text:
Mechanism for conversion of ketone to dichloride with phosphorus pentachloride

I would like to know the mechanism of this reaction leading to the formation of a geminal dichloride. I am unable to predict how the $\ce{C=O}$ breaks and how the 2 $\ce{Cl}$ atoms replace the oxygen atom. What kind of a reaction is it?
Note: This reaction is mentioned on page 310 of Solomon and Fryhle's  Organic Chemistry in the chapter "Alkenes and Alkynes". It's the precursor of the preparation of cyclohexylacetylene through E2 mechanism using 3 equivalents of $\ce{NaNH2}$

I would encourage you to think about how these molecules are likely to interact with each other and try outlining a possible mechanism.

Step 1: The carbonyl oxygen carries a partial negative charge due to its higher electroneegativity. It also has a high  affinity towards bonding with the $\text{III}^{\text{rd}}$ row atoms which have $\ce{d}$-orbitals, such as $\ce{Si}$, $\ce{P}$ and $\ce{S}$.
$\ce{PCl5}$ exists as $\ce{[PCl4+][PCl6-]}$. The lone pairs on the oxygen attack the central phosphorus atom.
Step 2: The negatively charged phosphonate complex has high energy (nucleophilic) $\ce{P-Cl}$ bonds. They attack the electrophilic carbonyl carbon. Note that these bonds are more nucleophilic than basic and would prefer substitution over elimination.
Step 3: The lone pairs on the oxygen are next to the $\ce{d}$-orbitals of the phosphorus. They can push out one of the chlorines attached to the phosphorus.
Step 4: The chloride pushes the $\ce{-OPCl3+}$ leaving group out giving the dichloride. As orthocresol pointed out, this step is likely to be an SN1 type of substitution owing to the excellent leaving group and the rather weak nucleophile $\ce{Cl-}$.
Note: Steps do not represent elementary reaction steps. The $\ce{PCl5}$ formed after $\ce{PCl6-}$ gives its $\ce{Cl-}$ is ignored. I have seen/heard Step 3 and Step 4 being described as concerted, but I have shown them here separately as I am not sure of those "rumours".

It's also possible that step 4 is some type of strange SNi type mechanism akin to what happens with thionyl chloride. @orthocresol

It is probably not such a complicated multistep reaction; initial formation of a chlorophosphate forms a rather weak C-O bond and strong P-O bond resulting in a fast transfer of the second Cl- and formation of the P=O bond. The first and second steps in the above mechanism are suspect, why involve second questionable molecules when there could be an ylid like first step and there is a Cl right there to react intramolecularly . This looks like it is almost a "click reaction". A good kinetic study should tell. [The remaining steps are wishful]


text:
Why are ethers treated as substituents and not as functional groups in IUPAC nomenclature?
Why are ethers not considered as functional groups in IUPAC nomenclature?
Let us consider the following example:

Image source
The preferred IUPAC name (PIN) for this compound is 2-ethoxy-1,1-dimethylcyclohexane. And we get this by first point of difference as ether is also considered as a substituent here.
But ethers do confer specific properties to the compound. They are studied as a separate functional group. Yet they aren’t given the preference in nomenclature in the absence of any other functional group too. Why is this so? Is it just a convention or is there a reason behind it?

PS: I checked this similar question on our site but it doesn’t answer my question as to why ethers aren’t functional groups when it comes to naming.

Names such as méthoxyéthane were already included in the Liége Rules in 1930 (the predecessor of the IUPAC nomenclature of organic chemistry). Early rules were usually descriptive rather than prescriptive.

Note that the genre of this post is is going to be dystopian speculative science fiction. (I'm not sure whether I should hope that in IUPAC commission, nobody reads it, or at least somebody reads it).

This compound could be named by our ancestors like e.g. amines, O-ethyl-2,2-dimethylcyclohexan-1-ol.
Or like e.g. phosphane derivatives. Note that in ancient times, they could be naming it like „(2,2-dimethylcyclohexyl)(ethyl)ether“ (beware archaic „2,2-dimethylcyclohexyl-“ or „ethyl-“), which would be silly, because by removing both substituents, you don't get ether at all! But, „(cycloalkyl)(alkyl)water“ would sound even more silly. Therefore, it could be named rather (2,2-dimethylcyclohexan-1-yl)(ethan-1-yl)oxidane (beware, „… oxane“ would be something completely different).
In the year 202X, funnily, e.g. $\ce{(C2H5)2O}$ was still being officially preferably named „ethoxyethane“ instead of oxydiethane, even though 1,1′-oxybis(2,2-dimethylcyclohexane) was already the correct name for our compound with ethyl replaced by the left-hand substituted cyclohexyl. So, one could obviously name it 1,1′-oxy(2,2-dimethylcyclohexane)(ethane).
But no, in the current century, we name it C₁₀H₂₀O/c1-4-11-9-7-5-6-8-10(9,2)3/h9H,4-8H2,1-3H3 (EIN) (some historic variants of the Enforced IUPAX Name were called InChI).

For nomenclature purposes, some characteristic groups (e.g., ethoxy, nitro, chloro, nitrazo, diazo, azido, etc.) are forever out of the limelight and are always cited as prefixes to the name of the parent structure (e.g., such as one described in Chapter P.2). Multiplying prefixes and locant can be added as necessary. Ether group is one of these forever suffix groups in substitutive nomenclature (Table 5.1, Section P-59.1.9, 2013 Blue Book). Thus, the PIN for the given compound is 2-ethoxy-1,1-dimethylcyclohexane (the parent hydride in here is cyclohexane ring).

The priority order for functional groups is as follows:
Radicals > Anions > Cations > Zwitterionic compounds > Acids (In the order -COOH, -COO(O)H, S-derivatives, Se-derivatives, Sulphonic acids, Phosphonic acids, etc)> Anhydrides > Esters > Acid halides > Amides > Hydrazides > Imides > Nitriles > Aldehydes > Ketones > Alcohols and Phenols > Amines > Imines > hydrazine > Phosphanes > Alkene > Alkynes > Alkanes> Ethers > peroxides >  Halo, Nitro, Nitroso, Alkoxy > Oxiranes
Functional groups such as Ethers and Nitro-groups are way down the order and will not be the primary functional group more often than not, the reason behind them be named as substituents.
In your question, the alkane group is given priority over the ether as per the above-mentioned priority order for functional groups. Thus the name given is 2-Ethoxy-1,1-dimethylcyclohexane.
Let's say instead of the 2 methyl groups over there, you had a nitro group, then the ether would have become the principal functional group and the nitro group being the substituent. There's no rule saying that ethers are always to be named as substituents as per my knowledge.

@J_B892 I'm not sure if I agree with this explanation. The question still remains as it was. See: alkanes are not a functional group, whereas nitro and ether are a functional group. Yet the former is given a higher priority over the latter. So, what was the need to separate ether and nitro from the rest of the functional groups (which are above alkanes), and place them below the alkanes instead?

“In your question, the alkane group is given priority over the ether as per the above-mentioned priority order for functional groups.” No. The name 2-ethoxy-1,1-dimethylcyclohexane is simply based on the usual substitutive nomenclature. The seniority order is not relevant in this case since no functional group is expressed by suffix or class. Compare also the analogous compound 1,1-diethyl-2-methoxycyclohexane.

“Let's say instead of the 2 methyl groups over there, you had a nitro group, then the ether would have become the principal functional group and the nitro group being the substituent.” No. That compound would still be named by usual substitutive nomenclature as 1-ethoxy-2-nitrocyclohexane.

Whats the difference in alkoxy and ethers in your list?

What is the source of your ‘knowledge’?

source: Stack Exchange

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

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
How to tell/control whether sodium ethoxide will react by a substitution or elimination pathway?
Why is the behavior of sodium ethoxide inconsistent?
In a reaction with 1-bromobutane, the anionic part of sodium ethoxide substitutes for $\ce{Br}$:
$$ \ce{CH3CH2CH2CH2Br + C2H5ONa -> CH3CH2CH2CH2-O-CH2CH3}$$
Conversely, in a reaction with 2-bromobutane, the anionic part eliminates $\ce{Br}$:
$$\ce{CH3CH2CHBrCH3 + C2H5ONa -> CH3CH2CH=CH2}$$
I simply don't see how changing the position of $\ce{Br}$ changes the type of reaction. I believe sodium ethoxide, being bulky, must always do elimination, consistent with Rule #3 here.
Given the above: can sodium ethoxide be forced to one type of reaction with a given reactant, by enforcing some reaction conditions?

I'm sure there are already several questions covering when a reagent causes E2 and when it causes SN2, did you have a look at them already? Thank you.

Think about the difference in steric hindrance in the two halides.(primary and secondary). Also, in the second reaction its not that only elimination occurs.

@GaurangTandon I except sodium ethoxide to show E2 only when a hallide is present. I've seen examples where even the primary hallide undergoes E2

On a side note, every single opening or closing parenthesis in your formula lines is redundant.

@IvanNeretin I think it adds readability. Although they are redundant.

@SmarthBansal Trust me, it doesn't.

SmarthBansal, if you dislike any of my edits, please feel free to change things back or edit further.

The above reactions belong to Williamson Synthesis of Ethers, probably the best of the alternatives for prepartion of ethers!
The Williamson Ether synthesis reactions follow SN2 mechanism. Since the SN2 mechanism proceeds through a single step where the nucleophile performs a “backside attack” on the alkyl halide, the only thing stopping this, is steric hindrance.
Methyl and primary alkyl halides are excellent substrates for this synthesis, as they provide no/very less hindrance for the approaching nucleophile.
Since alkoxides are strong bases, competition with elimination (E2) becomes a concern once the alkyl halide becomes more sterically hindered. This is exactly why, the first reaction that you’ve mentioned, proceeds by SN2 but in the second both elimination(major) and substitution (minor) takes place.
Also note that, when tertiary alkyl halides are used, there is no hope for substitution because of the bulky groups present around carbon. Hence, only elimination takes place in this case.

So the reactions are,

With primary alkyl halide,
$$ \ce{CH3CH2CH2CH2Br + C2H5ONa -> CH3CH2CH2CH2-O-CH2CH3}$$
With secondary alkyl halide,
$$\ce{CH3CH2CHBrCH3 + C2H5ONa -> CH3CH2CH=CH2(major)}$$

Moving onto conditions to favour a particular type of reaction:

One way to attempt to get the SN2 to be favoured over elimination is to use a polar aprotic solvent (such as acetonitrile or DMSO) that will increase the nucleophilicity of the alkoxide.
Increasing the temperature favours elimination over substitution.

Source for the above conditions: chem.ucalgary

I believe sodium ethoxide being bulky should always do elimination

In most cases, rather than the incoming nucleophile, the hindrance in the the alkyl halide plays a more important role.

The rule #3 mentioned in the link you stated clearly says that the elimination takes place for a tertiary carbon in presence of a strong base like $\ce{CH3CH2-O-Na}$.

Nice Answer! If I use polar aprotic solvent and the hallide is $2^0$ will it show SN2 or E2?

@SmarthBansal Thanks:). Majorily SN2 because of the aforesaid reasons.

