Why does two phenylhydrazine molecule react with glucose in osazone formation? From Wikipedia:

  It involves two reactions. Firstly glucose with phenylhydrazine gives glucosephenylhydrazone by elimination of a water molecule from the functional group. The next step involves reaction of one mole of glucosephenylhydrazone with two moles of phenylhydrazine (excess). First phenylhydrazine is involved in oxidizing the alpha carbon to a carbonyl group, and the second phenylhydrazine involves in removal of one water molecule with the formyl group of that oxidized carbon and forming the similar carbon nitrogen bond.

Why doesn't it first react with the formyl group and stop with just one phenylhydrazine? That is, why doesn't the nitrogen of phenylhydrazine attack the $\mathrm{sp^2}$ carbonyl carbon (a nucleophilic addition reaction)?

Why does it oxidize glucose first and then proceed?

As according to modern mechanism it has first reacted with formyl then goes for tautomers and to form an intermediate..which further form iminoketone ,as phenylhydrazine is in excess then it again react with the c=o formed on C2 and then with carbon first and give osazone .  And as further reaction stops by H- bonding between H and Nitrogen of each phenyl hydrazine which forms chelation and makes it stable.


Cyclohexene reduction reaction with hydrogen gas and ethanol I have a following reaction (see below). I found the section of reduction and oxidation states in my book but there is nothing about cyclohexenes. I know that the double bond will be reduced in this reaction but what I have no idea what happens to those substituents. Why do they move around and what are their orientations?

Note: This is a homework question, so I will not give a complete answer.

$\ce{H2/Pd}$ is a reducing agent, and it acts so by adding hydrogen to the double bonds. Alkenes are converted to alkanes. This proceeds via a concerted pathway, so rearrangements due to carbonation shifts do not occur. Now I think you can proceed from here.

As far as those substituents are concerned, think carefully on what would happen.

The reaction with $H_2/Pd$ is one of the ways to reduce alkenes to alkanes. This takes place over the Palladium surface which has $H$ adsorbed on it. Hence we can think of cis addition in this case.

It is clear through the mechanism that there is no formation of carbocations. So we need not worry of rearrangements. Now from the options up there it is quite clear I suppose. Hope it was helpful.


Acidic nature of boric acid when its concentration high Following statement was given for boric acid $(\ce{H3BO3}):$

  At low concentrations $(\leq\pu{0.02 M})$ essentially $\ce{B(OH)3}$ and $\ce{B(OH)4-}$ are present, but at higher concentration the acidity increases and pH studies are consistent with the formation of polymeric species such as $$\ce{B(OH)3 <=> H+ + [B3O3(OH)4]- + H2O}$$

Please explain the meaning of the italicised statement i.e. pH studies are consistent with the formation of polymeric species?

In the equation $\ce{H2O + H3BO3 -> H+ + B(OH)4-}$, one negative charge on the anion is shared among 4 electronegative oxygens. Yes, one positive charge on the proton is also shared with some water molecules, so the orthoborate ion probably shares some negative charge with some waters, but let's keep it simple.

Boric acid is known to complex with sugars to generate higher acidity. If boric acid complexes with itself, as in $$\ce{3 H3BO3 -> H+ + [B3O3(OH)4]- + 2 H2O}$$, the one negative charge on the anion can be considered to be distributed among 7 electronegative oxygens, or maybe still distributed mostly on the OH groups, but since they will be separated more, the stability of the ion will be greater, and the tendency to lose a proton will be greater: greater acidity. The electronegativity of boron is less than that of hydrogen (2.02 vs 2.2, electronegativity, Wikipedia), so when the atoms in the ions are counted, the first ionization equation has one hydrogen and one boron as the metallic atoms getting the positive charge, while the second equation has one hydrogen and 3 borons donating electrons - more sharing of the positive charge here, tho the hydrogen still gets the positive charge.

But this structure may be only a suggestion. Well-known polyborate anions include the triborate(1-), tetraborate(2-) and pentaborate(1-) anions. The condensation reaction for the formation of tetraborate(2-) is as follows Wikipedia

$$\ce{2 B(OH)3 + 2 [B(OH)4]− ⇌ [B4O5(OH)4]2- + 5 H2O}$$

The italicized text mentions an observation of pH that is consistent with the complexation of boric acid with itself. It doesn't prove the existence of the triborate ion only, and perhaps is only a way to keep the text shorter without ignoring complexation completely. Then along comes a student who looks into the sentence and sees that something was omitted or not explained thoroughly. (Keep asking questions!)

A more precise answer to the polymeric structure, at least at neutral pH, is found in Wikipedia on borate, to quote in part:

"At neutral pH boric acid undergoes condensation reactions to form polymeric oxyanions....
The tetraborate anion (tetramer) includes two tetrahedral and two trigonal boron atoms symmetrically assembled in a fused bicyclic structure. The two tetrahedral boron atoms are linked together by a common oxygen atom and each also bears a negative net charge brought by the supplementary OH− groups laterally attached to them. This intricate molecular anion also exhibits three rings: two fused distorted hexagonal (boroxole) rings and one distorted octagonal ring. Each ring is made of a succession of alternate boron and oxygen atoms. Boroxole rings are a very common structural motif in polyborate ions."

Also:

"Simple bonding theories point to the trigonal planar structure. In terms of valence bond theory the bonds are formed by using sp2 hybrid orbitals on boron. Some compounds termed orthoborates do not necessarily contain the trigonal planar ion, for example gadolinium orthoborate..."

My reading of the above is that it is possible, via a condensation reaction, for Boric acid to create polymeric oxyanions, and in particular Boroxole rings at neutral pH.


Decreasing order of basic strength I need to find the decreasing basicity of the following compounds(I, II, III, IV, respectively):

I know that to compare the basic strengths, we need to find the stability of their conjugate acids.
Following are the conjugate acids, I drew (I don't know if they're correct):

According to me, due to resonance of the benzyl groups or lack of resonance, the order of basic strength should be:

$$\mathrm{III > II > I > IV}$$

But the answer given is:

$$\mathrm{IV > I > II > III}$$

Where am I going wrong?

Basicity is how well an atom can donate electrons. More is the electron density more is the basicity of the compound.

In Aniline the lone pair of electrons present on the nitrogen are involved in resonance, so the electron density on N is less than expected.

