ozone if that is the primary mechanism.

What do you mean with "one step past ozone" ?

Might be the smell of ozone or that of nascent oxygen, if any. Same for H2O2. If any: https://en.m.wikipedia.org/wiki/Nascent_state_(chemistry)

Regarding our sense of smell, from the Wikipedia:

The perception of odors, or sense of smell, is mediated by the olfactory nerve. The olfactory receptor (OR) cells are neurons present in the olfactory epithelium, which is a small patch of tissue at the back of the nasal cavity. There are millions of olfactory receptor neurons that act as sensory signaling cells. Each neuron has cilia in direct contact with the air. Odorous molecules bind to receptor proteins extending from cilia and act as a chemical stimulus, initiating electric signals that travel along the olfactory nerve's axons to the brain.2

Regarding the odor of ozone, from the Wikipedia:

Ozone (/ˈoʊzoʊn/), or trioxygen, is an inorganic molecule with the chemical formula O
  3. It is a pale blue gas with a distinctively pungent smell.

From this site:

While oxygen is not detectable by the human nose, ozone has an extremely low odor threshold.  Humans have the ability to detect ozone [concentrations of ~10 ppb] before many expensive measuring devices.  Also, ozone in air at high levels (greater than 0.1 ppm) is hazardous to human health.
Ozone is a powerful oxidant (far more so than dioxygen) and has many industrial and consumer applications related to oxidation. This same high oxidizing potential, however, causes ozone to damage mucous and respiratory tissues in animals, and also tissues in plants, above concentrations of about 0.1 ppm. While this makes ozone a potent respiratory hazard and pollutant near ground level [...]

There are quite a few articles available online that discuss the "odor of ozone". It is described by some as "metallic" or as "similar to chlorine" (although note that "chlorine smell" is often "pool smell" caused by chloramines).
Given the reactivity of ozone (it oxidizes just about anything containing carbon to $\ce{CO2}$, which explains why it can cause inflammation and is generally regarded as deleterious to health), it is an interesting question whether there exists a distinct odor receptor that senses ozone or whether the response is nonspecific or mediated by pain receptors and registered as the "pungent smell" of ozone. Or perhaps it is other molecules altogether rather than the interaction of ozone with nasal receptors that causes the smell.
All of this is made even more complicated by the apparent loss of olfactory perception caused by exposure to ozone:

An acute exposure to 0.2 ppm ozone impairs odor detection threshold.
The cause is not clear, different modes of action might play at role.
Olfactory threshold seems sensitive to photochemical air pollutants’ effects.

The "The Curious Dr. Z" blog hypothesizes on the source of "ozone smell" (see also reference 1 cited in the blog):

It turns out that our olfactory receptors might be able to bind free radicals, and, since free radicals and ozone frequently occur together, that’s what we’re actually smelling.
In 1991, a group of patients undergoing irradiation for cancer all complained about a strong, pungent odor they later identified as ozone. This lead to a study where researchers discovered that humans can actually detect the irradiation of tissue by detecting free radicals trapped in mucus.
Another, older theory, is that ozone acts as the signal for irradiation. Ozone binds to and reacts with olfactory receptors the just as other odorants do, and effectively acts as a warning of more damaging chemicals nearby.
We don’t know too much beyond this, and as far as I can tell, ozone olfaction is not a very active area of research. Bummer.
Since we are so sensitive to the smell of ozone, and we don’t smell much else when we do smell ozone, I suspect that ozone/free radicals have many potential binding sites (a lot of those “hands” can grab ozone). When the ozone/free radicals does bind, it does so with high affinity, meaning it easily binds to receptors. This persistent and overwhelming smell experience is similar to other smells that indicate something noxious, including smoke and rotting tissue.

References

Lloyd M. Beidler. Olfaction, Handbook of Sensory Physiology, Chemical Senses. Springer Science & Business Media, 2012, ISBN 3642651267, 9783642651267


text:
Rate determining step in oxidation of alcohols using chromium-related compounds
I wanted to ask a question about the kinetics of an oxidation reaction using chromium reagents.
I was given a lecture today on the mechanism of alcohol oxidations using Chromate (VI) reagents:

My previous knowledge of Rate-Determining steps (RDS) were that on an energy profile diagram, the step with the largest "activation energy" is the RDS as it takes the most energy to go over the steps, so relative the other steps are faster.
For today's lecture, the professor mentioned the following:

If our $\ce{R}$ groups are bulky, which means that you have a bulky alcohol, the alcohol oxidises faster. This means that our Rate Determining step here is the second step and the first step is fast and reversible.

Unfortnuately, this point was not further explained.
I've tried to come up with a simple explanation to myself for two following points:

The R groups being bulky means the second step is the RDS
The rate determining step is the non-reversible step

but I can't seem to find any intuition for this. I only found the logic that there can be slow and fast steps in a reaction. My thought was that all reactions can be reversible, but maybe the second step has such a high activation energy and the reverse process involves so many unlikely species in the mechanism that it seems unlikely to happen.

Why does having a bulky R group prove the second step is the RDS?
Why is the non-reversible step the RDS, and not (say) the reversible step?

What is exactly X. Is it an atom or a radical ? Anyway, it should be charged. Furthermore, I thought chromate was an ion where Cr is bound to two Oxygen atoms through 2 double bonds, and to two Oxygen atoms charged 1 minus. In my opinion, Cr is not bound to X in a chromate. It is bound to 4 oxygen atoms, and not to X. I don't understand your development.


text:
Synthesis of tripeptide using Merrifield peptide synthesis procedure - Inclusion of asparagine
I am supposed to synthesize tripeptide, $\mathrm{Ala-Asn-Phe}$, using Merrifield peptide synthesis procedure. Accordingly, I have done followings:

Attached protected $\mathrm{Phe}$ as the first amino acid.
Got rid of protecting group ($\mathrm{BOC}$)
Added asparagine with 2 protecting groups - $\mathrm{BOC}$ for $\alpha$-$\ce{NH2}$ group and $\mathrm{FMOC}$ for the $\gamma$-$\ce{NH2}$ group.
Got rid of $\mathrm{BOC}$ and added $\mathrm{Ala}$ protected with $\mathrm{BOC}$.
Got rid of $\mathrm{BOC}$ and $\mathrm{FMOC}$, and used $\ce{HF}$ to release the peptide.

Is this going to work? The thing that worries me the most is the $\gamma$-amino group and using $\mathrm{FMOC}$ - is that a good idea?

Are you following a known protocol for solid-state synthesis? My question is because I wonder why you have to protect amide $\ce{NH2}$. Look, when you get asparagine attached to phenylalanine, you got anothe peptide (amide) bond with $\ce{NH}$ bond, but you didn't care to protect that.

@MathewMahindaratne well NH from peptide bond will not react with -COOH group of alanine that I am adding next... Or am I mistaken?

My point is it's not necessary to protect amide $\ce{CONH2}$ group. It is not competitive enough with readily available $\alpha$-$\ce{NH2}$ group for DCC coupling.

I agree, primary amides are not strong nucleophiles so it is not necessary to protect the Asparagine amide. I presume you are using coupling agents to attach the Asparagine and Alanine

You just don't need the FMOC at all.


text:
Free Electron-Bond Electron Interactions
I am reading an introductory textbook on electronics: Practical Electronics for Inventors by Paul Scherz and Simon Monk. In a section discussing the motion of electrons in circuits, the textbook mentions something called free electron-bond electron interactions:

It is likely that those electrons farther "down in" the circuit will not feel the same level of repulsive force, since there may be quite a bit of material in the way which absorbs some of the repulsive energy flow emanating from the negative terminal (absorbing via electron-electron collisions, free electron-bond electron interactions, etc.).

I have never heard of such bonds/interactions, and a quick Google search reveals nothing of the same description.
The description seems odd to me, since electrons repel rather than attract/bond.
I wonder if this could be in reference to metallic bonds? Metallic bonds are formed by the attraction between metal ions (of which circuitry components comprise) and delocalized, or "free" electrons (which is what is pumped out from the battery, as a result of chemical reactions within).
I would appreciate it if someone could please take the time to explain what these are, and/or please direct me to a source (Wikipedia article?) on this subject.

I have a feeling you may expect too much from a book called "Practical Electronics for Inventors".


text:
Reason for usage of Metal chlorides and Platinum or Nichrome wires in flame test
I'm studying about the s-block elements in my course of Inorganic Chemistry and I happened to come across the following text.

Electrons may be quite readily excited to a higher energy level, for example in the flame test. To perform this test, a sample of Metal Chloride or any salt of metal moistened with conc. HCl, is heated on a platinum or nichrome wire in a Bunsen burner flame.