So using aprotic guarantees it's SN2?

As per the link, SN2 will be favoured. But you cannot say that it will only be substitution. I can guarantee that  it surely won’t be Elimination only. ** Note: Substitution and elimination reactions are strongly influenced by many experimental factors**

You chose a good link but look at the not very helpful rule:
Your mentioned link clearly stated followings rules for the reactions of strong base/nucleophile with a primary, secondary, and tertiary halides, which are well acceptable:
Rule 1: A reaction of strong base/nucleophile with a primary halide (and methyl halide) will almost certainly be SN2.
Exceptions: (a) Bulky bases such as $\ce {(CH3)3CO-}$ (note that $\ce {CH3CH2O-}$ is not considered to be 'bulky') gives E2 product predominantly; and (b) some primary halides such as allyl and benzyl halides can form relatively stable carbocations (by resonance stabilization, for example) may proceed through either of SN2, SN1, or E1 pathway, based on other conditions such as solvent and temperature.
Rule 3: A strong base/nucleophile usually favors SN2/E2 pathways (Neither SN1 nor E1 should be considered).
Exceptions: Both Rule 1 exceptions apply here. Apart from them, tertiary halides give E2 product predominantly because of steric hindrance present in tertiary carbon prevents backward attack by the base/nucleophile, regardless of its size. The elimination products are, however, depend on the size of the base/nucleophile: (a) Bulky bases such as $\ce {(CH3)3CO-}$ would favor to give less substituted alkene ("Hoffmann" Rule), while (b) not so "bulky" bases such as $\ce {CH3CH2O-}$ would favor to give most substituted (hence most stable) alkene ("Zaitsev" Rule).
Rule 5: If all else being equal, a reaction of strong base/nucleophile with a secondary halide is based on other conditions such as solvent: For example, (a) polar aprotic solvents favor SN2 (substitution) pathway, while  (b) polar protic solvents favor E2 (elimination) pathway.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What does "the magnesium salt of EDTA" mean?
I am attempting to prepare an ammonium chloride/ammonium hydroxide buffer solution ($\ce{pH}=10 \pm 0.1$) for titrating water hardness with calgamite and EDTA.
In the 17th Edition of the Standard Methods for the Examination of Water and Wastewater, the reagent preparation section for titrating water hardness (Section 2340 C.) states:

Mix the $\ce{NH4Cl+NH4OH}$

Add $\pu{1.25 g}$ magnesium salt of EDTA (commercially available).

Dilute with distilled water.

After researching, I have assumed it to mean Magnesium Disodium EDTA (hydrate), CAS# 14402-88-1.

In reading the source material and searching online for synonyms of "magnesium salt of EDTA," I can't seem to find an exact hit on the term. Searches included PubChem, ChemSpider, Sigma-Aldrich, and Cole-Parmer, among others. Each seemed to point me toward "Magnesium Disodium EDTA Hydrate." I was hoping that with the amount of knowledge on this site, someone might recognize the term "salt of" and be able to help me out and clarify the term, or confirm that I am using the correct EDTA salt.

EDTA is an abbreviation commonly used for the ligand - ethylenediaminetetraacetate. The parent molecule is the corresponding ethylenediaminetetraacetic acid. So, an Mg salt of an acid.

The authors have some strange kind of humor... it's a riddle.
Below you might see

2) If the magnesium salt of EDTA is unavailable, dissolve 1.179 g disodium salt of
  ethylenediaminetetraacetic acid dihydrate (analytical reagent grade) and 780 mg magnesium
  sulfate (MgSO4⋅7H2O) or 644 mg magnesium chloride (MgCl2⋅6H2O) in 50 mL distilled water.
  Add this solution to 16.9 g NH4Cl and 143 mL conc NH4OH with mixing and dilute to 250 mL
  with distilled water.

From the molar mass of $\ce{MgSO4 * 7 H2O}$ ($\pu{246.47 g/mol}$) you can calculate the number of moles of magnesium the "magnesium salt of EDTA" should contain. With the help of the data sheet of your purchased magnesium salt of EDTA you can calculate the required quantity in grams.
You also could calculate possible stoichiometries of the imaginary "magnesium salt of EDTA" if you like. But be aware that not all commercially available magnesium salts of EDTA have a well defined stoichiometry.


text:
Why do we neglect the zero point energy in statistical mechanics?
Quantisation associated with translational modes can be modelled like a particle in a box (PiB).
$$E_n = \frac{n^2h^2}{8mL^2} \qquad n \in \mathbb{Z} \, | \, n \ge 1$$
By the equation above, the zero point energy associated with a PiB is $E_1 = \frac{h^2}{8mL^2}$. However, in my statistical mechanics lectures, we set this to zero by modifying the equation as below:
$$E_n = \frac{(n^2-1)h^2}{8mL^2}$$
Atkins does this too in Physical Chemistry, claiming to do so "for reasons that will soon become clear". But, alas, I'm still not sure why this is done.
Could somebody enlighten me?

Who cares about the zero point energy? It is always there, hence it will inevitably cancel out in the end.

@IvanNeretin, someone cared enough to remove it. What advantages does that bring?

OK, if you want a longer answer, here it goes. When Atkins and your lecturer mentioned these $E_n$, they most likely did that for some reason. They must have used these in some calculations later on. You have to read and actually repeat these calculations with pen and paper, like those ancient scientists you might have seen on YouTube. Do it all the way to the end, that is, until you derive some physically observable values. Then repeat everything again, but this time don't set zero energy to zero. See where this gets you. I claim that you'll get no difference in the end.

The reason might be that adding a constant to the energy of every state does not not affect the probabilities for the system being in a certain microstate.

The value of any  energy such electronic energy in a PIB or vibrational energy in a  harmonic  oscillator will obviously depend on whether the reference zero of energy is added in or not, but heat capacity and entropy are not dependent on the zero reference. In reactions, to be consistent, the zero energy  is taken as that of the molecule dissociated into atoms then the dissociation energy is $D_0$ if taken to the lowest level or $D_e$ if to the lowest point of the potential energy.


text:
What are the crystals that have formed around the top of a conc. nitric acid bottle?

I was looking through my chemicals cupboard and found that this decades-old bottle of conc nitric acid had grown a beard. I can't think of any reaction between (presumably) $\ce{NO2}$ and any airborne substance that could produce this. Any ideas?

You sure it's $\ce{NO2}$?

It might be ammonium nitrate. You could check this by triturating some of the crystals with baking soda. If you smell ammonia, it's ammonium nitrate.

... And please let us know your results, I am curious :)

Chemicals cupboard often produce quite extraordinary reactions, much like hospitals are home to some unusual and severe infections.
Nitric acid does not normally exhibit such a bizarre behavior. Yet in your case it did. How so? My bet is on some unknown additional component which came from the same cupboard, but from a different bottle. For instance, it could be some lower aliphatic amine (Upd. or simply ammonia, as was brought to my attention by other answers). They are volatile, and they produce crystalline salts when they meet with strong acids.
A similar thing is known to happen with $\ce{NH3}$ and $\ce{HCl}$:  if they are stored nearby, especially in poorly corked bottles, they both (as well as everything around) get covered in a white residue of $\ce{NH4Cl}$. It doesn't form nice crystals, though; it looks more like a layer of hoarfrost. But that doesn't mean much. Related compounds do not necessarily have related crystal structures, let alone related crystal shapes, so it looks perfectly plausible to me that (substituted or not) ammonia nitrate would grow into fine needles.
The prevalence of amine/ammonia in the air within the cupboard (which could happen if the other bottle was corked worse than this one) would naturally lead to $\ce{HNO3}$ molecules fighting this battle at their own doorstep, that is, at the very bottleneck of this bottle and nowhere else. This would explain the accumulation of residue at one spot.

Ivan can we collect this crystal and do some tests for ions that may be in this salt?

Yeah, I guess you can measure the melting point and compare it with the reference value.

Bases and acids should be stored in separate, hood-ventilated cabinets.

@user55119 Yes, should be. The reality of many school labs is that there is only one appropriate, heavy duty storage cabinet. This is a pretty common occurence in US schools.

There are several mentions of this on the interwebs, including this one on a science forum.
The consensus is ammonium nitrate. From working with the stuff and inheriting old bottle in schools, I have seen this a number of times. The fine crystal certainly looks like slow growth from airborne NH3 or amines as the other post said, but NH3 is far more common, depending on which chemicals are stored where in the lab.
Yes, ammonium nitrate is an explosive, but needs a pretty solid activation energy, so you can just slowly carefully rinse this away with lots of water.
You can certainly remove a small amount and test in a beaker with a few drops of water just to be sure.

Thank you for your comments, I do have ammonia stored in the same place. What would adding the crystals to water tell me?

A simple test is to wet a piece of red litmus paper and stick it on the bottom of a watch glass. Dissolve the crystals in water in a beaker. Add a couple of NaOH pieces. Put watch glass on top of beaker. The NaOH should release NH3 in air which would turn red litmus blue.

BTW, does the place smell of ammonia?

@WillHiggs If there were any concerns about safety of water dilution and cleaning.

The cupboard has an odd smell, but not strongly of ammonia. There are other reagents stored there including formaldehyde, HCl and acetic acid. I put some of the crystals on top of my wood-burner, which is probably at least 150', and they instantly melted and bubbled. Visible fumes were given off which may have been NO2. I also tried a lighted match and the crystals stuck to it and bubbled, as on the stove. I don't have any litmus handy just now, but I've ordered some.

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


text:
What does the small, steep curve at the very beginning of a weak acid strong base titration curve come from?
I understand that a buffer solution is being made, but I don't understand why the titration curve is different from a strong acid strong base one (apart from the steeper pH change). I understand the half-equivalence point and the rest of the titration curve. If Le Châtelier's applies here, please explain with it, as it makes more intuitive sense for me rather than the Henderson-Hasselbalch equation.
I included this animation if it makes it easier to explain, thanks!
http://www.chembio.uoguelph.ca/educmat/chm19104/chemtoons/chemtoons9.htm

This is part of estsblishing the buffer associated with the weak acid. To establish a buffer you need relatively large amounts of acid and conjugate base, and a weak acid can't produce the necessary conjugate base by itself.  You need to push it along with a little strong base to get into the buffering condition.
Say your solution is 0.1 molar acetic acid, with $K_\mathrm{a}=1.8×10^{-5}$. Before titration the pH is given by the Henderson-Hasselbach equation.  The logarithm of the above $K_\mathrm{a}$ value is about -4.74, therefore:
$\mathrm{pH}=4.74-\log_{10}\left(\frac{[\ce{HC2H3O2}]}{[\ce{C2H3O2-}]}\right)$
Now the amount of $\ce{C2H3O2-}$ in the solution is just what you get from dissociation of the acid, but acetic acid is only a weak acid and the actual amount of dissociation is small, in this case only about 1.3% of the acid you dissolved.  Merely 0.01 molar strong base reacting with the acid is enough to make a lot more acetate ion than just the acid dissociation alone, so the ratio in the H-H equation makes a big change and the $\mathrm{pH}$ follows suit.  Only when you have large amounts of both acid and conjugate base does the $\mathrm{pH}$ become buffered against slight additions of base or acid.