In compound II (Diphenylamine), since two benzene ring is present so the electron density will be less as now lone pair are involved in resonance of both the ring.
Thus electron density on Nitrogen is less than as compare to Nitrogen atom of Aniline

Similarly, in III compound (Triphenylamine), since three benzene ring is present the lone pair are more localized, and, thus electron density will be much less (least) as compared to the other compounds

In 4th structure, there is no resonance so, lone pair are present with nitrogen itself, and the electron density will be high, also due to inductive effect of the alkyl groups the electron density is more, and thus it can easily donate the electrons as compared to all other compounds. So the basicity is the highest for compound IV.

Therefore, the order of basicity is IV>I>II>III.

i guess i could help out on this one..as i have completed the book(for JEE)
The reason for the abnormal order is that the presence of more than one phenyl groups dosent actually increase the stability of the conj acid...rather it just increases the steric crowding around 'N'....The reason why more phenyl groups dont add to stability is that the phenyl grups due to repulsions between themseles keep rotating about the axis and are not always in the same plane for conjugation...thus do not really help in stabilising the N+ ion.....and i guess youd know that cyclohexyl amine is more basic than anilene...so for that simple reason.....4>1>2>3

P.S:The answer was provided to me by a teacher at FIITJEE...so you can be sure its authentic


Why is there a need for resonance? Why are we able to show structures of compounds with different position of electron, but with same position of them? Shouldn't the structure become unstable due to this?

Let's take benzene for the example of resonance.We know that in reality benzene has no pure single or double bond. How can this be possible?

The bond length for $\ce{C-C}$ and $\ce{C=C}$ bonds in benzene is same which is $1.39~\mathring{\mathrm{A}}$. This is due to partial single and double bond between carbon atom, which is due to resonance. Why is there a resonance? Why can't benzene be happy with three alternate single and double bond between each carbon atom?

As I mentioned in my comment, the Wikipedia article on resonance provides a rather nice description of the subject where the Resonance in quantum mechanics section is of particular interest for the question asked.

However, I decided to somewhat clarify the matter emphasising few important points. And I will start by quoting the OP.

  We know that in reality benzene has no pure single or double bond.

Now please, take a deep breathe and repeat with me: there are no bonds. In reality we do have electrons and nuclei that constitutes a molecule, but a chemical bond is merely a theoretical concept introduced in the framework of a scientific theory. In other words, chemical bonds do not exist in the same sense as electrons and nuclei do.

For instance, the concept of resonance was introduced by Linus Pauling in the framework of the so-called valence bond theory, according to which

  a covalent bond is formed between the two atoms by the overlap of half
  filled valence atomic orbitals of each atom containing one unpaired
  electron.

So, this is the way the concept of chemical bonding enters the scene in the valence bond theory: a chemical bond is used to describe overlapped atomic orbitals of two atoms in a molecule. Note that atomic orbitals also do not exist in the same sense as electrons and nuclei do, but an orbital has at least some strict mathematical and physical meaning, while a chemical bond, as introduced above, does not. Read the above quote carefully one more time: it is not even a definition! What we just say is that whenever half filled valence atomic orbitals of two atoms overlap a covalent bond is formed, but we do not say what exactly is a covalent bond. It is just something; something which is formed whenever orbitals overlap. It is a concept, purely theoretical construction. It is practically helpful in description of a reality, but no one will say that it is actually a part of it.

Do not be afraid of all this business, because it is the way science generally works. Scientific theories basically tell us how nature could be described, and they do so in the language of mathematics, since it is the only language nature can unambiguously "speak" with us. However, it is often so tempting to force it to speak with us in a different language, in a language of our theoretical concepts (such as, for instance, the concept of a chemical bonding), no matter how fuzzy they are. This is a very dangerous road, because, after all, who are we to tell nature which language to speak? Nature will likely refuse to "speak" with us in that language, and that would be it.

That is what actually happened with the valence bond theory as it was originally formulated. Indeed, all carbon-carbon bond lengths are equal in benzene, but the valence bond theory were not able to account for that simple fact. Nature refused to "speak" with us in the language of chemical bonding. At that point chemists had two choices: either abandon the valence bond theory or introduce another theoretical concept into it, introduce a new word in our language making it possible to communicate with nature in this language. The concept of a chemical bonding was already so tremendously useful in chemistry that the choice was actually predetermined, and in 1928 Linus Pauling introduced the concept of resonance between several valence-bond structures of a molecule.

What is resonance? Is it a physical process? No way! It is just a concept, a word, used to describe

  delocalized electrons within certain molecules or polyatomic ions
  where the bonding cannot be expressed by a single Lewis formula.

The word "resonance" was probably not the best choice by Pauling, since it gives an unwanted feeling that the molecule actually do oscillate back and forth between the structures, which is, as I said, not the case. But putting aside its somewhat misleading name, the concept of resonance is as good as the concept of chemical bonding and many other concepts in chemistry and other sciences.

One way to look at it is the Electron Clouds.

A chemical becomes reactive when it has either excess or insufficient electron cloud. For example, in case of Ammonia $\ce {NH}_3$, Nitrogen has a lone pair of electron. Thus, it has excess of electron cloud. It can donate these electrons. Hence, Ammonia is reactive. Similarly, Aluminium is highly electropositive and has empty p-orbital. Hence, it can accept electrons and is also reactive.

In case of Benzene, the electrons are delocalised, that is, the electron cloud is spread over the entire ring of Benzene. Thus, there is no accumulation of charge in the ring at one place. This gives extra stability to Benzene. And every molecule wants to attain maximum stability.

If resonance didn't exist in Benzene, then it would react easily with many reagents. Since each of the carbons is $\ce {sp}^2$ hybridised, that is, each carbon is quite electronegative, and has the capacity to react.

Maybe there is a better answer to your question. This is what I can think as the reason.


Z-matrix of a series of 5 colinear atoms How can I write a Z-matrix for the following complex?

  O==C==O  H--O
               \
                H

I have managed to write:

O
C   1  a
O   2  a     1  180
H   3  d     2  180    1  ?
O   4  b     3  180    2  ?
H   5  b     4    t    3  ?

but I don't know how to specify the dihedral angles indicated with ?. Are they all zero? Or completely unspecified? If so, will I be able to do a coordinate optimization with these variables or will it be a problem?

You cannot have bond angles of 0 or 180 degrees in a Z-matrix. This is because the dihedral becomes degenerate. To solve this, you can add 'dummy atoms' which provide an auxiliary point of reference to remove the 180 degree angle. These atoms are usually denoted $\ce{X}$ or $\ce{Xx}$ in computational chemistry suites. Usually, dummy atoms are placed at 90 degrees to a pair of atoms and the third collinear atom is then defined with a bond angle of 90 degrees and a dihedral of 0 or 180 degrees with respect to the previous 3 atoms.