Why are metal chlorides specifically used and also what is the role of the Plantinum or Nichrome wire?

The platinum wire is not attacked by the hot chlorides, nicrome is so cheap you dont care.

Metal chlorides are usually much easier volatilized than oxides or sulfates. They melt or sublime at temperatures much lower than 800°C. So they are volatilized in any flame like Bunsen burners.
Platinum is not attacked by hot chlorides or acids, and is not oxidized in hot air. So it can be used for a great number of consecutive tests in the flame. It will never wear out in the flame.

Adding to the first answer: NiCr wires are used because they are cheaper as platinum wires and still have a good chemical resistance and high melting point. Even tough platinum is much superior to NiCr.


text:
How much CO2 can I fit in a 0.5 L plastic bottle?
Let's say I have a 50 lb CO2 tank which can probably contain about 1000 L of the gas at 122 atm. Suppose I want to use the tank to fill a 0.5 L plastic bottle which only has air in it.
We know that gases always assume the shape and volume of their container. If that is true then that would mean that I can transfer the whole 1000 L of CO2 into the plastic bottle.
What happens in reality is that the pressure in the 0.5 L would have to rise to about 244 thousand atm, so how do I know how much CO2 to put into that 0.5 L bottle before it would presumably explode?

The pressures will equilibrate at roughly 122 atm.  That means that it will contain 0.5 L of CO2 at  122 atm (~1800 psi), neglecting the air.

@ChetMiller "Most two liter bottles begin to fail at pressures around 1,034 kilopascals, or 150 pounds per square inch. " (source: https://education.seattlepi.com/much-pressure-can-two-liter-bottle-handle-6313.html)

At 122 atm the CO2 should mostly be a liquid unless the temperature is more than 31 C. CO2 would liquefy at 5.1 atm or about 72 psi.

See the comment of @Karsten Theis. What is the pressure tolerance of your bottle?

If the temperature is at a level where the CO2 is gaseous at 122 atm, I supposed that p1V1 = p2V2 was valid.
Then I calculated p2 from it and it turned out a value of 244000 atm at 0,5 L. Where did I go wrong?


text:
What all information can be gathered qualitatively from the radial distribution functions of a molecule?
Apart from knowing the radius of coordination shells and the probability of finding, is there any other thermodynamic  information we can gather by looking at radial distribution functions of a molecule?
For instance, If I plot the radial distribution functions for 'O' atom of H2O, once for pure H2O and then once in the presence of ethanol. Will i be able to relate the excess molar volume through the radial distribution functions?
What other information can I get?

You can get alot from the RDF... an amazing amount actually. It is true though that people like printing out the RDF and thinking their paper now has substance, but there is so much more to the pair correlation function.
Integral equations are based on g(r). Kirkwood-Buff solution theory is too. Kirkwood-Buff solution theory will make you dizzy with what it can calculate.
See wikipedia for more on Kirkwood-Buff solution theory. See Arieh Ben-Naim's Molecular Theory of Solutions for the real deal explanation. What a good read. I get excited everytime I get to link it. Ben-Naim, amazing book

I'm not sure this answer is talking about the same RDF...

It sure is the same.

If you know the RDF and bulk density, then you know how many particles are at every distance from any given atoms center. If you then have a function for calculating energy that depends on distance, you can calculate the energy at every distance... or, you can... integrate over the distance (Integral Equations). You can calculate alot of things :)


text:
Solution of very small particles with neutral buoyancy and distinguishable to the naked eye
I'm trying to study the Tea leaf paradox, and asked a question about it on physics.stackexchange.com. To get further on the problem, I want to make a solution of very small objects in water that are distinguishable to the naked eye, and have (very close to) neutral buoyancy. I teach at a high school and have access to the chemistry lab here. Suggestions?

This is not called a solution. Solution means that whatever is in the water has been "dissolved", that is, broken and dispersed into individual molecules. What you are describing is a suspension - meaning the particles are much bigger than individual molecules.

I think that the cheapest and fastest way is that of working on natural products, like actual tea leaves, or ground coffee (I just tried it with coffee, and it can be done):
If you want only particles with neutral buoyancy, or, if I understand well, floating in the solvent, you can use said natural products, which contain particles of different shapes and densities, and discard by fractionation the unwanted ones.
What I did:

Place the ground coffee in a glass of water and stir a lot.
Let the suspension sit. You will notice that part of the parcticles are on a top layer, some are floating in water, and some are at the bottom.
Put the glass in a sink, and gently add water to it, so that the top layer eventually falls out of the glass, together with some water
Let the suspension sit: if needed repeat the last step, otherwise go on
To discard the bottom layer, gently pour the solution in another glass, by leaving some water in the first glass: the first glass will contain the heavier bottom layer, and the second one will contain only floating parcticles
If needed, repeat the last step

Brilliant! I will try that.


text:
Ionic content of different samples of water
So I was reading about impurities of in water and I have a question in mind:
So say we have distilled water, deionized water, tap water and soft water which was softened using dowex resin or any other cation exchange resin.
So my question is which samples would contain anions like $\ce{Cl^-}$ and $\ce{PO4^-}$ and which would contain cations like $\ce{Ca^2+}$ and $\ce{Mg^2+}$?
My speculation is that tap water would contain all the stated ions, deionized water wouldn't contain any, distilled water wouldn't contain any, and softened water would contain the anions but not the cations since $\ce{Ca^2+}$ and $\ce{Mg^2+}$ would be replaced by $\ce{Na^+}$ or so.
Thank you!


text:
PP05 plastic "not dishwasher safe" but does not deform? Leaching chemicals?
It's plastic shaker cups, they came with instructions saying they are not dishwasher safe but they did not deform in the dishwasher. Does this mean it's probably unsafe to clean them using the dishwasher due to stressing the plastic, unleashing chemicals?
Thinking of the study showing additives in PP5 plastic had estrogenic activity.
Sorry, probably really hard to give a straight forward answer to this question.

Probably far more about prolonged dishwasher exposure damages the material and makes it more likely to break.

"dishwasher safe" also depends on water temperature in dishwasher and temperature of drying cycle (if used).


text:
Can butan-1-ol be written as butanol, and is it connected to butan-1-ol having Isomers?
Can butan-1-ol be written as butanol, and is it connected to butan-1-ol having Isomers?

Depends where you are writing it. There's a difference between butan-1-ol and butan-2-ol. If you are describing something where that difference is important then you cannot just write butanol.

Yes, it's convention to write butan-1-ol as butanol, just as it's likely to write propan-1-ol as propanol (whenever it's attached to the first carbon, we usually just omit the 1)
I'm not quite sure about what you mean with isomers, but we name isomers in different ways so that we can distinguish them. For example, butan-1-ol and butan-2-ol are considered to be positional isomers (same functional group, same carbon skeleton, but the functional group OH is attached to different carbons on these two different isomers) and we use the number to distinguish between them. Hope this makes sense?


text:
How does a mass spectrometer treat ions?
I always thought that in a mass spectrometer all the ionized fragments were going in the quadrupole at the same time, and then this quadrupole was sorting them by ratio.
But last week I learn that a quadrupole mass spectrometer does not work like that, it apparently works by ratio range letting only specific ion enter the quadrupole first and then continuing for others range
I learn that you can even specify a specific mass/charge ratio in the programs for the MS to analyze more these specific ones.
How exactly does it work?

Can you Google "quadrupole mass spectrometer"?

A quadrupole is a bundle of four parallel metallic bars or rods, fastened horizontally. They must be neighbors, but they must not teach one another. I will call N the highest of all (like North), S the lowest of all  (like South), and E and W the two rods in the middle E at right and W at left.
The whole is in a vacuum. The ions to analyze are sent in the direction of the central axis, Suppose that the two bars E and W are charged minus in the beginning, and the two bars N and S are charged plus. The bundle of ions entering the quadrupole are attracted by the bars E and W. Their trajectory is deviated to the left or to the right. But they will not touch these bars, because the sens of the charges suddenly changes. Before the ions have touched the W or E plates, these plates have been quickly transformed and are positive, and the bars N and S become negative. The trajectories of the ions is deviated in the direction North or South. But here again, the ions can not touch the N and S electrodes, because the polarity changes once again, repelling the trajectories in the E and W direction. If you choose your frequency with care, the ions having a certain mass follow a helicoidal trajectory and get out of the quadrupole without having been intercepted. The lighter ions touch the bar and are eliminated in the very beginning of the bundle. The heavy ions have a helicoidal trajectory that opens out and touch one of the bars before getting out of the quadrupole.
It means that a quadrupole is a device for selecting ions of a certain ratio m/z. All the other ones are eliminated. This choice is dependent on the choice of the frequency.


text:
What is the distinction between alkyloxy and alkoxy prefixes, if any?
The IUPAC nomenclature guidelines reference alkyloxy and alkoxy prefixes at different points. Is there any distinction, at the level of a principle or rationale, between the use cases for the two? For example, is there an underlying reason why cyclopentyloxybenzene is or isn't preferable to cyclopentoxybenzene?