So if the Ka value is smaller, less of the weak acid is dissociated, and the pH change at the beginning is bigger because of a larger logarithmic change in the concentration of acetate ion? (Assuming the molarity of such a solution is also 0.1 molar). Or is this canceled out by the higher initial pH?

You get both effects.  As long as you're above $10^{-11}$ in 0.1 M aqueous solution (to keep water autodissociation at bay), a factor of 10 decrease in Ka raises the initial pH by one half unit and raises the increment into the buffering range by another half unit.

If the low amount (high OH- concentration) of NaOH is enough to significantly raise pH, why does the increase slow down then? I understand that a buffer is being established, but how does it differ from the very initial solution of titrant being added? Please use Le Châteliers principle if it applies here, I find it much more intuitive than the HH equation (which I don't really understand).

I can't fit this well to Le Chatelier's Principle.  The principle is qualitative and we are dealing with a quantitative effect.

The steepness occurs when the neutralization point is reached. And it occurs early on in the case of weak acid-strong base titration because the acid is weak, and less base is required to neutralize it.
Another point to be noted is that weak acids do not dissociate to large extents as in the case of strong acids. In a solution of a strong acid, almost all of the $\ce{H+}$ ions are already liberated and very few remain connected with the anion (say $\ce{A-}$) as $\ce{HA}$. On the other side, in the case of weak acids, only a few $\ce{H+}$ ions are released and most of them remain as $\ce{HB}$ (let $\ce{B-}$ be the conjugate base of the weak acid)
So lets see what happens as you add the base to these solutions:
Case 1: Strong acid
Once the base is added into the solution, the $\ce{OH-}$ ions combine with $\ce{H+}$ ions to form water. The $\ce{OH-}$ ions continuously deplete the $\ce{H+}$ pool in the solution, increasing $\ce{pH}$.
case 2: Weak acid
Here, it's similar to the previous case, but there's a catch: There still remains some undissociated acid molecules in the solution. Once the $\ce{OH-}$ combine with $\ce{H+}$ and remove some of these ions from the solution, the equilibrium of dissociation of weak acid is shifted to the right. The undissociated acid molecules now dissociate and replenish some of the $\ce{H+}$ ions lost due to base addition. This makes the change in $\ce{pH}$ quite less as compared to the strong acid case.

Thanks for the quick answer :) but i meant the steepness at the very beginning, when just a few milliliters of titrant is added (before the half-equivalence point)

@Jesper2k I don't understand what you mean. I see the steepness at the equivalence point here: https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/1941/2017/05/30163125/tit3.png

The small steepness at the "Initial" point in the image

Ahh I see, looks like Oscar Lanzi has pointed out that in his answer already :)


text:
Approaches to cap/modify the terminal 3' OH of DNA polymer
I am looking for a procedure that would cap/protect the 3' hydroxyl of DNA molecules with high efficiency. I have tried to perform a similar step enzymatically using dideoxynucelotide triphosphates but have found the efficiency to be unacceptable for my application, so I am exploring non-enzymatic approaches.
If this process also has side reactions that modify other hydroxyls (including 2' hydroxyls on RNA and/or other moieties on proteins), that would also be OK as long as the internals of the DNA backbone or the bases are not modified.
There is a standard part of some in-situ nucleic acid hybridization protocols that reacts tissue sections with acetic anhydride in a triethanolamine solution. The stated purpose of this step is to acetylate the 2' OH on RNA, and to acetylate some amino groups on proteins, and reduce background in subsequent hybridization steps.
1) Is there any reason to think that these conditions would NOT efficiently acetylate at the 3' hydroxyl of DNA (and RNA)?
2) Would this approach (acetic anhydride in basic conditions) cause any modifications to the standard DNA bases? I am specifically concerned that the primary amines critical for base pairing would be acetylated, which would be undesirable.

Hi Yashka, welcome to Chemistry.SE! We hope you enjoy your stay, please take the [tour] to familiarize yourself with the site!

Are you tring to protect only $\ce {3'-OH}$ and/or $\ce {2'-OH}$, but not $\ce {5'-OH}$, correct? I'm just tring to understand your need.

If there is an exposed 5' OH,  (for most biologically derived DNA there would be a phosphate there) it would not matter for my experiment if it was protected or not.

I understand that, but wasn't sure you want to cap $\ce{OH}$s after treat them with resricted enzymes.

Check this light sensitive protecting group (o-nitrobenzyl halide): Proc. Natl. Acad. Sci. USA, 1994, 91, 5022-5026 (May issue).

PDF available online: http://www.pnas.org/content/pnas/91/11/5022.full.pdf

I am not sure how this applies to my situation. I am looking for an irreversible/stable cap, and I’m starting with fully sysnthesized DNA , not being synthesized from 3’ to 5’ as Olivia, as in the paper.


text:
Covalent vs Ionic bond in Silicon dioxide
I was wondering why silicon dioxide tends to be a covalent bond instead of an ionic bond. Based on lewis structures, the octet rule seems to work for silicon dioxide so shouldn't it also be an ionic bond?

Losing four electrons to form a $\ce{Si^{4+}}$ is quite unlikely and requires a lot of energy to do so.

How does covalent bonding require less energy?

Because silicon and oxygen are nonmetals on the period, hence they form covalent bonds. For ionic bonding, it must be between a metal and a nonmetal.

You might want to add a word or two about electronegativity.

"It must be a metal and a nonmetal", like ammonium chloride?


text:
Are sp, sp2, and sp3 hybridisation only relevant to the carbon atom?
When we are talking about $\mathrm{sp, sp^2}$ and $\mathrm{sp^3}$ hybridisation, is it only relevant to the carbon atom only?
For the following molecules:
Acetone (propanone), acetic acid (ethanoic acid), tert-butanol (2-methyl-2-propanol), benzene, cyclohexane, 2-butyne.
Which of these contain at least one $\mathrm{sp}$ atom?
I am confused because for acetic acid and tert-butanol, the oxygen atom in the two molecules are bonded to a carbon and a hydrogen - that means 2 bonds.
Would this be and $\mathrm{sp}$ hybridisation? But doesn't $\ce{O}$ forms $\mathrm{sp^3}$ hybridisation as an alcohol function?

Not only $\ce{C}$,but also other elements (e.g., $\ce{B}$, $\ce{N}$, $\ce{O}$, etc.), which have partially filled $\ce {p}$ orbital can undergo $\ce {sp^3}$, $\ce {sp^2}$, or $\ce {sp}$ hybridization.
To understand how it works for molecules other than carbon, which do not show four bonds, let's take a water molecule as an example. Oxygen atom has $\ce {6 e-}$s in its outer most energy level (n = 2), which has electron configuration: $\ce {2s^2 2p^4}$. When water molecule is formed, it was quantum mechanically argued that all these 4 orbital ($\ce{2s, 2p_x, 2p_y, and 2p_z}$) undergo hybridization to form four equally energized $\ce {sp^3}$ orbitals (the name, $\ce {sp^3}$ is given to the newly formed hybridized orbitals because they were made up by $\ce {1\times s + 3\times p}$). Since they are equally energized, they would occupy the space in such a manner that they have equal distances and equal angles to each other. The only way it can arrange to achieve that is in tetrahedral shape. That four orbital can have maximum of eight electrons. When you put six electrons, which are belong to oxygen in its four $\ce {sp^3}$ hybridized orbitals, you end up with two fully filled and two half-filled orbitals, because you are supposed to avoid pairing them whenever possible to avoid pairing energy so that you keep the atom in its lowest possible energy state. Now, two hydrogen atoms can occupy these two half-filled orbitals with one  $\ce {s}$ electron of each, making 2$\sigma$ bonds. We called it $\sigma$ bonds because there are two nuclei involved to make the bond and it was head-to-head overlapping (precisely, $\ce {sp^3-s}$ bond). What about other two $\ce {sp^3}$ orbitals, which are already filled with electrons from oxygen? We called they are lone pairs.
Now $\ce {O}$ of $\ce {H2O}$ molecule is $\ce {sp^3}$ hybridized. Therefore, the theoretical shape of $\ce {H2O}$ molecule is tetrahedral. Yet, actual shape is angular because if you get a snap shot of $\ce {H2O}$ molecule, you can't see the electrons. Only you see is three nuclei, one from oxygen and other two from hydrogen, arranged in angular manner.
Similar to water molecule, oxygen atom in tert-butanol (2-methyl-2-propanol), in addtion to all four carbon atoms, is $\ce {sp^3}$ hybridized. The $\ce {O-H}$ bond is $\ce {sp^3-s}$ bond while the $\ce {O-C}$ bond is a $\ce {sp^3-sp^3}$ bond.
Now, you can look how it work for $\ce {sp^2}$ and $\ce {sp}$ hybridization: Only three $\ce {sp^2}$ hybridized orbitals can be made since only three orbitals used to make them. Only way three can arrange in space with equal energy is with trigonal planar geometry (say they are in $xy$-plane). Thus, the $\ce {p}$ orbital not used for the hybridization occupy as $z$-axis to that $xy$-plane. For example, oxygen and middle carbon it attached to in acetone are $\ce {sp^2}$ hybridized so that the $\ce {O-C}$ bond is a $\ce {sp^2-sp^2}$ bond. Similarly, one $\ce {O-C}$ bond in acetic acid is $\ce {sp^2-sp^2}$ bond while the other $\ce {O-C}$ bond is $\ce {sp^3-sp^2}$ bond (both oxygen attached to $\ce {sp^2}$ hybridized carbon).

Mathew, your statement “Not only $\ce{C}$,but also other elements (e.g., $\ce{B}$, $\ce{N}$, $\ce{O}$, etc.), which have partially filled $\ce {p}$ orbital can undergo $\ce {sp^3}$, $\ce {sp^2}$, or $\ce {sp}$ hybridization.” seems to me to be to weak. Are there any molecular bonds from a these elements which will not undergo the hybridization?

@HolgerFiedler: My point is those elements undergo similar hybridization like carbon as well. For example, $\ce{O}$ in acetone is $\mathrm {sp^2}$ hybridized, while  $\ce{O}$ in carbon monoxide is $\mathrm {sp}$ hybridized. Similarly,  $\ce{B}$ in BOBIPY is $\mathrm {sp^3}$ hybridized.

Mathew, I’m curious, are there any molecular bonds from a these elements which will not undergo hybridization?

@HolgerFiedler: There is a better explanation of hybridization can be found here. Enjoy the reading.


text:
What chemical compound makes colour change to copper into brown like this?

What chemical compound should I use to change the colour to make copper statue like this shown in picture

$\ce{CuO}\mathstrut$.

Are you sure it is copper, not bronze?

See https://www.sciencecompany.com/Patina-Formulas-for-Brass-Bronze-and-Copper.aspx ... information available after cursory online search

It's copper statue, brushed with soluble chemical. Sodium Phosphate change colour like little bit black, but I need brown in colour as in picture


text:
What is the relationship between ortho, meta and para-dimethyl benzene?