Example

To illustrate what I'm talking about, here's a schematic of your pair of molecules, with three dummy atoms:

This could have a Z-matrix like:

O   1
C   1    a
Xx  2  1.0  1  90
O   2    b  3  90     1  180
Xx  4  1.0  2  90     3  0
H   4    c  5  90     2  180
Xx  6  1.0  4  90     5  0
O   6    d  7  90     4  180
H   8    e  6  104.5  7  180

Notes

There are no hard and fast rules about how many you should use (obviously we could simply define subsequent dihedrals in terms of the leftmost dummy atom) however it's best to strive for locality so that small changes in some variable don't correspond to large displacements elsewhere, in which case rounding errors or large gradients can ruin your day. You could also have defined all of the dihedrals in your molecule with respect to that convenient non-collinear hydrogen, however this is the Z-matrix equivalent of spaghetti code.

You will note that I use different letters for each variable. This is because forcing the $\ce{H-O}$ and $\ce{C=O}$ bonds to be of equal length is probably an unrealistic constraint for a geometry optimisation of this system.

Finally, note that dummy atoms do not take part in the electronic structure of a molecule, so feel free to put them in anywhere and set the dummy atom 'bond' length to whatever you want. 1.0 Ångström is traditional as far as I know.

P.S.

Molden has a complete, if idiosyncratic graphical Z-matrix editor.

You can use SMILES in Avogadro or openbabel and it produces 3D coordinates that
you can edit. And you can create Gaussian, NWChem, gamess... inputs.


What is the Difference Between the Conjugate Effect and the Inductive Effect? As the title suggests, what is the difference between the conjugate and the inductive effect in organic chemistry.

Conjugate effect (or delocalization) is an effect in which molecular orbitals (MOs) are conjugated to new molecular orbitals that are more delocalized and therefore generally lower in energy (the amount of MOs stays the same of course). The electrons can move freely in these new extended orbitals. As an example consider this MO diagram of butadiene.

Note that the lowest energy MO is the MO with the least amount of nodes.

Inductive effect on the contrary does not increase delocalization but just pulls electrons towards the inductive (polarized) atom, this, for example, makes lone pairs less available decreasing the nucleophility of a molecule and increase its acidity. A lot of the time this inductive effect will be achieved trough a $\ce{C-X}$ (fluorine and chlorine) bond and you can observe its inductive effect simply by looking at the acidity of a series of more and less substituted molecules. Compare for example trifluoroacetic acid (pKa of 0.23) with regular acetic acid (pKa of 4.76).

I'm not exactly sure though what effect induction has on the size of a MO (but it won't match the effects of conjugation for sure). It is clear however, that the electron distribution will shift toward the spatial direction of the inductive atom.

The inductive effect refers to the shifting of electrons through σ bonds , the conjugations effect through pi bonds


Why does hand sanitizer heat up? While washing my hands with soap I added some hand sanitizer just because it was right there on the counter and couldn't resist mixing them... Ha!  However, when I added the hand sanitizer it heated up for a moment and then cooled back down.  Why does this happen?

Hand sanitizers are generally ~50% - 70% ethanol or isopropanol. When these alcohols mix with water, they may release heat. Starting at room temperature, mixing ~100% ethanol with an equal volume of water will raise the temperature of the mixture ~8 0C.

It sounds like some type of exothermic reaction happened between the hand sanitizer which is primarily alcohol, and the soap.


Does any known substance ignite on cooling? As the title says, I'm interested in knowing if there is any substance — or combination of substances — that ignites (or even increases its chance of spontaneous ignition) when cooled.

I've never heard of such a thing, nor can I find it in Atkins' Physical Chemistry, but I might easily have overlooked it, since I don't know what to call it. Googling gives me information on substances that lower the freezing point or ignition point of explosives/fuels.

It seems that this should obviously be forbidden on thermodynamic grounds, but I can't quite rule out a phase change allowing such behaviour, for instance.

Actually... yes! Iron(II) oxide is thermodynamically unstable below $848~\mathrm K$. As it cools down to room temperature (it has to do it slowly) it disproportionates to iron(II,III) oxide and iron:

$$
\ce{4FeO -> Fe + Fe3O4}.
$$

The iron is in a form of a fine powder, which is pyrophoric (it may catch a fire when exposed to air). You can see it in action here.

It depends on how you cool it.  For example, if you try to cool an alkali metal by immersion in water, you can get a rapid exothermic reaction.


Nucleoside/nucleobase chemical synthesis I'm interested in the chemical synthesis of nucleoside phosphoramidites, but all the procedures I can find have "start with a protected nucleoside" as the first step. Searches for chemical synthesis of nucleosides all seem to find process that start with existing nucleobases, or in vivo processes, or speculative origin-of-life processes. Searching for chemical synthesis of nucleobases ONLY seems to find speculative origin-of-life processes.

What I'm looking for (just out of curiosity) are processes for chemical synthesis of nucleosides from a non-biological basis (not using pre-existing nucleobases) as used in typical commercial/industrial supply of nucleoside phosphoramidites. I'm starting to wonder if they even are chemically synthesized, or if they are instead harvested in some way, but it's hard to find a smoking gun that outright says it one way or the other without access to whitepapers anymore (no more university account).

Does anybody know?

Finally found what I was looking for, especially paired with Mithoron's comments.

"Currently, purine nucleotides are obtained at industrial level mainly by microbial fermentation".

https://microbialcellfactories.biomedcentral.com/articles/10.1186/s12934-015-0234-4

Ledesma-Amaro, Rodrigo & Jiménez, Alberto & Santos, María de los Ángeles & Revuelta, José. (2013). Biotechnological production of feed nucleotides by microbial strain improvement. Process Biochemistry. 48. 1263-1270. 10.1016/j.procbio.2013.06.025.


Why is Na+ reduced instead of H+ in the electrolysis of dilute NaCl(aq) with Mercury cathode? In a class of electrolysis, my instructor told me that Hg forms Na-Hg in the electrolysis of dilute NaCl aqueous solution. For this reason, sodium cations are reduced in the cathode instead of hydronium ions. However, he did not give me any reason for how creating an amalgam affects the reduction of the metal or, why Hg creates amalgam in the first place.
So, I would like to know why Hg creates amalgam whereas other metals like Pt don't form them ? And also how this affects the reduction of Na.