I have ever seen only methoxy-, not methyloxy-. The latter may be formally more correct and consistent, the former get used as a rational abbreviation for some short alkyloxy groups. It may differ between alkyls due historical reasons. I guess octyloxy may be preferred to octoxy.


text:
Can a compound with indirectly bonded carbon and silicon atoms within the same aromatic system be called organosilicon compound?
Wikipedia says that organosilicon compounds are organic compounds with carbon-silicon bonds.
Does the class of organosilicon compounds include compounds in which the carbon and silicon are only indirectly bonded, for example the (hypothetical) molecule 1,3,5-trihydroxy-4,6-dioxa-2-lambda3-oxa-3,5-disilacyclohex-1-ene(3+), in which the carbon and silicon atoms are not formally directly bonded but are part of the same aromatic (six π-electrons) ring, sharing the same aromatic bonding orbitals?
Here's what the molecule looks like (direct uploadd won't work for some reason...)

Is there the possibility to include a picture of this hypothetical compound. Not all of us might be familiar with this type of naming scheme. (Yes, I really have not a single clue, what this compound might look like.)

Organosilicon compounds fall under the broader category of Organometallic compounds. It is a requirement that in organometallic compounds, atleast one $\ce{C-M}$ bond should exist. Since Si is only bonded to oxygen atoms in your compound, the given compound is not an organosilicon compound.
The following is also mentioned in the wikipedia page for Organosilicon compounds, under the image:

A carbon–silicon bond present in all organosilicon compounds


text:
What is the most acidic neutral hydrocarbon?
This is a question I've been thinking about quite frequently. I had seen that this hydrocarbon has a low $\mathrm pK_\mathrm a$ with reference to the highlighted H-atom

Loss of the central gives aromatic character to the three 5-membered rings in the compound, contributing to its acidity.
Are there any other neutral hydrocarbons which are stronger acids? How do their $\mathrm pK_\mathrm a$'s stack up against, say another organic acid such as acetic acid?
EDIT: As pointed out in the comments, $\ce{CH_5^+}$ (methanium) is a superacid, but is not a neutral hydrocarbon.

$\ce{CH5+}$ is cheating?

See also https://en.wikipedia.org/wiki/Methanium. But the OP may mean a neutral molecule.

@KarstenTheis yep ;) Nice try though. I will read about methanium. Also edited the question accordingly.

I think you can get much further by drawing a neutral hydrocarbon which is initially non-aromatic, but becomes aromatic when deprotonated. Cyclopentadiene is extraordinarily acidic for a hydrocarbon, with a pKa comparable to alcohols. With some imagination, it's likely possible to set up a structure with cascading aromatization upon deprotonation.

@NicolauSakerNeto "cascading aromatization"  . Nice to imagine in my head.

@NicolauSakerNeto That's exactly what I have drawn! The question is: are there any compounds that are more acidic than what I've drawn.

@AniruddhaDeb Well, what I had in mind was for the initial structure to be completely non-aromatic. For reference, cyclopentadiene is approximately as acidic as 9-phenyl-9H-fluorene, according to Hans Reich's pKa tables in DMSO, even though the latter has far more extended conjugation. It's better for aromaticity to suddently appear, instead of existing aromatic systems merely being conjugated.

related https://chemistry.stackexchange.com/questions/9212/can-a-organic-compounds-such-as-hydrocarbons-contain-an-ionic-bond https://chemistry.stackexchange.com/questions/20283/whats-the-strongest-known-organic-acid

In this answer a hydrocarbon anion called Kuhn's anion (1) has the formula $\ce{C67H_{39}^-}$ and reported $pK_b=8.1$, which would correspond to the neutral hydrocarbon having $pK_a=5.9$.  The anion, together with several hydrocarbon cations with which it forms stable salts, is shown below (taken from the answer referenced above; primary reference J. Org. Chem. 1990, 55 (3), 996–1002):

Ehhm, but the question is about neutral hydrocarbons…

But the neutral hydrocarbon is the acid.  It's mentioned here as the conjugate of the anion.

Right, but the illustration showing only ions looks confusing to me. Why not to draw alongside the actual compound you are posting $\mathrm{p}K_\mathrm{a}$ value for?

Not sure how to modify the anion, which would involve using CH in the center instead of the carbanion.  Would need a desktop I think and I cannot get to one this weekend.

How is this hydrocarbon so acidic? I understand that the $pK_a$ was derived via a practical experiment but how would one go about explaining it theoretically? Is it merely because of $\pi$-resonance? I don't see any aromatic rings forming/breaking to contribute to such a pKa.

Conjugation through the central carbon of the anion.

@AniruddhaDeb It does aromatise three cyclopentadienyl rings, only partially though.

When we put the proton on the central carbon, saturating it, we fix a pendant pi bond to one of the atoms in each five-membered ring, so aromatic coupling cannot extend to those rings. With deprotonation to form the anion, the central carbon can donate a pi electron pair, which delocalizes those double bonds and thus allows aromatic coupling through all the rings.


text:
Why is Aluminium oxide amphoteric?
I was reading my book, in which it mentioned that aluminium and gallium oxides are amphoteric and and indium and thalium oxide are basic in their properties.
But no explanation was given.
My main question is we know that mostly metal oxides are basic and non metal oxides are acidic. But we see some metal being amphoteric, what is the cause of this difference?
I tried to search online but found no reliable piece to read.

If anyone can suggest better tags, please do thanks

These are fundamental questions. I am afraid nobody knows yet. By that I mean that science has not progressed enough to predict from $ab$ $initio$ calculations that an element with this number of neutron, protons, and electrons will have an amphoteric oxide. Yes one can predict from periodic trends like Mendeleev, but what is the fundamental cause? It is an open question I guess.

I suggest you first find out what exactly "amphoteric" means, and then look for examples. Why is sodium oxide basic, why do nitrogen oxides react acidic, etc. You know, start with the simpler cases. Then look for reactions that show how Al oxide can do both.

@M.Farooq We can certainly predict the stability of different Al (hydr)oxides/ions. "Why" questions have a tendency to become philosophic, which is why chemists prefer to ask "how".

"Your Delta Tau Chi name is "Pinto".  "Why Pinto?"  (Belch) "Why not?"  There's my answer.

I don't think aluminium cation bonds directly to Cl- as Mr Yusuf shown above. Because, it is said that Aluminium ion is a 6 hydrated ion. By the way has anyone read this paper?
https://sci-hub.hkvisa.net/10.1021/la025651i In this article it is said that Al2O3 changes to Al(OH)3 in water.
Isn't it possible that the basic or acidic reaction of Al2O3 is multistaged and the first step is hydration of Al2O3 into Al(OH)3?

I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements
To start off, please go through my answer to this question: Why is chromate stable in basic medium and dichromate stable in acidic medium? to read about the preface to this method, including the potential concerns.
Essentially, the book talks about determining the type of reactivity of inorganic molecules by looking at their structures. While this may not reflect with 100% accuracy the actual reaction pathway, we can take it to be a pretty good approximation of what happens in reality. While the completely correct mechanism can only be found out experimentally, we can get a probable mechanism by employing an arrow-pushing approach, and that should at least mirror the actual pathway to a sufficient extent, if not entirely.
Now, if we look at the chemical formulae of Aluminium oxide and Gallium oxide ($\ce{Al2O3}$ and $\ce{Ga2O3}$), we can propose that the units possess a similar bridged dimer-kind of structure as shown below for aluminum oxide(the same structural skeleton is valid for gallium oxide as well, just by replacing aluminum with gallium below) :

This structure now possesses some unique properties. It contains an $\ce{Al-O-Al}$ bridge with an $\ce{Al=O}$ group on each side.Now,the $\ce{Al=O}$ group can be said to be structurally similar to the carbonyl group from organic chemistry.
Hence, like the carbonyl group, it can be said that there will be high partial positive charge on the $\ce{Al}$ atoms due to the electron-withdrawing effect of the $\ce{Al=O}$ groups. Hence, they can serve as the ideal electrophilic site for the attack of a nucleophile like $\ce{OH-}$ in basic medium, and lead to bond cleavage(and further reaction) in alkali as follows:

In a basic medium, the $\ce{AlO2H}$ moiety will also be deprotonated, and thereby you will end up with two moles of $\ce{AlO2^-}$ . If the base being used was $\ce{NaOH}$, then these products are better reported as $\ce{NaAlO2}$ for aluminum. In an analogous case, you would get $\ce{NaGaO2}$ if the we had started out with gallium oxide.
In excess of alkali, the reaction can proceed further, as further additions of hydroxide are still possible at the $\ce{Al=O}$ site as follows:

The reaction can either finish after the formation of the hydroxide, or a continued process may lead to the addition of yet another $\ce{OH^-}$ to form the complex ion $\ce{[Al(OH)4]^-}$ , which might be called the aluminate ion here.In another analogous case,we would have ended up with $\ce{[Ga(OH)4]^-}$ (called the gallate ion) if we had started out with gallium oxide.
Now, for the acidic medium, let us suppose that $\ce{Al2O3}$ reacts with $\ce{HCl}$. Here, the central oxygen of the $\ce{Al-O-Al}$ bridge represents a weak link that can be easily cleaved after protonation(because it has moieties that can function as good leaving groups on either side). Also,the counter-ion of the acid(that is, $\ce{Cl^-}$), can serve as a nucleophile to break down the dimer skeleton after protonation as follows:

The $\ce{AlO2H}$ can again be protonated to form an oxonium ion that can expel water and take up a $\ce{Cl^-}$ to form another mole of $\ce{AlOCl}$(shown in the next diagram).
Following that, the $\ce{AlOCl}$ can be engaged in a repetitive process of protonation, attack by chloride and removal of water to yield the final products as follows:

The products for these reactions can be verified for aluminium and gallium oxides from their Wikipedia pages respectively.
So we can see that one major reason for the amphoteric nature of oxides can be their bridged structures, which allows them to react with both acids and bases as illustrated above.
Now, coming to indium oxide, the formula of this compound as stated here is $\ce{In2O3}$, which bears a resemblance to the formula of $\ce{Al2O3}$ and $\ce{Ga2O3}$ which we stated above. So, I would expect this oxide to atleast have some amphoteric properties. This is indeed the case, as stated here for indium hydroxide(since hydroxides are simply hydrated forms of their oxides, we can expect a similar nature for indium oxide as well):

The result indicated conclusively that In(OH)3 is only slightly amphoteric, being much more basic than acidic.

Also, on the Wikipedia page for indium oxide it clearly states that this oxide is soluble in acids but not in alkali. This is pretty much expected because when you move down a group, the elements tend to become more metallic in nature, and hence their oxides tend to become more basic as well. Hence, an amphoteric oxide slightly tending towards the basic end seems reasonable for indium(III) oxide. The reaction between acid and $\ce{In2O3}$ should follow the general scheme that is illustrated above for $\ce{Al2O3}$ and $\ce{HCl}$, and should yield similar products. This indeed is the case as can be seen here
As for thallium, due to the inert pair effect, it generally forms an oxide in which thallium is in the +1 oxidation state, with the formula $\ce{Tl2O}$ as stated here. This doesn't fit in with the general formulation of $\ce{Al2O3}$ and $\ce{Ga2O3}$ that we discussed above. Hence, one would expect this oxide to simply become more basic than the other group members(as it is an oxide of the lower-most element of the group) and not show much signs of an amphoteric nature. This matches with what is stated here.
Before wrapping up the answer, I would like to re-iterate a point I made at the beginning of the answer:
All these pathways are simply possible mechanisms, and there is no way of knowing whether they are completely correct or not without performing experiments. However, they are highly probable propositions, and you can expect these pathways to represent the actual mechanisms to a good approximation

The wikipedia article you linked on $\ce {Al2O3}$ clearly states the structure to be ionic with lattice structure. Yet in your explanations, you invoked a bridged structure for the ionic solid? From which source did you find aluminium oxide to exists as covalent molecules with a bridged structure?

While what you write seems to make a lot of chemical sense, it made use of a false premise (assuming the stated metal oxides to be covalent molecules) and thus is invalid

@TanYongBoon I took my inspiration from this discussion here which stated that although aluminum oxide has 61% ionic and 39% covalent character, yet it does have a polar dipole moment. So yes, a major contribution is ionic, yet I would argue that the actual final mechanism will have similar ionic interactions as shown with the covalent diagram, as the dipole moment obtained from a covalent description matches with the one seen in the molecule as well

@TanYongBoon Also, I have never said that this mechanism is providing a completely accurate description, it is just a proposition. However, the interactions found in the actual experimental procedure should resemble what we have sketched here, as the molecule does have almost 40% covalent character, and the assumption is attempting to model completely ionic charges as partial charges. But a 100% accurate description was never expected anyway

Metal hydroxides are basic in water. Non-metal oxides are acids in water. The limit between metals and non-metals is a sort of staircase going through the periodic table from the middle of the first line to the lower corner at the right-hand side. Aluminium touches this staircase. This is why Aluminium behaves like a metal and like a non-metal, and why aluminium oxide can react with acids and with bases.
Tin Sn has the same property.

What criteria does aluminium or the other similar satisfy, is it the EN diff, the charge density, ? I know it cant be pin pointed but any help would be regarded


text:
Can we break double bonds by using laser and heat?
If it is possible then give an example for it. That chemical reaction is in this process.
step 1: {Reactant + heat(sunlight)} gives weakened bonded product1
step 2: {product1 + laser} gives {one solid product + one gas product}

You should search for solid state photochemistry, there are many examples.

You're going to need to be more specific


text:
Acidity of 2-sulfopropanedioic acid in water
Sulfuric acid is a strong acid ($\mathrm{p}K_\mathrm{a}$ value below that of $\ce{H3O+}$: completely dissociated in water) at its first deprotonation, but is a weak acid ($\mathrm{p}K_\mathrm{a}$ value over that of $\ce{H3O+}$) in its second one.
Does it also hold for 2-sulfopropanedioic acid too, being a strong acid at its first deprotonation and being weak acids at its second to fourth deprotonations?

Probably you get a weak second stage.  Second deprotonation has to come from a carboxyl group -> expected to be a weak acid.

Yes, the rest of pKa values should be only a bit lower then for malonic acid. There's no 4th value to speak of though.

I think the 4th deprotonation should be deprotonation from the alpha-carbon.


text:
Is contribution to pH by an acid or base always greater than salt hydrolysis?
I was solving a titration question and the reaction I came to was this:
$$
\begin{array}{lcccc}
 \ce{&BOH(aq) &+ &HCl(aq) -> &BC(aq)  &+ &H2O(l) }\\
\text{Initial} & x & & & 0 & & 0 \\
\text{Final} & x/4 & & & 3x/4 & & 3x/4
\end{array}
$$
Titration was done with $\ce{HCl}$ and $\ce{BOH}$ is a weak base. I got the answer treating this as a buffer but then I thought why would the salt formed i.e. $\ce{BCl}$ not get hydrolyzed and have an impact on the $\mathrm{pH}.$ Tried thinking about it and could only come to the conclusion that that $\ce{HA}$ being acid would have major contribution to the $\mathrm{pH}$ though I am not satisfied.

In a buffer, the concentration of both the weak base and the conjugate weak acid influence the pH. Could you edit your question to indicate states of matter and whether chloride is a spectator ion or covalently bound in the product? Then, it becomes possible to write the equilibrium constant expression and answer the question.

Oh I have written titration guess that implies that all states are liquid/aqueous and that thing related to chloride being a spectator ion was not mentioned in the question

What is your question ? Are you trying to calculate the pH of the solution ? In this case, you simply apply the formula of a mixture acid + corresponding base. This formula takes care of the possible back-reaction.

Yeah well the question was about finding the pH at 3/4th of the equvivalence point and I found the answer using the formula you just mentioned. My question is how does the formula take care that the salt formed is not hydrolyzed back. Is it something like because it is in equilibrium??


text:
How to find the volume of the products in methane combustion?

The combustion of methane gas, $\ce{CH4}$, gives as products $\ce{CO2}$ and $\ce{H2O}$, both in the gas phase. If $1$ liter of methane is burned in the presence of $10$ liters of $\ce{O2}$. Suppose all volumes measured under the same temperature and pressure conditions and optimal behavior for all gases.
  $$\ce{CH4 (g) + 2O2 (g) -> CO2 (g) + 2H2O (g)}$$
  What is the volume of the products formed?