What is the stereochemical relationship between ortho, meta, and para-dimethyl benzene, given that interconversion between the three structures requires you to break and form new bonds?

I know that ortho relates to the 1,2 position, the meta position relates to the 1,3 position and the para position relates to the 1,4 position.
However, what does it mean by 'stereochemical relationship'? Does it just mean that if the methyl is placed at the different carbon numbers, they're different molecules?

There is no sterochemistry in any of the dimethylbenzenes. Benzene is a planar molecule, which rules out both chirality and geometrical isomerism.

They are constitutional (structural) isomers; there is no "stereochemical relationship" because they are not stereoisomers. [(E) and (Z)-butene are stereoisomers. The two are constitutional isomers of 1-butene.]

These compounds do not have a stereochemical relationship. Stereoisomers have the same connectivity of atoms (constitution) with different arrangement in space. The (1,2), (1,3), and (1,4) in the names of these compounds indicate that this is not the case. These are not stereoisomers of any kind.

Not that the qualifying text about having to break bonds to interconvert is unhelpful and irrelevant. All isomers require bond-breaking to interconvert.
So, what then are the relationships between these three compounds? Let's look at a decision flowchart for relationship between structures:

Do they have the same formula?

Yes.

Do they have the same connectivity?

No. Thus, they are constitutional isomers. Constitutional isomerism is not a type of stereoisomerism. The two main types of stereoisomerism are enatiomerism and diastereomerism.

It would be worth contacting your instructor and asking for clarification, noting that these are not examples of any kind of stereoisomerism. It could be an honest mistake in the wording of the question. As someone in the teaching profession, my experience is that most instructors are happy when a student understands the content at a level high enough to catch mistakes.


text:
Requirement of different reagents for bromination of different substituted benzene rings
Consider the bromination reactions of phenol and anisole. We know that the former can be brominated by aqueous bromine, while the latter requires the catalyst $\ce{FeBr3}$ instead.
Our teacher said that $\ce{-OH}$ is a stronger ring activator. So, polarization of bromine takes place easily and, hence, phenol doesn’t require a Lewis acid for bromination.
But, this related question had the conclusion that both anisole and phenol are practically equally activating. So, does aqueous bromine react with anisole too? If not, why is there a difference in the reagents? If yes, do we get 2,4,6-tribromomethoxybenzene?
We have a similar case with chlorobenzene also. Difference in electronegativity between $\ce{C-O}$ is more than that in $\ce{C-Cl}$. So, chlorine must be able to give one of its lone pairs and activate the ring more readily than oxygen. But even bromination of chlorobenze requires a Lewis acid. Why is this so?

PS: I checked this similar question on our site but a proper reason wasn’t stated to my question.

$\ce{-OH}$ is a better activator than $\ce{-OCH3}$ see this question

Possible duplicate of Why is a hydroxyl group more activating than a methoxy group in electrophilic aromatic substitution?

So even anisole undergoes the same reaction?

The accepted answer says they both are practically equally reactive.

It might be true that $\ce{-OH}$ and $\ce{-OCH3}$ are about equally activating. But phenol in water is partially deprotonated, and $\ce{-O^-}$ is a great deal stronger than both.

Interesting! And phenoxide ion is stabilised by resonance. Thats a good reason. Thanks.

@IvanNeretin I was taught that phenoxide ion is only stable in basic medium. Can it also form and remain stable in neutral medium?

@GaurangTandon "Stable" is a word that means nothing, so you may just as well stop using it. True, to deprotonate all phenol molecules we'd have to go into much higher pH. With neutral solution, at any given moment only a tiny portion of them is deprotonated. This is also the portion that reacts. Then they exchange protons, and we go on to brominate another portion. Then another.

@IvanNeretin Alright, makes sense. Still, your comment makes me feel that aqueous bromination of phenol will be a slow process. Is my interpretation correct?

@GaurangTandon Not particularly. See, the portions may be tiny, but they exchange really fast, and each portion brominates pretty fast, too.

Phenol is easily brominated by bromine in water. The reaction hereby proceeds to 2,4,4,6-tetrabromocyclohexadienone. The reason for this is, as Ivan already explained, the presence of phenolate anions. The product can be reduced to 2,4,6-tribromophenole by sodiumhydrogensulphite.


text:
Prins reaction mechanism
I am trying to figure out the product for the following reaction:

Its a Prins Reaction. My proposed mechanism for it is:

In my second step, that particular carbocation is formed since $\ce{Ph}$ is an electron-withdrawing group.
Mechanism looks neat enough, but the reaction given in Wiley's Solomon and Fryhle (Page 379) is:

I'd like to know why my mechanism is "wrong"? I think there's problem in the carbocation generation step but isn't the carbocation that I have formed more stable?

The carbocation will definitely form at the benzyl carbon. The fact that the benzene ring can stabilize the cation through resonance outweighs the disadvantages of negative inductive effect.

Your generation of electrophile is correct the subsequent primary carbocation formation by the attack of electrophile is incorrect.

In my second step, that particular carbocation is formed since $\ce{Ph}$ is an electron-withdrawing group.

The benzene ring  stabilizes the carbocation through mesomeric effect.
The electrophile will attack at $\alpha$ position rather than $\beta$


text:
Use of Δ (delta) to describe alkenes
In a lot of (mainly older) papers discussing organic synthesis,  Δ is used to describe alkenes. As far as I know, this is no longer (if ever) IUPAC recommended nomenclature, so I haven't been able to find any reference to its intended use.
The main area I'd seen it used was in steroid synthesis, where I assumed it had some specific use relating to that class of natural product (i.e. related to the biosynthesis etc.). However it does appear in other places, such as in the synthesis of Rifamycin:

 Scheme from Nicolau Rapamycin synthesis. Taken from Classics in Total Synthesis I
"… the Δ35,36 double bond …" is used in the text to describe the alkene that gets reduced. Given that this could easily be referred to as the (E)-alkene or even just 206, I suspect that there is a special reason why Nicolau chose Δ35,36 however that reason isn't apparent to me.

The current (2013) IUPAC recommendations still mention a method that, for ring assemblies with a double bond junction, the presence of a double bond is indicated by the Greek letter Δ and the point of attachment is given by superscript locant numbers. However, this method is not continued in the current recommendations.

@Loong In fact it is allowed even for the PINs (and not just for junctions), see my answer update ;)

Steroid nomenclature
The Δ notation was obviously in use in some fields of the chemistry of natural compounds and/or by some publishers, but probably never recommended by IUPAC. A mention in older specialized IUPAC rules is e.g. in IUPAC-IUB revised tentative rules for nomenclature of steroids (1969):

Rule 2S-4 (extended version of Rule S-4)
  …
  I. Unsaturation is indicated by changing terminal
  ‘-ane’ to ‘ene’, ‘-adiene’, ‘-yne’ etc., or ‘-an’ to
  ‘-en’, ‘-adien’, ‘-yn’ etc.; e.g. 5β-cholest-6-ene,
  5β-cholesta-7,9(11)-diene, 5-spirosten; see also the
  names of examples (22)-(25). [For uniformity with
  the IUPAC Rules cited above, the conventions of
  Chemical Abstracts are used also in the present
  Rules for the position of locants (positional
  numerals) and designation of unsaturation. In
  such matters, and in use of Δ (Greek capital delta)
  to designate unsaturation (which is not recommended
  by IUPAC), authors should respect the
  house customs of the journals to which their papers
  are submitted.]

Later it became just a footnote in (currently obsolete) IUPAC Definitive rules for nomenclature of steroids (1971/1972) (p. 303).
Pretty recent IUPAC Nomenclature of steroids (Recommendations 1989) states:

3S-2.5. Unsaturation
  …
  3) The use of Δ (Greek capital delta) is not recommended to designate
  unsaturation in individual names. It may be used, however, in generic terms, like 'Δ5-steroids'.

General natural product nomenclature
However, presently, the most widely known use of this delta notation is not in steroids, but in cannabinoids (to specify positional isomers of unsaturated compounds):

Δ9-tetrahydrocannabinol (Δ9-THC, THC)

Δ8-THC
(Please note that there are at least two different numberings for cannabinoids, so you might meet, especially in older literature, names like Δ1-THC and Δ1(6)-THC instead of Δ9-THC, resp. Δ8-THC, but this is a different issue.)
Recent organic chemistry nomenclature
In the recent Nomenclature of Organic chemistry by IUPAC, Recommendations and preferred names (2013), besides the non-preferred notation for double bond junctions already mentioned in another answer, the delta notation is allowed even for the preferred names (PINs):

P-25.7.1.2 Localized double bonds
If it is necessary to identify isomers that differ only by virtue of the location of localized double bonds, this differentiation is indicated by the use of the Greek letter Δ. The locant cited corresponds to the first cited locant of the localized double bond.
Examples:

  1,6-dimethyl-Δ1-heptalene (PIN)

  1,6-dimethyl-Δ1(10a)-heptalene (PIN)

(Howewer these won't occur very often, the structure in the examples above looks like conjugated system, but it's non-aromatic).
Comma usage
Also note that the coma was/is being used for multiple double bond position specification, as seen in excerpt from Fieser: Natural Products Related To Phenanthrene (1949)

Examples. Δ7,14-Cholestadiene (…), Δ7,9(11)-cholestadiene (…), Δ8(14)-cholestenol-3β

(But note that this comma usage somewhat collides with ring assemblies notation mentioned in another answer.)
In your example, the comma usage is rather incorrect, should be Δ35(36) (so called compound locant, see e.g. A: Numbering bonds in cyclic compounds), instead of Δ35,36. And here the locants differ by 1, so Δ35 should be sufficient.

@Loong's encyclopaedic knowledge of IUPAC nomenclature prompted me to have another look through the (irritatingly unsearchable) 2013 recommendations and the 2004 draft upon which they were built.
The only use of the uppercase Δ listed is when describing a double bond that links two cyclic systems:

P.28.2.2: Ring assemblies with a double bond junction
When two cyclic systems are linked by a double bond [...] the presence of a double bond is indicated by the Greek letter Δ, and the point of attachment of the ring is given by superscript locant numbers. This method is not continued in these recommendations

The example given is 1'1-bi(cyclopentylidene) (PIN) which should be used in preference of Δ1,1'-bicyclopentylidene.

Based on this, the example given in the question, using Δ to describe an unsaturated carbonyl would appear invalid, and used for convenience rather than to convey anything specific.


text:
Calculating the density matrix from a WFN file
I want to calculate the density for the $\ce{HCN}$ molecule. This molecule has 14 electrons and thus 7 occupied molecular orbitals. I have a .wfn file which includes 8 parts, where each of them has 14 rows and 5 columns (due to 5 basis functions). The first part belongs to the basis functions (exponents) and the 7 other parts contain coefficients for each of seven MO's. How can I calculate the electron density using this file? Generally I have used the electron density formula based on the Roothaan expansion introduced in Szabo's textbook, namely:
$$
P_{\mu\nu} =2\sum_{a=1}^{N/2} C_{\mu a}C_{\nu a}
$$
where $a$ runs over the number of electrons, though I have seen in several codes that $a$ is considered as the number of occupied MOs. $\mu$ and $\nu$ stand for the number of basis functions (5 here).
Should I use all 7 parts of the coefficients to obtain the electron density of the $\ce{HCN}$ ground state, or is the first part which has the minimum energy sufficient?