The mental conflict occurs because two processes could theoretically happen (electrodeposition of sodium, or electrodeposition of hydrogen), and the theoretically unfavorable one occurs - and not only occurs, but is so favorable that it was used commercially to produce NaOH as EdV noted.
So why is the theoretically favored process not favored in this instance? Hydrogen evolution is a multi-step process, and if you wave your hands a lot, you can call it overpotential and be off to other projects.
When (or if) sodium is deposited (on mercury), it can dissolve and be removed from contact with water. So this process is a slam-dunk if you push with a high enough voltage.
But the hydrogen evolution should take place at a lower voltage. Why not? If/when hydrogen is deposited, its first appearance is atomic: one H atom sticks somewhere on the mercury surface; the second step is deposition of another H atom somewhere else. The next step, for the two atoms to form a molecule, requires some mobility on the part of the hydrogen and the mercury surface. At this point, conventional wisdom suggests that the molecule could dissolve into the solution, but eventually the aqueous solution becomes saturated. Then the first molecule must meet another molecule, and another, to begin the process of forming a bubble which will eventually rise to the surface of the aqueous solution. Somewhere in there is an impediment to the formation of bubbles, and it may involve mobility of atoms or molecules, or attachment of hydrogen to mercury, but the process is difficult enough to allow the voltage to be raised high enough to allow deposition of sodium to proceed.
And then we say that mercury has a high overpotential for deposition of hydrogen.
Now suppose we do the same, but with gallium. Sodium should deposit and become dissolved as with mercury, if the voltage is high enough. (Sodium alloys with gallium; amalgamation is a chemical term that should be reserved for mercury alloys.) But the hydrogen overpotential on gallium is less than on mercury (Ref 1), and gallium and hydrogen are deposited together from a solution of $Ga^{+3}$ (Ref 2). So, in the case of liquid gallium, it seems that hydrogen would be more easily deposited than sodium (unless, perhaps you used a very high voltage, which would override the thermodynamic preference because diffusion couldn't keep up with the current demands).
Ref 1. https://pubs.rsc.org/-/content/articlelanding/1966/tf/tf9666203524/unauth#!divAbstract
Ref 2. https://link.springer.com/article/10.1007/BF01016217

If you heat metallic sodium with let’s say gold / silver / mercury / platinum etc.. it will form an amalgam.
I have tried this with a tiny bit of gold some years ago.
I did heat Na in an inert argon atmosphere to several hundred degrees in an quartz tube and drop the gold inside.
It did take only fractions of a second till the amalgam was formed.
The result is a green looking amalgam when it is hot - when it cools down it is a brown solid substance.
Given the fact that gold is a pretty inert substance, I guess that sodium will do the same with Pt, Rh, Ir, Ru, Os, Ag, Hg ...
I guess during electrolysis you have metallic Na formed. I don’t mean junks of Na but Na atoms and some of them may live long enough to amalgamate with the Hg.


Strength of hydrogen bonding in phenol or methanol I wanted to know whether hydrogen bonding is stronger in phenol or methanol.
I saw on Wikipedia about the enthalpies of hydrogen bonding in various cases.
but could not find the answer to this question of mine.

I also want to know whether the hydrogen bond strength depends upon the carbon atom to which oxygen is attached.

Let me answer the second question first, because the answer is so much easier: Of course the hydrogen bond strength depends among others upon the carbon oxygen is attached to. Consider phenol and methanol or methanol and formic acid.

The guess as to which hydrogen bond is stronger is — I have to admit it — my speculation. But I think that methanol should be able to undergo stronger hydrogen bonding.

Phenol, of course, is aromatic and capable of delocalising negative (or positive) charges across its entire ring system — the reason why its $\mathrm{p}K_{\mathrm{a}} = 10$, a good five units lower than methanol’s. That means that at the same pH value of the solution, there should be less strong a negative charge on phenol’s oxygen than on methanol’s.

But hydrogen bonds are, to a certain extent, electrostatic interaction. So the system which has a stronger negative charge on the oxygen (methanol) should be able to build up stronger hydrogen bonds by attracting hydrogen more strongly.

One could also picture it differently: Phenol gets deprotonated at a lower pH, so the $\ce{O-H}$-bond must be a lot weaker than methanol’s. But if the $\ce{O-H}$-bond is weak, the oxygen has a larger tendency to simply loose the hydrogen to another electronegative partner (deprotonation/protonation) rather than share it (hydrogen bond).

Phenol has week acidic effect due to the presence of benzene ring produces a partial positive charge on O atom that facilitates the separation of the H+ ion.


Is it possible to use the school supplied algorithm to build the Lewis diagram of the iodate ion? I am upgrading the grade 12 chemistry credit and have to use the school supplied algorithm to build the Lewis diagram of $\ce{IO3-}$ ion. It seems to me that iodine oxidation state in this ion should be 5, therefore it would have one single and two double bonds with oxygen. When I follow the steps below, I am arriving at a different formula.

I do not need to answer whether the ion is polar or not and what is its shape.

  Step 1: Count all of the valence electrons in the molecule or ion. In the case of an ion, add or subtract electrons to account for the ionic charge.

$$\ce{I = 7e^-, O = 18e^-}; \text{ionic charge} = \ce{1e^-}; \text{total} = \ce{26e^-}$$

  Step 2: Arrange the peripheral atoms symmetrically around the central atom. Use a pair oil electrons to form a bond that links these atoms to the central atom.

6 electrons placed on $\ce{I}$, $\ce{26e^- - 6e^- = 20e^-}$ remain

  Step 3: Add pairs of electrons to complete the octet of the peripheral atoms.

$3\cdot 6$ electrons placed on $\ce{O}$, $\ce{20e^- - 18e^- = 2e^-}$ remain

  Step 4: Place any unassigned electrons on the central atom.

2 electrons placed on $\ce{I}$, $\ce{2e^- - 2e^- = 0e^-}$ remain

  Step 5: If the octet of the central atom is incomplete, move a lone pair of electrons from a peripheral atom to a new position between the central and peripheral atom.

After completing step 4 of the rules above I am arriving at $\ce{I}$ having 8 electron dots. At this point all electrons have been used up, the octet of $\ce{I}$ is complete, and I can only create a structural diagram with two single and one double bonds.

In the course I have taken previously we were taught that $\ce{I}$ was an exception to the octet rule, but when I asked this teacher, I was told that I was not allowed to use the exceptions and had to apply the octets only.