I know that for a gas, the volume is proportional to the amount of substance, that is, maintaining the temperature and pressure conditions, doubling the volume doubles the volume of moles.
The answer to the exercise is 3 liters.

I agree that the question is different. Why did you mark it as a duplicate then? In any case, I decided to reopen it.

One mole $\ce{CH4}$ reacts with 2 moles $\ce{O2}$, producing 1 mole $\ce{CO2}$ and 2 moles $\ce{H2O}$.
So 1 liter $\ce{CH4}$ produces 1 liter $\ce{CO2}$ and 2 liter $\ce{H2O}$, if $\ce{H2O}$ is taken in the gaseous phase.


text:
Why are 9 g of hydrogen 4.5 mol?

A gas mixture consisting of $\pu{14 g}$ nitrogen ($\ce{N2}$) and
  $\pu{9g}$ hydrogen ($\ce{H2}$) is in a closed vessel exerting a
  pressure of $\pu{1.0 atm}$. The nitrogen and hydrogen partial pressures
  exerted on the walls of this vessel are respectively?

We know that $\pu{14 g}$ of nitrogen is $\pu{0.5 mol}$ and $\pu{9 g}$ of hydrogen are $\pu{4.5 mol}$.
We have a total of $\pu{5 mol}$ and a pressure of $\pu{1 atm}$:
\begin{align}
\pu{5 mol} &\mathop{\hat{=}} \pu{1 atm}\\
\pu{0.5}   &\mathop{\hat{=}}\\
x &= \pu{0.1 atm}.
\end{align}
I wonder why $\pu{9 g} = \pu{4.5 mol}$?

Wait, didn't you just say "We know that 14 g of nitrogen is half mol and 9 grams of hydrogen is 4.5 moles"?

@IvanNeretin This is a resolution I found on the internet. Have another solution?

Why bother with another when this one is good?

There exists an equation that you must know:
$$n = \frac{m}{M}.$$
Here $n$ is the amount of substance of a given sample of pure product, $m$ is the mass of the same sample, and $M$ is the molar mass of the pure substance.
Here, your sample has a weight equal to $\pu{9 g}$. The molar mass of $\ce{H2}$ is $2\times \pu{1 g/mol} = \pu{2 g/mol}$. So the amount of substance in the sample is: $$n = \frac{m}{M} = \frac{\pu{9 g}}{ \pu{2 g mol-1}} = \pu{4.5 mol}$$

I have removed the deprecated term "number of moles" and substituted the proper amount of substance as you have also used that in your post.


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What does it mean to balance a redox reaction in acidic or basic mediums?
Until now it was clear to me what balancing a reaction was. Balancing redox reaction also make sense to me. What does it mean to balance a redox reaction in acidic or basic medium?

Well, some reactions may turn out different ways depending on the medium.

I think you are confused by the word "medium." Most of the reactions are carried out in the liquid phase. When you are able to adjust the pH of a liquid phase, and redox reaction occurs, one can talk about an "acidic or basic" medium.

In acidic medium the equation must mention somewhere that some $\ce{H+}$ ions appear somewhere in the equation. In basic medium, the equation must mention somewhere that some $\ce{OH-}$ ions appear somewhere in the equation,
For example, you may say that permanganate ion reacts in acidic conditions to produce $\ce{Mn^2+}$. In basic conditions, it could not produce $\ce{Mn^2+}$, because this ion will react with $\ce{OH-}$ ions to produce a precipitate of $\ce{Mn(OH)2}$.
As an application, the permanganate ion reacts with oxalic acid $\ce{H2C2O4}$ in acidic conditions according to :
$$\ce{2 MnO4-  +  5 H2C2O4 +  6 H+ ->  2 Mn^2+  +  10 CO2  +   8 H2O}$$
Here, one sees that 6 ions $\ce{H+}$ are necessary for the reaction to happen. If no $\ce{H+}$ are present in the mixture permanganate + oxalic acid, this reaction cannot take place. Another reaction is produced without $\ce{H+}$. If you mix permanganate ion with oxalic acid in a basic solution, the first reaction that happen is the neutralization of the acid by the base namely :
$$\ce{H2C2O4  +  2 OH-  ->  C2O4^2-  + 2 H2O}$$
So, here, it is not oxalic acid that reacts with permanganate, but the oxalate ion $\ce{C2O4^2-}$. The reaction will be :
$$\ce{2 MnO4-  +  3 C2O4^2-  +  4 OH-  ->  2 MnO2  +  6 CO3^2-  +  2 H2O}$$
Here you see that some $\ce{OH-}$ are needed for the reaction to proceed. It cannot happen in acidic medium.

Protons or hydroxide ions must appear as reagents, if the solution is acidic or basic; not just somewhere.

Say we have a given redox reaction:
$$\ce{MnO4- + Fe^2+ -> Fe^3+ + Products}$$
This is an unbalanced redox reaction, just depicting how $\ce{Fe^2+}$ is oxidized to $\ce{Fe^3+}$ by $\ce{KMnO4}$. Now, depending on the medium, the following reactions may take place:

In an acidic medium, $\ce{MnO4-}$ is reduced to $\ce{Mn^2+}$ by accepting 5 electrons
$$\ce{MnO4- + 5Fe^2+ + 8H+ -> 5Fe^3+ + Mn^2+ + 4H2O}$$
In a neutral/weakly basic medium, $\ce{MnO4-}$ is reduced to $\ce{MnO2}$ by accepting 3 electrons
$$\ce{MnO4- + 3Fe^2+ + 2H2O -> 3Fe^3+ + MnO2 + 4OH-}$$
In a strongly basic medium, $\ce{MnO4-}$ is reduced to $\ce{MnO4^2-}$ by accepting 1 electron
$$\ce{MnO4- + Fe^2+ -> Fe^3+ + MnO4^2-}$$

This is just one example of how different mediums can affect a given redox process; other redox reactions may also be influenced, depending on whether the medium is acidic or basic.

In basic medium, $\ce{Fe^3+}$ does not exist. It is transformed in Fe(OH)3


text:
Is a liquid vitamin E supplement possible without additives?
Looking at E vitamin products sold commercially, all “pure vitamin E oil” products seem to be for skincare rather than consumption as a supplement.
I have read that d-alpha tocopherol is the form of vitamin E with greatest bioavailability, but even liquid supplement products often contain sunflower or other oils in addition, which I understand to be a source of but of course are not entirely vitamin E.
Is a pure liquid supplement possible or practical for consumption as a supplement, and if not, why?

To be honest, I am on the edge whether this is on topic for our site. It appears like medicinal advice that you are seeking; such questions are off-topic.

Thanks @Martin-マーチン I will review whether I can edit. I am not looking for advice but an understanding of the chemical practicalities and realities of these bioactive substances vis-a-vis how they are presented on the commercial market.

I have reworded the question so it hopefully reads better as seeking chemical understanding.

As a practical matter, vitamins generally can't be provided to consumers in their pure forms.  The problem is that we need so little of them. For instance, the RDA for d-alpha tocopherol is 15 mg.  Consumers simply don't have the training or equipment necessary to precisely measure out 15 mg (or the time or patience to do this daily!). Most likely they would overdose if they tried.
Thus the vitamin E has to be diluted with a much larger quantity of edible material into which it can dissolve (such as a vegetable oil), to make the size of a single dose large enough for a consumer to be able to measure out.   For instance, if we were to dilute the vitamin E 100-fold with vegetable oil, then 1 RDA would be 1.5 g (instead of 15 mg), which works out to about 1/3 tsp.  That's a dosage consumers can handle.  Of course, the typical approach is not to offer it as a liquid, but rather conveniently encapsulated.

This is a good commentary on the practicalities. In terms of chemistry, do you agree with Maurice's answer? It almost seems the two should be merged as they both tell half the story.

(1) Maurice's answer is plausible.  (2) Except in specific instances, answers on this site aren't merged, since they each stay attached to the person that created them; that's related to how the the points system works.

Vitamin E is not soluble in water. It is only soluble in oils. So it is no use swallowing pure Vitamin E, or this vitamin with water. It will poorly be absorbed by the intestine. You must absorb it in an oily solution.