What do you mean by "the first part which has the minimum energy"? And welcome on Chem.SE!

Thanks. My mean was in my original .wfn file that the energy of each part has been written on the first line of it. The most negative energy belongs to first MO and it increases as the MO's goes on.

Calculating a density matrix by hand from a wfn/wfx file is possible, but even for small molecules a time consuming work. The wfn-file (AIM traditional wavefunction data) is better read in by a program of choice (e.g. multiwfn, see http://sobereva.com/multiwfn/) which then can calculate the desired properties.

Do you just want to have a density matrix of your wavefunction or do you want to have a real electron density? Could you clarify this in your question, please.

@awvwgk Thanks for your suggestion but notice I don't want to calculate electron density value for a specific coordinate rather I wanna to obtain the electron density as a function of distance (R). Otherwise I could simply obtain electron density using Softwares such as AIMALL or AIM2000.

@NSR what do you mean by distance? Distance from what?

where a runs over the number of electrons, though I have seen in several codes that a is considered as the number of occupied MOs.

If it is closed shell (all spatial orbitals are either doubly occupied or empty), then running up to $N/2$ is the same as running over all occupied orbitals. Note that your formula is for closed shell systems described by a single determinant.
There are two representations of the first order reduced density matrix, or simply density matrix: the discrete one and the continuous one.
What you have in the formula is the discrete one, represented in the space of primitive functions. The only reason why it is discrete is because the original Schrödinger equation has been projected on a finite dimensional space. You need to locate the Molecular Orbital Primitive Coefficients ($C_{\mu a}$) in the file. You could as well calculate the overlap matrix ($S$) of the primitives and then get the density matrix in the basis of molecular orbitals
$$ D =  C^t S P S C.$$
You will see that it is a diagonal matrix with 2's.
You could also represent in real space the general density matrix,
which is continuous and spans $\mathbb{R}^3$. You could discretize the space and compute the values there. Then, the general formula
$$ \rho(\pmb{r},\pmb{r}') = \sum_{a,b} D_{ab}\phi_a(\pmb r) \phi_b^*(\pmb r'),$$
where $\{\phi_a\}_{a=1}^M$ represent the spatial canonical molecular orbitals, and
becomes
$$ \rho(\pmb{r},\pmb{r}') = 2\sum_{a}^{N/2} \phi_a(\pmb r) \phi_a^*(\pmb r')$$
for a closed-shell single determinant wavefunction.
Setting $\pmb r' = \pmb r$ you get the density.
If you plan to do it you will have to deal with some nuances, like the normalization used by the program that generated the .wfn file.


text:
Colour of the transition metal complexes
I read in my Chemistry AP class that the colour of the transition metal complexes is due to the wavelength of the visible light that is left unabsorbed, when the d subshell is split due to the repulsion between the ligand electrons and d subshell electron in the transition metal ion. When white light is passed through the complexes, the wavelength that corresponds to the energy ($\Delta E$) required to promote electrons in the split d subshell is absorbed. The part of the visible light wavelength that is left unabsorbed becomes the colour of the transition metal complex.

I was intrigued by a concept of absorption of light, relating it to absorption and emittion spectra. My question is: If we keep that complex isolated in the dark, where no light could reach it would the colour of the complex still be blue? I understand if that is done it wouldn't be visible to the naked human eye. I wonder if there is a way to find that out. Secondly, I was wondering if the colour elements obtain their colour in a similar way?

The concept of "color" would be meaningless in a pitch black room. Color depends on the wavelength (or group of) of light that reaches the eye. If there is no light to be detected, how can you decide on the color?

Pritt Balagopal, I did write that concern down. I was hoping it might be possible. I without percieving it through a naked eye, and rather through some technology.

Most colour in chemistry comes from selective absorption of light by electronic transitions of the appropriate energy
Colour is not limited to transition metal complexes. They have a particularly rich range of colours as the electronic transitions often match the energy range of visible light but also because they are easily altered by the nature of the ligands surrounding the metal (for example, Cobalt complexes can be almost any colour of the spectrum depending on the ligand).
But plenty of other compounds are coloured. Most dyes are purely organic and the colour comes from the absorption of light in transitions between molecular orbitals made from unsaturated chains in the molecules. Many elements are coloured because they, too, have some electronic transmission that matches the energy in some part of visible light. The mechanism for colour is the same in all these cases.
There are other ways to generate colour but they are less common. Some colours come from interference not selective absorption of light: some physical structures that contain features close to the wavelength of light  can reflect light but in a way that causes selective interference that cuts out some colours. This is why oil films on water produce rainbow colours but is also the source of unusual colours in some fruits, butterflies and birds.
Are things coloured if the light isn't present? That is a philosophical question like "does a tree falling the the forest make a sound if nobody hears it". From a chemical point of view the feature that creates colour is still present whether or not there is light. But in the absence of light nobody is going to notice it.

I was hoping to maybe learn if there is some sort of advance technology present to make the deduction of colour without light hitting them. I find it intriguing

@ShahSaud The chemical structures (and their electronic transitions) exist whether or not there is light present. They are a function of the molecular structure (which can be observed without light by a variety of techniques). The electronic property that creates light is there: but asking whether there is colour when there is no light is straying into philosophy not chemistry.

A bit further than than the tree in forest. At least in that case something happens, which is not the case for something standing in dark. Plus 1


text:
Difference between probability and probability density of finding a particle in space

What is the difference between probability and probability density of finding a particle in space?

I am at basics of the wave function and this thing confused me.
Probability = $[\psi(x, y,z)]^2(\mathrm{d}x,\mathrm{d}y,\mathrm{d}z)$
Probability density at point $(x,y,z)$ in space = $[\psi(x, y,z)]^2$.
The latter one is the probability of finding particle in an infinitesimally small volume element $\mathrm{d}V=\mathrm{d}x, \mathrm{d}y, \mathrm{d}z$ situated at $(x,y,z)$. Then what's the former one?

The probability of finding particle at any particular coordinate is zero.

The latter is the limit of probability per volume, as volume approaches zero, at a particular point in space.

What is exactly probability? I mean probability of finding particle where?

@DivyankaS.Chaudhari if an experiment is repeated a large number of times, the fraction of the outcomes the met condition X is the probability of X.  So for a hydrogen atom, the probability that the electron is within all space is 1.  And for any finite volume the probability of the electron being in that volume is a fraction that is less than 1.  The probability that the electron is at any specific point, is zero, as Ivan said.  But the probability per volume, in the limit that the volume approaches zero at that point, is non-zero.

So, the former is the probabilty of finding electron in a given volume and the latter is the same but for very small volume?

@DivyankaS.Chaudhari The former is a probability of being in an infinitesimal volume element, and is a pure number (no units).  The latter has units of per volume ( such as per cubic meter).  The former needs to be integrated over a volume to get a non-zero number.  The latter can be non-zero at a specific point.


text:
Calculating the change in pH upon diluting an acid

What volume of water is added to $\pu{400 ml}$ of $\ce{HCL}$ solution so that the pH of the solution increases by 1 unit

This is how far i went but the answer says $\pu{1000 ml }$ where am I wrong is there any simplet alternative  to this question?

You a wrong in applying the Henderson-Hasselbach formula in the first place. It is meant for weak acids, which HCl is not. Then again, the given answer of 1000ml doesn't look right to me either.

As Ivan said in the comments, $\ce{HCl}$ is a strong acid. It dissociates as follows,
$$\ce{HCl -> H+ + Cl-}$$
Now, let us assume the intial pH to be x. Since, $$\mathrm{pH} = \ce{-log[H+]}$$ $\ce{[H+]_i}$ = $\mathrm{10^{-x}}$, where subscript i means initial. Likewise, since final pH= x+1, $\ce{[H+]_f}$ = $\mathrm{10^{-(x+1)}}$ where f means final. Since $\mathrm{V_i = 400 mL}$, since total moles remain the same,
$$\mathrm{M_iV_i = M_fV_f}$$
we get $\mathrm{V_f = 4000 mL}$.
Hence amount of water to be added is $\mathrm{4000-400 = 3600 mL}$.


text:
Definition of water Plasticization
Can someone explain what is water plasticization? I tried to look for the definition but could not find one that explains it.

Is that term used in the scope of polymers? If so, it is "regular" plasticization, with water as the plasticizer.

Quoting Wikipedia,

Plasticizers (UK: plasticisers) or dispersants are additives that increase the plasticity or decrease the viscosity of a material.

As far as most research papers say, water is not plasticized, but rather it acts as a plasticizing agent for other substances such as starch films as mentioned in this article. Water makes these films permanently deformed upon external stress and resists reverting back to its original shape.


text:
Why is zinc not considered to have a noble gas configuration?
I am wondering why we say that, for example, zinc does not have a noble gas configuration?
I would like to say beforehand that this question may has its origin in a confusion about the word "shell" or "outer shell". Wikipedia states that:

The properties of the noble gases can be well explained by modern theories of atomic structure: their outer shell of valence electrons is considered to be full.

Ok, so argon is a noble gas and its configuration is: $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$. When we speak of a shell we mean the principle quantum number $n$. In this case, the highest principal quantum number is $n=3$. However, for $n=3$ there is also the $\ce{3d}$ subshell available, which is not filled by argon.
So I conclude that "filled shell" does not mean that all available subshells for a given principal quantum number (here $n=3$) have to be filled. Because the $\ce{3d}$ subshell is empty here. Instead, I conclude that "filled shell" means, that whenever we start to fill a subshell (here $\ce{3s}$ or $\ce{3p}$), they must be filled. If there is another subshell with $n=3$ available, it has to be completely empty (like $\ce{3d}$).
But when we look at zinc, we find the configuration: $\ce{[Ar] 3d^{10} 4s^2}$.
Again, for the highest principal quantum number (here $n=4$) we have a filled subshell (here $\ce{4s}$). Additionally, all shells of the $n=3$ are also filled. There is no subshell (neither for $n=3$ nor for $n=4$) which is partially occupied. So, according to the definition from Wikipedia, what is missing for zinc to be in a noble gas configuration?
Again, let me stress, that I am not asking for what a noble gas is (I know that zinc is neither a gas nor a noble gas). I am asking what I am doing wrong, when I conclude that zinc obeys the definition of a noble gas from Wikipedia?

Maybe I'm misunderstanding your point, but wouldn't by this definition every alkaline earth element (beryllium, magnesium, etc) also have a noble gas configuration since they are filling there $2s$ subshell?

Yes, exactly. So what is wrong with this argument?