Am I missing anything? Is it possible to comply with the algorithm above and come up with the correct answer?

When I follow your school's algorithm for iodate I get a central iodine atom with three single bonds to oxygen.  I've shown my work below.  This is problematic, as you well know, but I can potentially see where your teacher is coming from.

When I teach formal charge and resonance structures and the like, I often have students use their intro chemistry rules to determine the structure of a polyatomic ion such as this.  If they follow the rules they get a pretty structure that looks proper - I then introduce the idea of formal charge by pointing out the wild imbalance of charges in the ion and how energetically unfavorable that would be, then we work out a way to rearrange the electrons to minimize that problem.

I also have a lot of students who immediately go to google to find Lewis structures rather than drawing them themselves, so when a student tells me that they think the structure should be something different, it takes an actual argument from them before I believe that they know why that should be.  I don't know that that is the case here; it also seems entirely possible that your instructor is just wrong.


Orientation in benzene rings with more than one substituent Why is it that the chlorination of 1-chloro-3-nitrobenzene (18) gives 1,4-dichloro-2-nitrobenzene (19) as the major product, 1,2-dichloro-3-nitrobenzene (20) as a minor product and 1,2-dichloro-4-nitrobenzene (21) is not formed at all?

I know $\ce{-NO2}$ is a deactivating, meta-directing group and hence it decreases the electron density at positions 2, 4, 6 while position 5 is unaffected. On the other hand, $\ce{-Cl}$ is a deactivating, o-p directing group and hence it directs the incoming group to positions 2,4 or 6. Taking into account the fact that $\ce{-NO2}$ is a more powerful deactivating group than $\ce{-Cl}$, the incoming $\ce{Cl+}$ should be attached at position 5 as it has more electron-density than 6,2 or 4 position. But the product with $\ce{Cl}$ at position 5 is not formed. Why so? I am unable to rationalise this.

Source: page 583 of "March's Advanced Organic Chemistry, 7th ed." under the heading "Orientation in benzene rings with more than one substituent".

I quote March's Advanced Organic Chemistry, Reactions, Mechanisms and Structure, 6th edition, page 666.

  Some substituents have a pair of electrons (usually unshared) that may be contributed
  toward the ring. The three arenium ions would then look like this:

  The stability of these two ions is increased by the extra form not only because it is another canonical form, but because it is more stable than the others and makes a greater contribution to the hybrid. Every atom (except of course hydrogen) in these forms (C and D) has a complete octet, while all the other forms have one carbon atom with a sextet. No corresponding form can be drawn for the meta isomer.The inclusion of this form in the hybrid lowers the energy , but also because it spreads the positive charge over a larger area—out onto the group Z. Groups with a pair of electrons (e.g., as the halogens) to contribute would be expected, then, in the absence of field effects, not only to direct ortho and para, but also to activate these positions for electrophilic attack. (my emphasis)

In your example, if chlorine electrophile attacks at C-4 or C-6, the resulting intermediate  is stabilized by an extra resonating structure, lowering the energy of this intermediate.Hence electrophillic aromatic substitution takes place.

If chlorine electrophile attacks at C-5 , the resulting intermediate is not stabilized compared to attack at C-6 .Hence the resulting electrophillic aromatic substitution is not seen.

This is the possible reason  that  the product with Cl at position 5 is not formed.

Quoting from chemistry.msu.edu:

  The influence a substituent exerts on the reactivity of a benzene ring may be explained by the interaction of two effects:

  The first is the inductive effect of the substituent. Most elements other than metals and carbon have a significantly greater electronegativity than hydrogen. Consequently, substituents in which nitrogen, oxygen and halogen atoms form sigma-bonds to the aromatic ring exert an inductive electron withdrawal, which deactivates the ring.

  The second effect is the result of conjugation of a substituent function with the aromatic ring. This conjugative interaction facilitates electron pair donation or withdrawal, to or from the benzene ring, in a manner different from the inductive shift. If the atom bonded to the ring has one or more non-bonding valence shell electron pairs, as do nitrogen, oxygen and the halogens, electrons may flow into the aromatic ring by p-π conjugation (resonance).

  The second factor that becomes important in reactions of substituted benzenes concerns the site at which electrophilic substitution occurs.

  The exact influence of a given substituent is best seen by looking at its interactions with the delocalized positive charge on the benzenonium intermediates generated by bonding to the electrophile at each of the three substitution sites.

Therefore chlorine is an ortho para directing group with minor meta product.

References:

Advanced Organic Chemistry 5th edition Part A: Structure and Mechanisms, Francis A. Carey and Richard J. Sundberg
March's Advanced Organic Chemistry, Reactions, Mechanisms and Structure, 6th edition


Maximum temperature attainable by thermite reaction In the thermite reaction
  $$\ce{2Al(s) + Fe2O3(s) -> 2Fe(l) + Al2O3(s)}$$
  what is the maximum temperature attainable by the products? Assume reactants at $25\ \mathrm{^\circ C}$ and $1\ \mathrm{atm}$ and that all heat capacities remain constant.

I started by using Hess's law to find the enthalpy change of the reaction. I couldn't find a heat of formation for liquid iron so I assume I have to find that out. I just wrote out the equation
$$\ce{Fe(s) + heat + enthalpy of fusion -> Fe(s)}$$
and figured the heat of formation for liquid iron at $25\ \mathrm{^\circ C}$ is around $-302.71\ \mathrm{kJ/mol}$.

To find the max temperature I used
$$\Delta H_\mathrm r=\Delta T\cdot C_\mathrm p(\text{products})$$
and came out with an equation like
$$-14445\ \mathrm{kJ/mol}=0.147\ \mathrm{kJ/(mol\ K)}\cdot T_\text{max}$$
This gives me a negative temperature, so I assume it is wrong. Even if I were to take the absolute value, the answer seems awfully high. I was expecting something around a few thousand degrees not near ten thousand.

According to Wikipedia the maximum temperature is approximately 2500 °C, limited by the boiling point of aluminum, which is 2519 °C.  The article provides some discussion of how the temperature is limited, and how to calculate it using Hesse's law.

There are many different types of thermite compositions although the cheapest form is aluminium based, while you could have magnesium, titanium, zinc, silicon and boron as the fuels source. The oxidzers can also vary from bismuth (III) oxide, boron (III) oxide, silicon (IV) oxide, Chromium (III) oxide, manganese (IV) oxide, iron (III) oxide,iron(I,III) oxide, copper (II) oxide or lead (II, IV) oxide.