This is good information on the chemistry; do you agree with theorist's answer about the practicalities?


text:
Anomaly in van der Waals constants of noble gases
In van der Waals equation for a real gas, the equation is
$$\left(P + a\frac{n^2}{V^2}\right)(V - nb) = nRT$$
where $P$ is the pressure of the real gas, $n$ is the number of moles of the gas, $R$ is the ideal gas constant, $T$ is the temperature, $a$ is a constant to correct for the intermolecular forces of attraction, and $b$ is a constant to correct for the size of the molecules.
It follows that gases with large intermolecular forces have large $a$ values and gases with large molecular size have large $b$ values.
Below is a table of the $a$ and $b$ values of the ideal gases:
\begin{array}{ccc}
\hline
\text{gas} & a & b \\
\hline
\ce{He} & 0.0341 & 0.0237 \\
\ce{Ne} & 0.211 & 0.0171 \\
\ce{Ar} & 1.35 & 0.0322 \\
\ce{Kr} & 2.32 & 0.0398 \\
\ce{Xe} & 4.19 & 0.0511 \\
\hline
\end{array}
Why is the $b$ value for helium larger than the $b$ value for neon even though helium's atomic radius is smaller than that of neon? Does it maybe have to do with helium's small size, and thus higher electron density surrounding the nucleus, creating more repulsions with other helium atoms?

What's the source? Also did you check other sources?

This table is invalid, it lacks the temperature/pressure range in which the equation was fitted for each gas. If they used the same range for all of them, that would definitively make the results incomparable. $a$ and $b$ are not universal constants for each gas, and generally vdW is not a good model for any gas over a wider p/T range.

@Mithoron from the Wikipedia page: https://en.wikipedia.org/wiki/Van_der_Waals_constants_(data_page)

@Karl Why do you say the table is invalid when almost all of the chemical data pages for van der waals constants give only "a" and "b", as exemplified here: https://www.engineeringtoolbox.com/non-ideal-gas-van-der-Waals-equation-constants-gas-law-d_1969.html and here: https://en.wikipedia.org/wiki/Van_der_Waals_constants_(data_page) and here: http://www2.ucdsb.on.ca/tiss/stretton/database/van_der_waals_constants.html

I know, its terrible. a and b are just empiric parameters, valid for a limited p/T range. If you make the range too big, you get a bad fit and nothing makes sense, if you make it small enough, you still cannot compare between different gases. Giving vdW parameters without a T range is simply nonsense.

@Karl (1) The VDW constants are not temperature-dependent. They are determined by the critical temperature and pressure, which are fixed: ${b =  \frac{R T_c }{ 8 P_c}}$ and ${a = \frac{27 R^2 T_c^2}  {64 P_c}}$.  Thus, in the VDW equation, they truly are constants. If you replaced the VDW constants with temperature-dependent functions a(T) and b(T), then you would no longer have a VDW model. (2)@DrPepper Given the above, your question can be reframed as: "Why is $b =  \frac{R T_c }{ 8 P_c}$ larger for helium than for neon?"  To answer this, we'd need to think about why $\frac{T_c }{P_c}\sim b$

It is practically impossible to draw conclusions without some error estimates and a better idea about how the coefficients were derived. It would be a good idea to look at the CRC and the original sources. However He being He it would not surprise me if there is an odd explanation to the trend.

@DrPepper What I wrote also explains why, as you've correctly noted, all the online reference tables only give single "a" and "b" VDW values for each gas.  Contrary to what Karl wrote, those tables are not invalid.  Interestingly, most of the more sophisticated gas models listed at  https://en.wikipedia.org/wiki/Real_gas likewise use non-T-dependent consts., and instead attempt to improve accuracy w/ more sophisticated functional forms (&/or more consts.).  Two exceptions, among those listed, are Peng-Robinson, which uses a T-dependent "a" (but const. "b"), & the virial eqn.

@theorist Sorry, that is complete bullshit. You can derive a and b with that additional theorem, however this assumes that the gas behaves as an "ideal van der Waals gas". Which no real gas does. Ergo those values are basically always worse than what you would get by fitting a/b with a set of p/T values in a reasonably limited range. And a list of a/b values that does NOT say how the values were determined is just a laugh. ALL "constants" in any gas equation have a residual T dependence, because there is no such thing as a final, perfect gas law. Even ten parameters are not enough!

The values in the table in this post come from the CRC handbook, and that book explains that they were computed from critical constants. I side with Karl in that there is no guarantee that the vdw constants properly describe the behavior of the gases. In other words, while van der Waals gases sharing the same critical points as the noble gases have to possess the tabulated constants, this does not mean that noble gases are well described as vdw gases having that particular set of constants (except at the critical point).

@Karl You don't understand how the VDW model works, and are peddling falsehoods as result (such as when you said a non-T-dependent table of VDW constants is invalid).  Of course the VDW eqn. has serious limitations modeling real gases. Everyone knows that. That's not the issue here. The issue is why, within the constraints of the VDW model (which is a model in which, by defintion, the constants are not T-dependent, b/c they're based on critical pts.), we get an exception in the trend.   A way to understand that is to look at how the constants are derived, which is what I offered in my comment.

To put it another way, consider, ideal gas (IG) eqn, PV = n R T.   The IG model, which contains the constant "R", is an even worse model for real gases than VDW.   That doesn't mean that a table of physical constants that shows the IG R w/o a T-dependence is invalid!!   Anyone who thinks R is invalid without a T-dependence b/c the IG model is a limited model for real gases doesn't understand that model. And insisting that the VDW a and b are invalid without T-dependencies reflects the same lack of comprehension as insisting the IG R is invalid w/o a T-dependence.

There seems to be a reasonable explanation for this phenomenon here: http://www.madsci.org/posts/archives/1998-02/887638249.Ch.r.html. The author concludes that polarizability will affect the b parameter, appealing to ionization potentials to explain this trend.

I'm not sure I really want to wade into this argument, but... a and b are indeed constants defined by critical parameters in the vdW model, as theorist has described. This might not describe real gases all that well, as Karl has stated, but it remains a property---indeed, a necessary property---of the model for consistency. Karl's statements make more sense for virial coefficients, where no such restriction holds and a temperature range should indeed be stated.

The Quantum Gases
I will venture a guess, and hopefully a rational explanation. As discussed in the comments, the van der Waals equation parameter can be derived from mathematical considerations at the critical point and can be calculated as $\displaystyle b = \frac{RT_\mathrm{crit}}{8P_\mathrm{crit}}$.
The critical temperatures of the first three entries from your table are
$$\begin{array}{cr}
\text{Gas} & T_\mathrm{crit}/\pu{K} \\ \hline
\text{He} & 5.2 \\
\text{Ne} & 44.5  \\
\text{Ar} & 150.7
\end{array}$$
From these, I will offer hand-waving "quantum effects" as the culprit affecting the parameters of this classical equation of state. At the critical point of helium, quantum effects dominate. As things warm up to neon's critical temperature, quantum effects become less important, and by the time we hit argon's, many (if not most) quantum things can be ignored. Because the quantum effects will also scale inversely with mass, they are again important in the order: helium, neon, argon.
So I imagine because hydrogen, helium, and to some extent neon are "quantum gases"
their van der Waals equation parameters will not be directly comparable to those of "classical gases".


text:
Sum of a rate constant gives the total decay rate
I want to ask a question about the sum of a rate constant to yield the total decay rate.
From previous studies, the rate of a reaction can be expressed evidently as
$$\ce{v_{rate} = k[A]}$$
where $\ce{A}$ is a molecule in a reaction.
I studied today about unimolecular photochemistry and that a molecule in its excited state $\ce{S_1}$ can do a number of processes:

Fluoresence (emit a photon and return to its ground state), $\ce{F}$
Intersystem crossing (convert to a triplet state and then relax), $\ce{ISC}$
Internal conversion (release heat to the environment and relax back to the ground state), IC
Photochemistry (be involved in a reaction), PC

and according to the lecture material, the total decay rate can be expressed as:
$$\ce{k_F + k_{ISC} + k_{IC} + K_{PC}}$$
i.e. the sum of the rate constants.
But I thought that from my studies, the total decay rate is expressed as:
$$\ce{v_{total \ decay \ rate} = (k_F + k_{ISC} + k_{IC} + K_{PC})[A]}$$
so why do I not include the concentration term as part of the total decay rate?
My thought was that $\frac{v_{\ total \ decay \ rate}}{[A]}$ is expressed as the total decay rate and hence it can be expressed as the sum of the rate constants, but I want to check that my logic is correct.
Why is the total decay rate expressed as a sum of all the rate constants of the expressions above?

Because it's not exactly a first rate reaction. More like pseudo-zeroth order, depending on illumination.

@Mithoron can you explain this a bit further - I'm a bit confused here apologies!

It looks to me like an example of the common misuse of "rate" when they mean "rate constant".