The difference hinges on the word "valence" in "their outer shell of valence electrons is considered to be full".
That's the difference between a $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$ configuration and a $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2}$ one. It's not just filled principle quantum numbers or filled subshells, but (un)filled valence subshells.
The definition of valence is a bit circular here. A valence subshell is one which can react to form a bond. For argon, all the valence subshells are filled because there's a substantial stability gap in adding that last $\ce{3p}$ electron and the first $\ce{4s}$ or $\ce{3d}$ electron. (If you think about potassium, the $\ce{3d}$ orbital is even less easily filled than the $\ce{4s}$ one.) That's why the $\ce{3d}$ orbital doesn't count as a "valence" orbital for Argon -- it takes too much energy to fill. Even though it has a principle quantum number of 3, it (re)acts like it has one of 4.
Contrariwise for zinc, there's a relatively small energy gap between the $\ce{4s}$, $\ce{3d}$ and $\ce{4p}$ orbitals. You probably already know this for $\ce{4s}$ and $\ce{3d}$, as they trade back and forth during the transition metals, but $\ce{4s}$ and $\ce{4d}$ are also relatively close in energy. This is part of the reason why atoms like carbon can "borrow" a $\ce{2s}$ electron to make four one-electron-donated bonds, rather than being forced to make two one-electron-donated bonds and one two-electrons-borrowed one.
So the definition of valence orbitals doesn't depend on their quantum numbers, but on the energy required to fill them. That's why zinc is not a noble gas - the $\ce{4p}$ orbitals count as valence (reactive) orbitals for zinc even while the $\ce{4d}$ don't.

Thanks for this great answer. So, in a pictorial way, am I correct when I say, that we have a noble gas, whenever all orbitals in these blue bubbles (see link) are completely filled? https://i.imgur.com/NaNnScV.png
The blue bubbles could be interpreted as the valence orbitals of an element, whenever at least one electron enters a bubble. One can also see why this is the case, since the energy levels of the orbitals inside a blue bubble are relatively close to each other, why the jump from one bubble to another costs more energy.

@thyme Yes, that's about right.

Albeit few excellent answers are already provided, I’d like to add a few additional points to the argument for the benefit of the readers:
Each shell in an element can be referred to a collection of subshells (e.g, $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, and $\mathrm{f}$) with the same principal quantum number ($\mathrm{n}$). Filling up of electrons in subshells takes place according to Aufbau Principle and the Madelung Energy Ordering Rule. The Aufbau Principle states that in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. The Madelung Energy Ordering Rule is based on the total number of nodes in the atomic orbital, $\mathrm{n + ℓ}$, which is related to the energy (here $\mathrm{n}$ is the principal quantum number representing shells and $\mathrm{ℓ}$ is the azimuthal quantum number, which give us information about subshells). Accordingly, orbitals with a lower $\mathrm{n + ℓ}$ value are filled before those with higher $\mathrm{n + ℓ}$ values.  In the case of equal $\mathrm{n + ℓ}$ values, the orbital with a lower $\mathrm{n}$ value is filled first (https://en.m.wikipedia.org/wiki/Aufbau_principle). Basically, two subshells with the same principal quantum number essentially have totally different energy levels. For example, for $\mathrm{2s}$ subsell: $\mathrm{n = 2}$ and $\mathrm{ℓ = 0}$, therefore $\mathrm{n + ℓ = 2 + 0 = 2}$, while for $\mathrm{2p}$ subsells: $\mathrm{n =2}$ and $\mathrm{ℓ = 1}$, therefore $\mathrm{n + ℓ = 2 + 1 = 3}$. That means electrons will occupy in $\mathrm{2s}$ first before $\mathrm{2p}$.
According to the Madelung Energy Ordering Rule, the $\mathrm{4s}$ orbital $\mathrm{n + ℓ = 4 + 0 = 4}$ is occupied before the $\mathrm{3d}$ orbital $\mathrm{n + ℓ = 3 + 2 = 5}$, because the lowest value of orbitals is filled first. According to the Aufbau principle, the $\mathrm{4s}$ sublevel is filled before the $\mathrm{3d}$ sublevel because the $\mathrm{4s}$ has lower in energy. The $\mathrm{4s}$ sublevel is only lower in energy if there are no electrons in the $\mathrm{3d}$ sublevel.  As the $\mathrm{3d}$ sublevel becomes populated with electrons, the relative energies of the $\mathrm{4s}$ and $\mathrm{3d}$ fluctuate relative to each other, and at the end, $\mathrm{4s}$ would have higher in energy as the $\mathrm{3d}$ sublevel fills. This is evident in transition metal elements when electrons are lost from their orbitals, they are lost from the $\mathrm{4s}$ first because it is higher in energy.
In above account, it is clear that why ‘the $\mathrm{4p}$ orbitals count as valence (reactive) orbitals for zinc even while the $\mathrm{4d}$ don't’ (the point made by @R.M.). According to the Madelung Energy Ordering Rule and Aufbau Principle, the $\mathrm{4p}$ orbital $\mathrm{n + ℓ = 4 + 2 = 6}$ is occupied before the $\mathrm{4d}$ orbital $\mathrm{n + ℓ = 4 + 3 = 7}$, based on the lowest value of energy. As a matter of fact, electrons go into $\mathrm{5s}$ even before $\mathrm{4d}$.
Historical Remarks about noble gases:
$\ce {He}$ (1895), $\ce {Ne}$ (1898), $\ce {Ar}$ (1894), $\ce {Kr}$ (1898), and $\ce {Xe}$ (1898) were discovered in Earth by an English group led by Sir William Ramsay during their pioneering work (1894-1898), while $\ce {Rn}$ was discovered separately by German scientist, Friedrich Ernst Dorn  in 1900. Naturally, these gases were given zero oxidation number because of their inertness with compare to other elements. That concept changed completely in 1962 when the first nobel gas compound, $\ce{XePtF6}$ was synthesized.


text:
Why is LiAlH4 a stronger reducing agent than DIBAL-H
Asides the different number of hydrogens, why is LiAlH4 a stronger and less selective reducing agent than DIBAL-H

How did you try to solve the problem? It will help the answerers to answer your question! (I was going to say welcome to chem.SE, but it seems like you have been here a while)

If you consider $\ce{LiAlH_4}$ and $\ce{DIBAL-H}$(Di-isobutyl Aluminium Hydride), other than different no. of Hydrogens attached to them, they also have significant structural dissimilarities between them.
 First of all, in $\ce{LiAlH_4}$, the reducing agent is $\ce{H^-}$, which comes from ionic $\ce{AlH_4^-}$. As, the compound is ionic it is easier for the anion to donate $\ce{H^-}$, and after donation also, the remaining $\ce{AlH_3}$, can get high stabilisation by dimerising to $\ce{Al_2H_6}$. So, the rate of formation of $\ce{H^-}$ is very high from $\ce{LiAlH_4}$, which makes it stronger reducing agent. On the other hand, Di-isobutyl Aluminium Hydride is a pure covalent compound. It is difficult for it to donate reducing $\ce{H^-}$, and after donation also, it can't get any extra stabilisation. Thus, eventually, it becomes weaker reducing agent.  Now coming to selectivity, if you have a high concentration of reductant (here $\ce{H^-}$), the probablity of forming all possible products automatically increases, because the kinetic control of the reaction overcomes the thermodynamic control.But, on the other case, less concentration of the reductant doesn't allow the kinetic control to exceed thermodynamic control, and thus the more selective product(which is actually thermodynamically more stable) is formed. Thus, huge concentration of $\ce{H^-}$ makes $\ce{LiAlH_4}$ less selective and more reactive.

Thanks, but in terms of the reduction of cinnamaldehyde, why is LiAlH4 able to fully reduce it as appose to DIBAL reducing it to an allylic alcohol. Except the availability of a second hydrogen for the hydroalumination of the C=C?

Do you know the fact that $LiAlH_4$ can reduce double bonds only if it is highly conjugated??


text:
Does MEK-toluene system create a negative deviation from Raoult's law?
From the experimental data obtained in our lab, when pressure is plotted against mol fraction and partial pressure, it exhibits a negative deviation from Raoult's law which must mean that the forces between MEK and toluene are attractive. But all the info I've found on the internet says that the deviation is positive.
Is MEK soluble in toluene? And could you please give me chemically similar systems.

Are you saying that there is literature data that differs from you own experimental data?


text:
Most activated position for Electrophilic addition in substituted biphenyls

I know the first structure will go fastest electrophilic addition due to +mesomeric effect but I was wondering what would be the product(major) if a electrophile attacks
Will the electrophile attack both ring or the ring near to $\ce{OH}$ group? Also the negative charge due to resonance will resonate on both the rings then how will I decide which ring to attack to give me major product?
According to me, the ortho positions of the ring near the $\ce{OH}$ group will have a negative charge - as well as the para position - but the para position is hindered. So, the electrophile will attack the ortho positions.
If this the correct explanation of the question, then I have one more question as well, that: when negative charge delocalises to the para position, it can delocalise to the other ring as well. Then, the electrophile could attack in the second ring also, where the negative charge would be generated by resonance. So, will the electrophile attack-the first ring or the second ring?

Addition not substitution.

@Mithoron sorry my bad

Note that electrophile could attack alkene as well.

@MathewMahindaratne so will there be multiple major products

It's based on the nature of your electrophile.

@user190625 I hope you understand that  EAS and electrophilic addition are different. Electrophilic addition will take place on the pi bond alone.

Dipole moment is what matters in substituted biphenyls.
The dipole moment in the first case is from left ring to right ring due to -OH groups predominant +M effect to the ring.(the alkene being an EWG(electron withdrawing group)), further enhances this effect. So effectively, electron density is more in the right hand side (I know it is counter-intuitive but it is how it is).
So EAS takes place preferably/majorly in the right side ring.
In the second case, NO2 shows -M effect and this dominates that of the alkenyl group. So dipole moment is from right to left ring and EAS takes place preferably/majorly in the left side ring.
Note: If there is any group(which is symmetric, or else we need to consider electronic effect due to assymetric grp also) between the C-C bond linking the two rings together, then the answer is as it "intuitively seems/as you expect". Try to rationalize this yourself.


text:
Hydroxyl group to free amine in polyethylene glycol using CDI?
For converting $\ce{OH}$ of methylated polyethylene glycol. The first step is converting hydroxyl group to imidazole carbamate as an active intermediate. But what about the next step?
If I will add primary amine containing group to form carbamate linkage, how can I convert it to a free amine?
Reference: Zhou, S.; Wang, J.; Wang, G.; Jiang, Z.; Ren, H. An approach to developing high dielectric constant nanocomposites based on polyrotaxane derivative. European Polymer Journal 2017, 90, 312-322. DOI: 10.1016/j.eurpolymj.2017.03.037.

I don't think it will convert to a free amine. If you add an amine,  you will make a carbamate where PEG has acted as a carrier or intermediate. For what it's worth,  if you want a hydroxyl to amine conversion, I would try diphenylphosphoryl azide. Then convert the azide to the amine by hydrogenation or triphenylphosphine.