Most thermite compositions burn at around 4,000 degrees, however once started the reaction can't be stopped until all the fuel and oxidizers have been spent.


Why is one lobe of an sp3 hybridized orbital smaller than its other half? A hybrid sp3 orbital is drawn with one lobe smaller than its other half, the latter which is of equal size when drawing the p orbital. Why is it so?

When combined at a given atomic center, any two atomic orbitals add in a vectorial
manner. For example, consider the orbital $\phi$ defined by $\ce{p_{x}}$ and $\ce{p_{y}}$ atomic orbitals as

\begin{align}
\phi = c_1 \ce{p_{x}} + c_2 \ce{p_{y}}
\end{align}

The orbital addition can be pictured like this

for the two cases $c_1 = c_2 > 0$ and $c_l = -c_2 > 0$, respectively. The ratio $c_1 / c_2$ controls how much the orbital $\phi$ is tilted away from the $x$ (or $y$) axis.

The mixing of atomic orbitals with different angular momentum quantum number is
also controlled by a vectorial addition and leads to various types of hybrid orbitals
some examples of which are shown here:

The example a) would be an $\ce{sp}$ hybrid orbital. It gets its sense of direction from the $\ce{p}$ orbital used to construct it because the $\ce{s}$ orbital is isotropic. Its shape comes about because the parts of the orbitals that have an opposite sign (this is hinted at by the shading in the pictures) cancel each other out to some extent. So, if the left lobe of the $\ce{p}$ orbital is negativ and the right one is positive and the $\ce{s}$ orbital is also positive then the left half of the $\ce{s}$ orbital overlaps with the left lobe of the $\ce{p}$ orbital in a destructive manner and they cancel each other out, such that only a small portion of the initial $\ce{p}$ orbital lobe remains. But the right half of the  $\ce{s}$ orbital overlaps with the right lobe of the $\ce{p}$ orbital in a constructive manner and so the right lobe of the initial $\ce{p}$ orbital becomes larger.

'p' orbital has an 100% p-character , but 'sp3' has 75% p-character and 25% s-character in addition . Therefore, it is the presence of the s-character which gives one of the lobes a kind of spherical shape , thus, making it look comparitivly smaller as to the other lobe.

(P.S.- s-round shape ;  p-dumbbell shape)


How to RAISE the melting point of water? It is common that adding common salt into water will lower its melting point. But, is there any way to RAISE the melting point of water? Does it lower the boiling point of water as well? Why?

source

One way to raise the melting point of water is to increase pressure beyond about 635 MPa.  By raising pressure you could get the melting point to be even greater than the normal boiling point.

A second way is to lower the pressure, but this can only increase the melting point by 0.01 K.

A third way would be to added enough of a high melting point solute that you pass the eutectic point. In other words, generally when you add a solute to water, the melting point decreases, but only until you add enough solute that the eutectic point is reached, beyond which adding more solute increases the melting point.

source: http://www.phasediagram.dk/binary/CaCl2.PNG

For example, in the diagram, above, A is the eutectic point.  Beyond this point, adding more CaCl2 increases freezing point.  Then beyond 38% CaCl2, melting point exceeds ) degrees C.

Water is a liquid, thus already melted. Your question should either be about raising the melting point of ice or raising the boiling point of water.

To raise the melting point of ice, meaning trying to keep ice as a solid at a temperature as high as possible, reduce impurities.

To raise the boiling point of water, meaning trying to keep water liquid at a temperature as high as possible, increase the pressure, that is, do it in a pressure cooker instead of with a surface exposed to atmospheric pressure.


Sign of enthalpy for exothermic and endothermic reactions In Cambridge Chemistry Coursebook [1, p. 94] it’s written that

A rise in temperature is given a positive sign. So the value of $\Delta H$ is negative for an exothermic reaction. A fall in temperature is given a negative sign. So the value of $\Delta H$ is positive for an endothermic reaction.

Why is the sign of enthalpy for exothermic reaction negative? Doesn’t exothermic reaction rise the temperature of surrounding?
Why is the sign of enthalpy for endothermic reaction positive? Doesn’t endothermic reaction take in energy and cool down the surrounding?
Reference

Ryan, L.; Norris, R. Cambridge International AS and A Level Chemistry Coursebook, 2nd ed.; Cambridge International Examinations; Cambridge University Press: Cambridge, 2014. ISBN 978-1-107-63845-7.

Reason lies in definition of enthalpy of reaction. Enthalpy of reaction is heat exchanged between our system in which reaction happens and surroundings when reaction is carried at constant temperature and pressure. If reaction is exothermic, it releases heat and increases temperature of our system and so to keep it at the same temperature you need to give that heat to surroundings. If you do so, than since our system lost heat its enthalpy decreases (first law of thermodynamics) and because of that enthalpy of exothermic reaction is taken as negative. The opposite reasoning holds for endothermic reactions. You need to bring heat from surroundings into system to keep it at the same temperature and that heat increases enthalpy of the system.

The reaction is said to be exothermic or endothermic according to the sign of the Enthalpy change. To obtain Enthalpy you subtract the enthalpy of formation for reactants from that of products. So, if the final enthalpy of the system is less than the initial one the result is negative and this difference manifests it self in the system as heat and temperature increases. The opposite is true for endothermic reactions.


What is the driving force for imine formation? I know this that when a primary amine is reacted with a carbonyl compound, an imine is formed:

I'm wondering why the position of this equilibrium lies to the product side / right-hand side. Why is the formation of the C=N double bond favoured?

In practice, a drying agent is sometimes added to the reaction: I've used molecular sieves before, but a cursory database search brings up a few other options, e.g. magnesium sulfate, silica gel, or Montmorillonite (this list is probably not exhaustive). These remove water from the system and pull the equilibrium position over to the right-hand side.

That's one possibility, but I don't think that's all there is to it - I think a lot of imine formations proceed perfectly ok even without dehydrating agents being added to them. However, you'll have to hear about these from somebody else, as I don't know what would drive those reactions. It could well be that simply adding an excess of either amine or carbonyl drives the reaction to completion.

Interestingly, a 2009 paper by Saggiomo and Lüning1 describes their investigations into supposed imine formation in water. Apparently, both the carbonyl compound (benzaldehyde or salicylaldehyde) and amine (aniline) do not dissolve in water - not particularly surprising. The imine did not form appreciably in water and only started to form when water was removed during workup and purification.

P/S This is just the top hit I found on Google, it's hardly a complete literature search.