In your usage v is rate (conc /time) and k is rate constant in your case 1/time and as @Andrew writes you are confusing them. And yes rate constants do add  as all process start from the same excited state. As you write $k=k_F+k_{isc}$ etc

I am simply confusing two different variables here, rate and rate constants.
In this situation, the sum of the rate constants is the sum of the individual rate constants of the processes occuring in the reaction i.e.
$$\ce{k_F + k_{ISC} + k_{IC} + K_{PC}}$$
This is consistent with my current knowledge.
It should be referred to as "total decay rate constant"
Of course, the rate equation can still be written here as:
$$\ce{v_{rate} = k[A]}$$


text:
Do SN1 reaction always result in Racemisation

In the following reaction the products formed are not enantiomers but are diastereomoers.So there is no racemisation occured.So is it always correct to say that SN1 result in racemisation?

This doesn't qualify as racemisation, it is epimerization. Racemization would require both stereocentres to invert to form the enantiomer and - as you rightly suppose - the stereocentre bearing the methyl group is untouched by this reaction.
It is also unlikely to lead to a 1:1 mixture of products since the two diastereomers (and the transition states leading to them) are different in energy, so under either thermodynamic or kinetic control one would expect a non-1:1 mixture. This contrasts with the case of a molecule undergoing true racemization in which the two enantiomeric products will be equal in energy so will be formed in equal quantity.
Your title also poses an interesting and more general question - do SN1 reactions always lead to racemisation even in molecules containing only one stereocentre? I was winding up to write an answer to this question, before finding the outstanding answer previously given by NotEvans here - well worth a read. (hint: the answer is - not necessarily).

It results in racemisation if only one chiral carbon atom is present in the compound.  If there are many,  it merely leads to formation of compounds with both R and S configurations of the chiral carbon on which substitution takes place.


text:
Selectivity of metal nitrates towards ammonium bicarbonate
I am performing coprecipitation reaction, and in the precipitating step, I have a bunch of metal nitrates(ferric nitrate, aluminium nitrate and barium nitrate)in a  solution reacting with ammonium bicarbonate.
I need to know the selectivity of these metal nitrates towards the ammonium carbonate and the order in which these metal nitrates will start reacting with the base. I know the Ksp of barium and aluminium nitrate but not of the ferric nitrate. Is there any way of knowing the order in which these compounds will react?
How do I proceed with the analysis? Any link to a book or any article will be appreciated.

In other words, you are asking for the solubility products for the carbonates and hydrocarbonates of all the metals (91) out there, aren't you? If so, the question seems quite broad.

I have Ferric nitrate, barium nitrate and aluminium nitrate reacting with ammonium bicarbonate. I know the Ksp of barium and aluminium nitrate but not of the ferric nitrate. Is there any way of knowing the order in which these compounds will react?

It may be better to update the question, instead of keeping it as a comment appendix.

When dissolving for exemple baryum nitrate in water, there is no baryum  nitrate any more in the solution. The solution is a mixture of Baryum ions and of independent Nitrate ions. So there is no "selectivity of these nitrates" as you say. Your question has no meaning. The only order of reaction is the order of the solubility products of the corresponding carbonates or bicarbonates of these metals. The trouble is that aluminium and ferric ions do not make any carbonates or bicarbonates. These compounds are decomposed into hydroxides. So basically, the answer of your question depends strongly on the pH of the solution.

My operating pH is between 7.5-8.


text:
Best method to increase the chain length of allyl to butane-1,2-diol
Best method to increase the carbon chain length of allyl. and also add 2 OH  carbon after reducing double bond give (S)-4-R-butane-1,2-diol and (R)-4-R-butane-1,2-diol .
R= Any Ak, Cb, Cy, Hy

Notwithstanding the convention that you should offer you ideas before receiving an answer here is a viable route.
Step 1 - hydroborate the alkene then oxidise the resulting alkylborane with N-Methyl Morphiline-N-Oxide/Tetrapropyl Ammonium Perruthenate to give the aldehyde as described here
Step 2 - Wittig reaction with Methyltriphenylphosphonium Iodide/LDA to give the chain extended alkene
Step 3 - Syn hydroxylation of the alkene with Osmium Tetroxide/N-Methyl Morphiline-N-Oxide details here.  You can buy micro-encapsulated OsO4 which is much safer to handle than the neat liquid

Is their any method which can make aldehyde together with the addition of one carbon in one step? and then we can make 1,2-diol by using asymmetric catalyst. I see it but forget this method after long time.

Thank you so much. I find that paper, also going to mention here.https://onlinelibrary.wiley.com/doi/full/10.1002/anie.201001972#

You can also carbonylate the alkylborane to give the aldehyde, but you need the specialised equipment to do it - pressurised reaction vessel and CO gas supply - and these are not cheap.

Yes, you are right, I am going to follow your procedure as you said. It's a convenient method to do.

I used your method but it's giving a mixture of two products, and also yield is only 35%. Any other method that first I convert alkene to alcohol then into aldehyde without decreasing too much yield.

i think you need to open a new question. Give full experimental details and then I'll try to answer

I take terminal olefin and make epoxide which is rarely stereoselective in the absence of a directing group. But luckily i got pure R enantiomer, now regioselective ring opening of epoxide with Li2NiBr4 in THF afforded the bromohydrin as a single regioisomer. Then i will add R groups at OH and Bromine position to get the desired product.


text:
How do Lewis acids interact with 1,4-diketones and amines to form an alkaloid?

The following double cyclization was used to generate the ring system of the lycoanitine alkaloids in 2005. Demonstrate your understanding of this C−C bond-forming process by providing a mechanistic rationale (using arrows that represent
  electron movement) for the boxed reaction.

From what I know in my current synthetic organic chemistry class, this question appears to me like a Paal–Knorr reaction, but the product isn't a pyrrole. I don't even know where to start, and I'm not entirely certain how and where the Lewis acid interacts.
Ignore the boxed part, that's for me to do; I'm just curious to know how the carbons attached to the methoxy group on the tertiary amine are able to bond with the C2 carbons on the 1,4-diketone.

This is a variant of the Mannich reaction. The methoxymethyl groups on the amine are a form of captured Mannich intermediates or masked formaldehyde equivalents. The Lewis acid unmasks them by binding to the oxygen of the methoxy group, creating the functional equivalent of H2C=N+. These are reactive electrophiles which react with the active methylenes of the B-ketoester to give the alkaloid.
More on the Mannich reaction here


text:
which oxygen atom make intramolecular hydrogen bond in salicyclic acid?

When I googled about intramolecular hydrogen bonding in Salicylic acid, I found that there are two type of hydrogen bonding. Which one is more appropriate?
And another question, if there ARE two types of hydrogen bonding, why doesn't the 15.2 ppm peak of NMR specturm doesn't split in two? It seems that the two form of hydrogen bonding has diffrent energy level.


text:
Stable physiological buffer to make NAD+ solution
I was wondering what would be the most stable physiological buffer for nicotinamide adenine dinucleotide (NAD+), for potential storage as frozen aliquots - 6 months stability or around there would be ideal. A buffer that is physiologically compatible is essential (e.g. suitable for in vivo use).
I would also be interested in potential solubility limits. I've seen reports of people achieving up to 200 mg/mL solutions, but others stating that they made 50 mg/mL.
I have read from Sigma that

NAD is soluble in water at 50 mg/mL, and that an aqueous solution may be made by adjusting the pH using NaOH or a sodium bicarbonate buffer.

Does this not imply that a carbonate buffer would be usable? They then also state that

As per external source information β-NAD is soluble in PBS pH 7.2 at 10mg/ml.

This to me implies that a phosphate buffer would be suitable. However, they finally said the following:

Aqueous solutions between pH 2 - 6, stored as single-use aliquots at -70 °C, are stable for at least 6 months. Neutral or slightly acidic solutions are stable at 0 °C for at least 2 weeks. Solutions are rapidly degraded upon heating and are very labile in alkaline solutions, especially in the presence of phosphate, maleate, or carbonate.

The last part of this sentence makes me think that phosphate or bicarbonate buffers would thus be inappropriate. I'm not sure why they say conflicting things, and they were not able to provide me with any further information.
They do confirm that neutral or slightly acidic pH should be fine for long-term storage.

My current thoughts are that I can just try something like PBS (10 mM phosphate buffer) but ensure the pH is between 5 and 6. I assume it could then potentially be stored at -20 (not -70) C for 6 months...
Not sure if 10 mM buffer is sufficient to maintain a 200 - 300 mM NAD+ solution, or if I should go with 100 mM. Storing it at -20 may slow any reactions sufficiently so as to make 10 mM appropriate - happy to hear any thoughts on this or any of the post in general!


text:
Finding percent weight
I have a mixture of $\ce{NO2(g)}$ and $\ce{N2O4(g)}$ at $63\ \mathrm{^\circ C}$ and $750\ \mathrm{mmHg}$. I need to find the weight percent of the first compound. Density $d=1.98\ \mathrm{g/L}$.
I assumed that we have $M=aM_1+bM_2$ where $M_1$ and $M_2$ are the molar masses of the first and second compound. I know that $m=nM$ so the weight percent would be $$\frac{m_1}{m}\cdot100\ \%=\frac{aM_1}{aM_1+bM_2}\cdot100\ \%$$
My problem lies in finding the coefficients. Can someone give a hint please?