Conversion of Hydroxyl group to free amine in polyethylene glycol: This is a case of converting a primary alcohol to a primary amine
$$\ce{R-CH2OH -> -> -> R-CH2NH2}$$
To my understanding, converting hydroxyl group to imidazole carbamate as an active intermediate is not the way to do this. First thing come to my mind was a named reaction called the Mitsunobu reaction. The Mitsunobu Reaction allows the conversion of primary and secondary alcohols to esters, phenyl ethers, thioethers and various other compounds including amines. The reaction uses triphenylphosphine ($\ce{Ph3P}$) and diethylazodicarboxylate ($\ce{DEAD}$) to activate hydroxyl group. Then, activated alcohol can undergo substitution reaction with a suitable nitrogen nucleophile (either phthalimide or hydrogen azide can be used) followed by reduction (in the case of azide formation) or acid or base hydrolysis (if phthalimide were used as nucleophile) to give primary amine groups.

Late edit:
Actually, to answer OP's curiosity, N,N'-Carbonyldiimidazole has been used to convert terminal hydroxyl group to amine group. For example, see Ref.1. However, the reference given by OP (Ref.2) has an error on the product structure, which probably made the OP confused about the reaction. Reference 2 has used diamine (1,2-ethylenediamine) as the sunstrate, but the product structure have missed one amine group, which is the one replaces terminal hydroxyl group:

Otherwise, this method (CDI) can be used to convert terminal $\ce{OH}$ group to terminal $\ce{NH2}$ group without using Mitsunobu reaction. Yet, I'd use one $\ce{NH2}$ protected  diamine to avoid unnecessary byproducts.
References:

Rajappa Vaidyanathan, Vikram G. Kalthod, Duc P. Ngo, Jerad M. Manley, and Sean P. Lapekas, "Amidations Using N,N'-Carbonyldiimidazole: Remarkable Rate Enhancement by Carbon Dioxide," J. Org. Chem. 2004, 69(7), 2565–2568 (ODI: https://doi.org/10.1021/jo049949k).
Shuwei Zhou, Jingwen Wang, Gaoqiang Wang, Zhi Jiang, and Hua Ren, "An approach to developing high dielectric constant nanocomposites based on polyrotaxane derivative," European Polymer Journal 2017, 90, 312-322 (ODI: https://doi.org/10.1016/j.eurpolymj.2017.03.037).

I actually know this method but i am asking about the other using CDI in the attached paper ?


text:
How to make a pH 6.8 Tris buffer?
To preface this, I am a biologist.
I need to make a $\mathrm{pH}=6.8$, $\pu{1 M}$ Tris buffer. We have access to Tris base and Tris Hydrochloride. (Unfortunately, none of the fancy Tris compounds will work in my situation.)
I tried making a straight Tris base buffer, but before I could lower the $\mathrm{pH}$ to 6.8 (using concentrated $\ce{HCl}$), it crashed drastically. I assume that I got past the buffering capabilities.
Are there any recommendations for this? Thank you.

You need $\mathrm{pH}$ of your Tris buffer @ 37 degrees, I believe.

From Wikipedia...

The conjugate acid of tris has a pKa of 8.07 at 25 °C, which implies that the buffer has an effective pH range between 7.5 and 9.0.

So you're outside the range where TRIS is an effective buffer.
I assume by "crashing" you mean that a precipitate formed. A buffer is also rated based on its "buffer capacity." Effectively this depends on the the concentration of the buffer. Thus solubility of the buffers comes into consideration.

Tris is short for tris(hydroxymethyl)aminomethane. Tris has a $\mathrm {pK_a}$ of $\mathrm {8.1}$ and a $\mathrm {pH}$  level between $\mathrm {7}$  and $\mathrm {9}$. Although that range may change at physological temperature, I doubt it go as low as $\mathrm {pH} = 6.8$ without sacrificing  $\mathrm {[B]/[A]}$, which may go as low as $\mathrm {\approx0.05}$.
Thus, I think you should use $\mathrm {BIS-TRIS-propane}$ instead. $\mathrm {BIS-TRIS-propane}$ (1,3-Bis[tris(hydroxymethyl)methylamino]propane) is two Tris molecules connected by 1,3-diaminopropane. It has $\mathrm {pK_a}$s of $\mathrm {6.8}$ and $\mathrm {9.0}$, and effective $\mathrm {pH}$ range of $\mathrm {6.3-9.5}$.
I'm not sure how soluble it in deionized water, but worth trying ($\mathrm {$1/g}$). Probably a good idea to keep your stock solution in $\mathrm {0.5 M}$ range.

Another choice: Ypu may use N,N-Bis(2-hydroxyethyl)-2-aminoethanesulfonic acid (BES) (CAS Number: 10191-18-1), instead. It has p$K_\mathrm{a}$ of $7.193$ at $\pu{25 ^{\circ}C}$ (Ref.1) and useful pH range of $6.4-7.8$. BES is therefore better suited for pH control in the buffer region of biochemical interest (Ref.1,2). Sigma-Aldrich sells it in 99% pure form (~$1.0/g).
BES is soluble to the extent of $\pu{3.2 M}$ at $\pu{0 ^{\circ}C}$ (Ref.2) and can be easily purified from aqueous alcohol. Two equimolal buffer solutions consist of BES and sodium besate (sodium salt of BES), each at a molality of $\pu{0.02 mol kg^{-1}}$ and $\pu{0.10 mol kg^{-1}}$, respectively have shown p$K_\mathrm{a}$ value of each solution as $6.97$ at $\pu{37 ^{\circ}C}$ (Ref.1). On the other hand, a solution prepared with $\pu{0.03 mol kg^{-1}}$ of BES and $\pu{0.02 mol kg^{-1}}$ of sodium besate has shown p$K_\mathrm{a}$ value of $6.76$ at $\pu{37 ^{\circ}C}$ (Ref.1).
References:

R. N. Roy, E. E. Swensson, G. LaCross, Jr., C. W. Krueger, “Standard buffer of N,N-bis(2-hydroxyethyl)-2-aminoethanesulfonic acid (Bes) for use in the physiological pH range 6.6 to 7.4,” Anal. Chem. 1975, 47(8), 1407–1410 (DOI: 10.1021/ac60358a076).
N. E. Good, G. D. Winget, W. Winter, T. N. Connolly, S. Izawa, R. M. M. Singh, “Hydrogen Ion Buffers for Biological Research,” Biochemistry 1966, 5(2), 467–477 (DOI: 10.1021/bi00866a011).


text:
Crazy glue vs Nail glue
So I have a nail repair kit, consisting of nail glue(ethyl cyanoacrylate), powder and a nail buffer. I'm almost of out glue yet stick have a bunch of powder. One of the crazy glue tiny bottles has almost the same name(ethyl 2-cyanoacrylate). Can I refill the nail glue with the crazy glue?
Short version -
Nail glue - ethyl cyanoacrylate VS ethyl 2-cyanoacrylate - super glue
Are these the same?

they are the same chemical yes - (the 2 just refers to the position of the side branch.)
I Would not expose the glue to air unnecessarily as any moisture will make it harden . I can not say for you using it on your nails for insurance reasons ( but scientifically they are the same )


text:
Major product monochlorination of 2-methylpropane

What is the major product of monochlorination of 2-methylpropane and for 2-methylbutane?

My attempt: as in 3 degree position, monochlorination is favoured (due to free radical formation), so according to me, it should be 2-chloro-2-methylpropane and same as in other 2-chloro-2-methylbutane.
But, it is stated to be incorrect. The correct answer is 1-chloro-2-methylbutane. Why's that the correct answer instead?

1-chloro-2-methylbutane makes no sense to form as the major product. I highly suspect you are right, and the key might be wrong.

For each example consider the relative reactivity of each type of CH bond and the number of each one. http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#Radical

@PrittBalagopal on the contrary, Cl isn't selective enough (contrary to bromine)

@Mithoron I believe that would simply reduce the % of 2-chloro-2-methylbutane in the product mixture as compared to other products, but I don’t really think it could actually flip the scenario (make 1-chloro more preferable).

And here I wonder why no one in 9 hours marked it as dupe of first question on the "related" list...


text:
What effect, if any, would each of the following actions have on the calculated concentration of the NaOH in this experiment?
What effect, if any, would each of the following actions have on the calculated concentration of the NaOH in this experiment? (e.g. would the calculated concentration be higher, lower or the same?) Treat each action separately and explain your answer.
1-You overshoot the endpoint.
2-The beaker that you put the NaOH in already had a small amount of distilled water in it.
3-You add a small amount of water to the conical flask containing the KHP.
4-You forget to add indicator.
5-There is an air bubble in the tip of the burette before you start the titration but it isn’t there after the titration.

Is this supposed to be a homework for us? Explain your efforts that you made to solve your own question, or else the question will be closed due to lack of such. We are not a homework help service.

I'm assuming you have the sodium hydroxide in the flask and not in the burette.
1) if you over shot you will have more of titre in the flask that you want to it would be assumed that you had to add more titire to reach the end point so the concentration of the NaOH would be mistaken to be stronger than it actually is.
2)if the NaOH has added to a flask with water in it, it should make a difference as you will still be placing the same number of moles in the flask for the titration.
3) dont know what KHP is
4)if you didn't add any indicator then it wouldn't affect anything as long as the reaction still reached the end point. but you wouldn't know where the end point is so the results are pretty useless unless you knew how much to titrate.
5)if there is a bubble in the burette then you would have added less of the titre than you think you have added. this will mean that calculations will so the NaOH to be stronger than it actually is.

Hi,  and welcome to our site! Although we appreciate the time you took to answer this, we typically don't answer questions that are likely to be closed so as not to encourage further low quality questions.

ok, ill try to remember that. although it did only just read that it as was likely to be closed down

no the naoh iis in the burette and the khp is in the flask

well you should have said that in your question as it wasn't clear so I took it to be the typical way round.


text:
Is the bond dissociation enthalpy of Cl-F greater than that of Cl-Cl?
In my book, it is mentioned that $\ce{Cl-F}$ has greater bond dissociation enthalpy than $\ce{Cl-Cl}$. Is it true? I know that interhalogen molecules usually have weaker bonds than dihalogen molecules except $\ce{F-F}$ molecule due to high electron-electron repulsion.

"I know that Interhalogen molecules usually have weaker bonds than dihalogen molecules" - in an X-Y molecule where X is the lighter halogen what are you comparing to, the dissociation energy for X-X, or Y-Y, or Sqrt( X-X * Y-Y ), or something else?

Comparing dissociation energy of XY with that of XX and YY

From this source, it can be seen that the $\ce {Cl-F}$ bond (BDE $\approx\pu{251 kJ/mol}$) is indeed stronger than the $\ce {Cl-Cl}$ bond (BDE $\approx\pu{243 kJ/mol}$). This can be explained by the partial ionic character of the $\ce {Cl-F}$ bond. The Pauling electronegativity value of fluorine ($\chi = 3.98$) is appreciably higher than that of chlorine ($\chi = 3.16$). Thus, there are significant partial positive and negative charges on the fluorine and chlorine atoms in the $\ce {ClF}$ molecule. Although the difference in orbital energies and size would mean that the covalent interaction between the chlorine $\ce {3p}$ and fluorine $\ce {2p}$ orbital would be less strong compared to that between 2 chlorine $\ce {3p}$ orbitals, the ionic character of the bond overcompensates for this and in fact, results in the $\ce {Cl-F}$ bond being stronger than the $\ce {Cl-Cl}$ bond.


text:
Good books in inorganic chemistry?
I would like you to reccomend me a book (books) in inorganic chemistry , with good explanations , problems and solutions. Actually, i want to study it profoundly, but my theory isn't good enough.