Saggiomo, V.; Lüning, U. On the formation of imines in water—a comparison. Tetrahedron Lett. 2009, 50 (32), 4663–4665. DOI: 10.1016/j.tetlet.2009.05.117.

See,We compare any reaction thermodanimaclly and kinetically
Let me put the thermodynamical aspect first which is that the formation of water is the very good sign of thermal stability for a reaction and a energy of 572 KJ/mol is realesed and hence helps to meet with the enrgy demands for formation of carbon and nitrogen bond
And suprisingly carbon and nitrogen bond is stable due to presence of lone pair on nitrogen and high density of electrons around it.
Now coming back to thermodynamics the carbon nitrogen energy is comparatively very high 614KJ/mol for this carbon and nitrogen bond. Also important thing note oxygen is more electronegative but yet as wr see oxygen is ready to leave with hydrogen to form carbon nitrogen bond double bond.

Now Kinetics
The product hence formed is by first order Kinetic nad hence also it is easy for hydrogen to migrate to alchol group and hence give water and with it leave it as to provide a negative charge to nitrogen and henceforth the water is left to make the way for carbon and nitrogen bond to form.Also notice  there will be a positive charge on oxygen so would leave easilyand hence there will a negative charge on "NZ" leaving formation of the double bond.
See this for data  https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Bond_Energies and
http://witcombe.sbc.edu/water/chemistryelectrolysis.html


Why are there peaks in electronegativities in d-block elements? Looking at the Pauling electronegativities in the Periodic Table (below, from ChemWiki):

Asides from the overall trend of increasing electronegativity across and up the Periodic Table (towards fluorine), there seems to be a minor peak in the first 3 periods of the $d$-block elements, this peak becomes more pronounced as you go down the $d$-block, with the most noticable peaks in Periods 5 and 6 - seemingly the reverse of the overall trend. This is shown in the graph below (Image source):

In the ChemWiki page, there is also a reference to the Allred-Rochow Electronegativity Scale, which also has peaks occurring most prominently in Periods 4 and 6:

Image source

The Wikipedia page Electronegativity details several methods of calculation, which brings into question how well-defined the electronegativity values are; however, there still seems to be the peaks occurring, particularly in Periods 5 and 6, and to a lesser extent in Period 4.

What is the reason for these electrongativity peaks in the $d$-block?

You should know that closeness to filling a shell and atomic size are the reasons for the overall electronegativity trend. This also applies to the subshells, how close is an element to filling, or half-filling, a subshell? How large is the atom?

The atomic size trend for the d subshell is opposite to that of the overall trend. In period 4 the 3d subshell is deep in the core orbitals and the electron density is high, whereas in the lower periods the d orbital is more diffuse and can more easily accommodate more electrons, leading to a better energy payoff for completing the subshell. For the elements that take s electrons to fill the d subshell, this energy payoff is even greater from then filling the s subshell as they are very diffuse.

The peaks in electronegativity for elements may be named as groups as next:

V Cr Mn has a peak and then Mn drops down. Iron is the key as shown next.
Zr Nb Mo has a peak and then a a plateau. Zirconium is the key as shown next.
My research is displayed :
Periodic Table of Elements : Shape of Each Nucleus affecting electronegativity

http://pyramidalcube.blogspot.com/2017/08/periodic-table-of-shapes-of-nuclei.html
The shape of the nucleus affects the properties of the element. Notice how
Manganese is the last in a series from the carbon core cube-2, then iron starts the cube-3 shape. That transition across the Table causes the abrupt change in electronegativity.
Alan Folmsbee


Is the Beckmann Rearrangement technically an "SN2" reaction at an "sp2" center? Below is the mechanism from this link of the Beckmann Rearrangement.

In the alkyl migration step, the alkyl group migrates simultaneously as the $\ce{H2O}$ leaving group is expelled. This seems as though it is an $\mathrm{S_N2}$ reaction at the $\mathrm{sp^2}$-hybridized nitrogen atom, but I thought that it was geometrically impossible to have an $\mathrm{S_N2}$ reaction at an $\mathrm{sp^2}$ center, per this link.

What am I missing in my reasoning?

When you were taught that we don't see $S_{\mathrm{N}}2$ reactions at $sp^{2}$ centers, what evidence or rationale was given? In organic chemistry, very few things are definite. When they are, it's usually because of being forbidden by quantum mechanics (e.g., symmetry in pericyclic reactions). Otherwise, they are just very unlikely, unless you somehow magically make the conditions right.

Greg Fu used to give this example: J. Am. Chem. Soc. 1998, 120, 2275-2282.

In it, a trans-substituted alkene undergoes nucleophilic substitution to become a cis-substituted alkene. Using that fact and other data, the authors conclude that the mechanism most consistent with the observed reaction data is a vinylic $S_{\mathrm{N}}2$. If you have a better mechanism, consider writing a paper and submitting it to JACS.


Is Hoffman elimination anti or syn elimination? I read an amazing PDF from nptel on elimination reactions...but I was really confused when it used used $\ce{^+NMe3}$ as a leaving group in the reaction pathway via:

$\mathrm{Ei}$ (syn-elimination)
$\mathrm{E2}$ (anti-elimination)
And also, in an illustration for $\mathrm{E1_{cb}}$.

My organic chemistry teacher also did syn-elimination but in the problem solving book and on various online sites it took the $\mathrm{E2}$ elimination pathway.

There is a range of elimination reactions with E1cb at one end, E1 at the other end and E2 in between. It is not uncommon for these different reaction pathways to compete with one another. For example, in some elimination reactions, the E1 and E2 pathways can operate in competition with one another. Activation energy is associated with each of these 3 reaction pathways. Whichever pathway has the lowest activation energy will be the major pathway followed. By changing solvent, reaction temperature, the relative strength of the nucleophile, the relative strength of the base, leaving group stability, etc., we can raise or lower the activation energy for each of these 3 pathways and shift a reaction towards one side of this mechanistic range or the other.

(Taken from Ron's answer, in order to start an explanation)(1)

From Wikipedia:

  The first step of an E1cB mechanism is the deprotonation of the of β-carbon, resulting in the formation of an anionic transition state, such as a carbanion. The greater the stability of this transition state, the more the mechanism will favour an E1cB mechanism.