You need to know the equilibrium constant for this temperature, and then use a similar approach as shown here: How can I calculate the percentage of dissociated dinitrogen tetroxide?

@andselisk I still haven't learnt that, so the question doesn't require that approach. thank you though :)

There is an equilibrium between both components which dictates the fraction of each, and without $K_p$ I doubt you'll be able to do much.

but seriously, we are supposed to do this without using the equilibrium.

Personally, I don't see how this could be done with the data you posted. Could you please add a source for this problem?

From my university professor. Since this is leading nowhere.. In the solution he uses mole fractions $X$. He takes the molar mass found as an average molar mass, so that $M=XM_1+(1-X)M_2$ and then finds $X$ using given data. Then he considers mass fractions and gets $%w=\frac{XM_1}{XM_1+(1-X)M_2}*100%$

I don't understand this, however, $X$ is dimensionless, $XM_1$ makes no sense to me. Or maybe by weight percentage he means molar mass not mass?

There is no given average molar mass of the mixture, so you stuck with one equation and two unknowns, and in order to use the second equation, you also need to find molar fraction $x_1$ (or $X$, according to your notations), which is normally done using $K_p.$ From the dimensional analysis everything looks fine, I don't understand why $x_1M_1$ doesn't make sense to you.

We can find the actual molar mass with the density formula involving pressure and temperature, then consider it as the average molar mass. I know understand when I look at the full equation: the total number of moles will cancel each other from nominator and denominator and we will end up with (implicitly, since we won't actually be simplifying) $\frac {XM_1}{XM_1+(1-X)M_2}=\frac {n_1M_1}{n_1M_1+n_2M_2}=m_1/m$.

I assume you mean ideal gas law $\bar{M} = \frac{ρRT}{p}$ ($\bar{M}$ is the average molar mass), but there is no given density of gaseous mixture $ρ$ either. So how is this going to be helpful here?

My bad, I forgot to include it. It is $1.98,g/L$.

Why do we consider that as average molar mass, though? Doesn't it give the actual one of the system?

Well, now you know how to solve this, too bad you didn't show the complete data from problem from the beginning as this would save time. What do you mean by "actual one"? $\bar{M}$ is the actual average molar mass of the mixture.

I meant the real not average.

The average one is also very real. At this point I suggest you edit the missing density into the question, and you can also post an answer. Lengthy comments are going to be purged (or moved to the chat by the system or us mods) in a matter of hours/days.

Please don't use number of moles, this is an ancient term dating back before a proper definition of amount of substance.

We can find the average molar mass with the density formula involving pressure and temperature $\overline M=\frac{\rho TR}{p}$.
We also know that, in this case, $\overline M=\frac{n_1M_1+n_2M_2}{n_1+n_2}$. Since we only "care" about $M_1$ we can write the average molar mass as $\overline M=\frac{n_1}{n_1+n_2}M_1+(1-\frac{n_1}{n_1+n_2})M_2$ and then find $\frac{n_1}{n_1+n_2}$.
$$\frac{\frac{n_1}{n_1+n_2}M_1}{\frac{n_1}{n_1+n_2}M_1+(1-\frac{n_1}{n_1+n_2})M_2}^{(*)}=\frac{\frac{n_1}{n_1+n_2}M_1}{\frac{n_1}{n_1+n_2}M_1+\frac{n_2}{n_1+n_2}M_2}=\frac{n_1M_1}{n_1M_1+n_2M_2}=\frac{m_1}{m_{total}}$$
Compute $(*)\cdot100\%$. The development of the fraction was to show why it works.
Another possible solution is this :
$$\%w_{m_1}=\frac{m_1}{\rho V}*100\%\Leftrightarrow m_1=\%w_{m_1}\rho V/100\%$$
$$\%w_{m_2}=\frac{m_2}{\rho V}*100\%=100\%-\%w_{m_1}\Leftrightarrow m_2=(100\%-\%w_{m_1})\rho V/100\%$$
$$n=\frac{m_1}{M_1}+\frac{m_2}{M_2}=\frac{\rho V}{100\%}\left(\frac{\%w_{m_1}}{M_1}+\frac{100\%-\%w_{m_1}}{M_2}\right)=\frac{PV}{RT}$$


text:
How to remove putrescine odour
Does anyone know how to remove the odour remaining from a decomposed corpse which has fallen onto a wooden floor? Aparently the two main chemicals are putrescine (tetramethylenediamine) and cadaverine (pentamethylenediamine).

Both of those are amines so repeated washes with dilute acid (maybe vinegar) should help.

Thanks. So using bases such as Sodium hydroxide would not be the thing to do.

No, not at all. Bleach is another possibly taking the amines to the amine oxides

https://chemistry.stackexchange.com/questions/6290/what-causes-the-smell-in-rotten-meat-grease-and-how-can-i-get-rid-of-it-thorough


text:
Equivalence point vs inflection point in the titration curve of a weak acid
Let's suppose we want to titrate a solution containing an unknown monoprotic and weak acid. We use a strong base, such as $\ce{NaOH}$. When the number (and moles) of hydroxide ions coming directly from the base is equal to the amount of hydronium ions due to the acid, here we have the equivalence point.
When titrating a strong monoprotic acid the equivalence points coincides with the inflection of the titration curve, and pH is seven.
In our case, however, the aqueous solution results to be slightly basic at the equivalence point. Therefore, does the inflection point on our curve still coincide with the equivalence point or it is to be found slightly before?
In other words, what does the inflection point indicate, that $\mathrm{pH}$ is seven (so the inflection is distinct but may coincide with the equivalence point, and shows where the rate of increase in $\mathrm{pH}$ is maximum) or the equivalence point itself?

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

@Poutnik I saw where the mistake was; I edited now, this is what I meant

Let's suppose we want to titrate a solution containing an unknown monoprotic and weak acid. We use a strong base, such as $\ce{NaOH}$. When the number (and moles) of hydroxide ions is equal to the amount of hydronium ions, here we have the equivalence point.

The equivalence point is, when the molar amount of the spent hydroxide is equal the molar amount equivalent to the originally present weak acid. At this point,  $[\ce{H3O+}]<[\ce{OH-}]$, so $\mathrm{pH} \gt 7$.

When titrating a strong monoprotic acid the equivalence points coincides with the inflection of the titration curve, and pH is seven.

For the approximative experimental chemical approach, it is true.
But mathematically, even this is not exactly true, as the curve is slightly asymmetric. When there is the same molar amount of hydroxide as was the acid, the solution volume is 3 times bigger. And hydroxide concentration 3 times lower.
It would be symmetric, if we constructed the graph for the constant total volume and complementing strong acid and hydroxide solution volumes, as shown in the formula below.
$$\mathrm{pH}=f\left(\frac{V_\mathrm{acid}}{V_\mathrm{acid} +  V_\mathrm{hydroxide}}\right)$$
And the concentrations of both should be the same.

In our case, however, the aqueous solution results to be slightly basic at the equivalence point. Therefore, does the inflection point on our curve still coincide with the equivalence point or it is to be found slightly before?

For weak acids, the coincidence is mathematically even less true, as the curve shape before and after equivalence differs.
But for the chemical experimental point of view, the difference is still negligible, as the equivalence/inflection coincidence error is much less then experimental errors.
Both inflection and equivalence points ( approximately identical ) lay in the alkalic region.

In other words, what does the inflection point indicate, that $\mathrm{pH}$ is seven (so the inflection is distinct but may coincide with the equivalence point, and shows where the rate of increase in $\mathrm{pH}$ is maximum) or the equivalence point itself?

For weak acid/strong base ( and vice versa ) titrations, the neutral $\mathrm{pH}$ does not play any special role, so this point on the curve is insignificant.
Neither is there at the neutral point any inflection point, unless by chance $\mathrm{p}K_\mathrm{a}=7$. In such a case, the inflection point  at the maximal acid/salt buffering capacity at the half equivalence would be at $\mathrm{p}K_\mathrm{a}=7$.
For further reference, see e.g  Titration_of_a_Weak_Acid_with_a_Strong_Base