I think J.D Lee is good book to get a Good grip on  concepts Of inorganic chemistry.(Its the type of book which is recognized by the name of its Author)


text:
How to crystallize a solution of N,N-dimethylphenylalaninemethylesterhydrochloride without heating it?
I have an aqueous solution of N,N-dimethylphenylalaninemethylesterhydrochloride and I need to crystallise it as a salt. I'm afraid of increasing the temperature or the concentration of acid due to the possibility of ester hydrolysis.
What crystallisation method would be the most efficient in this situation?

How dilute is the solution?

This molecule is an ammonium anion with an ester side chain as a chloride salt.
The first option you have is to undertake the vacuum evaporation of the solvent at room temperature. This would should prevent any heat degradation. I do understand your point about increasing the acid because in effect you have a buffer solution.
Your second alternative is to partition into a  warm amine-ophilic non-aqueous/organic layer and crystallize from there. The method is briefly described in Phase-Transfer Catalysis - Fundamentals, Applications, and Industrial Perspectives[1].
An excerpt from the book:

b. Simple Transfer Across the Interface    A second mechanism for anion transfer is simple anion interchange across the interface, as
represented in the following diagram:

This mechanism does not require catalyst to actually leave the organic
phase, to the extent that the interfacial area is still considered a
part of the organic phase, but it does require the quaternary salt to
have a dominant presence at the interface. Large water-insoluble
quaternary salts, such as tetrahexyl- or trioctyl-methylammonium
salts, which are predominantly (>99%) partitioned into the organic
phase, function as catalysts mostly by anion transfer across the
interface. However, if the quaternary cation is extremely large, the
cation will not be frequently in the interfacial region, and then the
reaction rate will become transfer-limited. On the other hand,
quaternary salts that are good surfactants, such as
$\ce{C16H33NMe3+Cl-}$, are dominantly positioned at or in the
interfacial layer. These accessible quaternary salts provide for
higher rates of anion transfer across the interface.

This research paper[2] goes into further details about the process.
References:
[1]: C. M. Starks; M. Halper. Basic Concepts in Phase-Transfer Catalysis. Phase-Transfer Catalysis - Fundamentals, Applications, and Industrial Perspectives, First Edition; Springer Netherlands: Dordrecht, 1994; pp 8.
https://doi.org/10.1007/978-94-011-0687-0
[2]: Hull, R.L. and Biles, J.A. (1964), Physical chemical study of the distribution of some amine salts between immiscible solvents. II. Complexation in the organic phase. J. Pharm. Sci., 53: 869-872.
https://doi.org/10.1002/jps.2600530804

If you have no solid to test solubility, my first attempt would be to use an anti-solvent (which you'll need to determine empirically).  The anti-solvent for your hydrochloride salt should be miscible with water, so for ease, I would suggest starting with simple alcohols; methanol, ethanol,  2-propanol. See which gives you some solid precipitating out of solution, check purity and then work from there on concentrations or temperature if required.


text:
Jahn Teller distortion and magnetic moments.
It is relatively well known that the discrepancy between the observed and experimental magnetic moments in the first row transition metals is attributed to spin-orbit coupling when it comes to metal ions with $A_2$ and $E$ ground states. For example, Cu(II) has an $E_{2g}$ ground state term has the observed moment is slightly higher (1.96 BM for a phenanthroline complex) than the theoretically predicted "spin only" value (~1.73 BM, attributed to 1 unpaired electron). I also know that this is calculated by a formula which makes use of the spin-orbit coupling coefficient:
$μ_{eff}$ = $μ_{s.o.}$(1-$α λ$ /$Δ_o$)
where, $α$ takes up the values 2 (for E ground state) and 4 (for $A_2$ ground state), $\lambda$ is the spin orbit coupling coefficient, $Δ_o$ is the crystal field splitting energy.
My question relates to the possible effect of a Jahn Teller distortion on the magnetic moment of, say, Cu(II). Intuitively, a distortion should reduce the orbital contribution to angular momentum by the loss of orbital degeneracy. On the other hand, for Cu(II), for example, the $e_g$ set is magnetically non active. So does splitting basically have no effect on the orbital part of the angular momentum? The $t_{2g}$ set should be magnetically inactive as well because of being fully filled for Cu(II).

Planar complexes may be affected by this effect- not orthogonal complexes .  ref-(2017)
https://aip.scitation.org/doi/pdf/10.1063/1.4974805

"The most striking result is that the planar copper(II) (3d9
    ) and nickel(II) (3d8
    )
    chelation afforded very strong ferromagnetic coupling often with 2J ≥ 300 K, owing to the orthogonal orbital
    arrangement between metal ion 3dx2-y2 and oxygen 2pz orbitals. Such a structure-magnetism relationship has
    recently been extended to coordination to rare earth metal ions."

Thanks for the reference! My confusion was solved quite a while ago. Didn't know about this though!


text:
Library / function for computation of molecular mass / molecular weight in C
We are looking for a function to be used for computing the molecular mass / molecular weight / formula weight in C, either in a library or standalone.  So far we have found the following implementations:

R language: shown here (limited implementation)
Matlab: shown here (full implementation)
Python: shown here (issue is discussed, preliminary implementation)
Perl: shown here (for protein molecular weights)

Ideally, the function would be able to handle a string input which could be either an element or compound.  In the latter case, the total number of each element forming the compound would be determined prior to atomic mass look-up.

For Python, periodictable should be used.

If you download AmberTools, under the AmberTools/src/sff directory, there is atomic_number.{h,c}. Note that this code is under the GPL.
atomic_number.h:
#ifndef ATOMIC_NUMBER_H
#define ATOMIC_NUMBER_H

#include "sff.h"

// Atomic number 0 has mass 0 and is a lone pair or extra point

#define NUM_ATOMS 119

const static REAL_T ATOMIC_MASS[NUM_ATOMS] =
   {
     0.0, 1.0079, 4.0026, 6.941, 9.0122, 10.811, 12.0107,
     14.0067, 15.9994, 18.9984, 20.1797, 22.9898, 24.3050,
     26.9815, 28.0855, 30.9738, 32.065, 35.453, 39.948,
     39.0983, 40.078, 44.9559, 47.867, 50.9415, 51.9961,
     54.9380, 55.845, 58.9331, 58.6934, 63.546, 65.409,
     69.723, 72.64, 74.9216, 78.96, 79.904, 83.798,
     85.4678, 87.62, 88.9059, 91.224, 92.9064, 95.94,
     98., 101.07, 102.9055, 106.42, 107.8682, 112.411,
     114.818, 118.710, 121.760, 127.60, 126.9045, 131.293,
     132.9055, 137.327, 138.9055, 140.116, 140.9077, 144.242,
     145., 150.36, 151.964, 157.25, 158.9254, 162.500,
     164.9303, 167.259, 168.9342, 173.04, 174.967, 178.49,
     180.9479, 183.84, 186.207, 190.23, 192.217, 195.084,
     196.9666, 200.59, 204.3833, 207.2, 208.9804, 209.,
     210., 222., 223., 226., 227., 232.0381,
     231.0359, 238.0289, 237., 244., 243., 247.,
     247., 251., 252., 257., 258., 259.,
     262., 261., 262., 266., 264., 277.,
     268., 281., 272., 285., 284., 289.,
     288., 292., 291., 294.
   };

// This routine will return the atomic number of
int get_atomic_number(REAL_T mass);
#endif

atomic_number.c:
#include "atomic_number.h"
#include <math.h>

/*
 * Get the atomic number by selecting the atom whose atomic mass is the closest
 * to the atomic mass that is used. This has the advantage of working for all
 * isotopes with the possible exception of tritium (which is closer to He).
 */
int get_atomic_number(REAL_T mass) {
   int i, atomic_num = 0;
   REAL_T min_diff = fabs(mass);
   REAL_T diff;

   for (i = 0; i < NUM_ATOMS; i++) {
      diff = fabs(mass - ATOMIC_MASS[i]);
      if (diff < min_diff) {
         min_diff = diff;
         atomic_num = i;
      }
   }

   return atomic_num;
}

Kind thanks for the comment.  One of the features that the matlab and python implementations above possess is the ability to parse a string representing the formula of a compound, for example C2H2, returning the formula weight.  I will edit my question to reflect this.

I am not aware of such an implementation. Typically you don't want to do string parsing in C, because you need to look at each character in the string individually and can't apply a regex. It would be easier if the requirement was for C++.


text:
electronegativity and hydrogen bonding
I have a basic idea why, but can someone just confirm, why chlorine isn't able to hydrogen bond despite the fact that it has the same electronegativity value as nitrogen.

H-bonding only occurs between a hydrogen bonded covalently to oxygen, fluorine or nitrogen, and another oxygen, fluorine or nitrogen bonded to a hydrogen.It does not matter that chlorine has the same electronegativity as nitrogen because (i assume you mean H-Cl) chlorine is larger than the O,F or N so its electron density is lesser than these elements. If you meant Cl-Cl, there is no difference in electronegativity between the two atoms so polarity = 0.

That is true of the strongest hydrogen bonds.  But, as other questions/answers under the hydrogen-bonding tag point out, there are also hydrogen bond interactions involving heavier halogens and even carbon.  Shakespeare:  "There are more worlds, Horatio, than are dreamt of in [routine] philosophy."


text:
Regenerating a Cobalt Acetate Catalyst
I have an eluent from an ion-exchange column containing Cobalt (II) ions and Chloride ions. In order for the cobalt catalyst to be recycled to the main process, it must be in its acetate form.
Is there a route from cobalt chloride to acetate that is feasible at an industrial scale?
Thanks

Quick Laboratory method:
$$\ce {CoCl2(aq) + 2NaOH(aq)->Co(OH)2(s) + 2NaCl(aq)}$$
$$\ce {Co(OH)2(s) + 2CH3CO2H(aq)->Co(CH3CO2)2(aq) + 2H2O(l)}$$
$$\ce {Co(CH3CO2)2(aq) + \Delta->Co(CH3CO2)2\cdot 4H2O (s)}$$
You should separate $\ce{Co(OH)2}$ by filteration and washed with water ($\pu {K_{sp} = 1.0 \times 10^{-15}}$) to remove all $\ce{NaCl}$ before subjected to react with $\ce{CH3CO2H}$.
This might work in industrial scale, if be careful. $\ce{Co(OH)2}$ is unstable to heat but no need to dry before react with $\ce{CH3CO2H}$.

Care should be taken in the presence of complexants as ammonia or too high pH values that might form soluble cobaltates.

@aventurin: You are absolutely correct. Thank you for the valuable point.

add sodium sulphide solution -> precipitate  cobalt sulphide complex ( highly stable ) -> filter - dissolve in acetic acid