So on further thinking about the Hofmann elimination, you notice the following

$\ce{NR_3^+}$ is a poor leaving group (Smith, March. Advanced Organic Chemistry 6th ed. (501-502))
Due to this, the rate of cleavage of the $\ce{C-H}$ bond cleavage exceeds the rate of the $\ce{C-N}$ bond cleavage
In fact, the three well known poor leaving groups namely $\ce{-F}$, $\ce{-SMe^{2+}$, $\ce{NR^{3+}$ all have the same observation

What's the conclusion?

Of course, these undergo $E1cb$ when heated to a suitable temperature

There are sources which infer the same -

Wikipedia
Organic mechanisms by Xiaoping Sun (3)

  Some examples of compounds that contain poor leaving groups and can undergo the E1cB mechanism are alcohols and fluoroalkanes.

  The $\ce{-NMe3}$ group is electronegative, it can activate beta hydrogen to make it acidic, vulnerable to attack by a base. This gives rise to carbanion  character (e1cb) for the transition state.

Hence, $E_1cb$ is the correct answer

Honestly, leaving groups usually do not determine whether it is syn or ante elimination. Hence, although the leaving group is an amine, it does not change how you view the problem. What determines syn or ante elimination is the type of reaction occurring, whether it is E1, E2 or E1cB.

E2 would always go through an anti elimination due to the requirement of the antiperiplanar transition state.

Whereas for E1 and E1cB, there is no fixed answer to whether it is syn or ante. The product that ultimately forms is dependent on the thermodynamic stability of the product.

Now, let us go back to your "E1" example. That, my friend, is not an E1 elimination! That is in fact an E2 elimination. Your E2 elimination has the requirement of having your hydrogen and leaving group be coplanar. Since the D and ammonium quaternary cation are coplanar, though it is fixed in a syn position, they are forced to go through a syn coplanar E2 elimination. The reason it is not E1 is that your leaving group is on a secondary carbon and therefore E1 is unfavourable.


How do I analyse this data? (optimization experiment) I conducted an experiment to find the optimal concentration for sugar in yeast respiration, and this is the data I found:

What do I do from here? Do I just fit a linear model on the increasing portion, another one on the decreasing portion, and see where they meet and say that's the optimal value? If not, do I try to fit another type of model? How do I justify that model?

Some materials (e.g. ionic salts) in certain solvents (e.g. water) actually do decrease volume of the solution in certain concentrations (obviously at some point it inevitably increases).

So you certainly can end up with non-linear functions of displacement from $\Delta_{conc}$.

That said, a negative quadratic may fit your data well, e.g. $ax^2 + bx + c$, with $a < 0$. Maybe even a damped exponential, e.g. $ae^{-bx}$. You can use Microsoft Excel etc. for a quadratic fit or fancy math skills for other functional forms.

Another Ben describes in more detail here but the image he used is worth repeating. Here the concentration of an added substance actually shrinks/increases volume displacement "oddly".

As you add sugar the displacement will only increase. In other words, if you add 300 grams of sugar to a liter of water, the volume will have increased (or been displaced) by about 150 ml (guesstimated)... the displacement is certainly not 25 millileters!

For what it's worth, you should have two (practically) straight lines. One relatively straight line before the saturation point (at around 200g sugar per liter at room temperature) and another relatively straight line after the saturation point. But believe me, the more sugar you add, the greater the volume will be (until the volumetric cylinder is full).

Displacement in a nutshell:


Cathode gained mass in zinc electrolysis I performed an electrolysis of zinc, using two zinc electrodes and a zinc nitrate solution. However, after leaving the electrolysis to run for 30 min, at about 0.4 A and 6 V, I discovered that the cathode gained more mass than the electrode lost. I know that ideally the anode oxidises zinc into zinc ions, and the cathode reduces zinc ions into zinc metal, suggesting that total mass should be conserved.

A key observation I made was that a white substance was observed on the anode, which probably led to the extra weight. Also, bubbles were observed on one of the electrodes. I am not sure which one though.

Edit: My chemistry teacher suggested this could be related to zinc oxide or zinc hydroxide, but I can’t find much information on this.

If the differences between the two electrode masses is small its probably due to excess electrolyte left on it or a measuring error. The white substance is  dried up zinc nitrate.

Bubbles should have been observed at both electrodes since that would have been the hydrogen, oxygen and nitrous oxide dissociating form the diluted solution of zinc nitrate.

ie: $$\ce{3e¯ + 4H+ + NO3- -> NO + 2H2O}$$ $$\ce{O2(g) + 4H+(aq) + 4e ->  2 H2O}$$

So I'm pretty sure the bubbles didn't have anything to do with the additional mass gain. $%edit$

The electrolysis of aqueous aluminum nitrate forms $\ce{N2O}$ and other products (including $\ce{OH-}$) along with some chemistry involving radical formation. I would expect similar with zinc nitrate. Note, $\ce{N2O}$ can form intermediate abducts with metal oxide (surface chemistry). Bottom line, be humble in the magnitude of complexity of this seemingly single experiment.

As background on a different forum (Sciencemadness.org) I shown how the net reaction ascribed to the reaction of $\ce{Cu(I)}$ aqueous, $\ce{H+}$ and $\ce{O2}$ could be express as the sum of several individual radical reactions. The associated quoted net cell reaction (which is actually a spontaneous electrochemical reaction as is the related better known reaction based on iron in place of copper) is also misleading, in my opinion, with even ascribed half cell reactions, which while depicting the electrochemical vantage and is certainly informative on energy measures, the mechanics like intermediate reactions and conditions (pH, temperature, ...) affecting the quoted net cell reaction are totally lost. Please do not associate such reductive simplicity with a true understanding of what is actually occurring, even though it is educationally convenient and you don't have to hire a true scientist to teach science.


Delocalised electrons and stability of benzene In most high school chemistry books, it is said that the delocalised electrons in benzene makes it particularly stable. However, how does it contribute to stability? Is there any simple way of understanding this?

In physics (or you can google it) you will learn about the particle in a box experiment.  At the atomic level, if you confine a particle to a box, it will have some energy.  It turns out, if you make the box longer, the energy of the particle will be lowered, if you make the box smaller, the particle will have a higher energy.  The same concept applies to benzene.  Basically you start with 3 separate acetylene molecules in which the pi electrons are confined to a 2 carbon segment.  Once you trimerize the acetylene and form benzene, the pi electrons can now roam over 6 carbon atoms, a much longer "box" if you will.  So the pi electrons in benzene will be stabilized (lower energy, longer box) compared to the pi electrons in the starting acetylene (higher energy, shorter box).  Molecular Orbital theory tells us the same thing in a slightly different manner.  In each acetylene (let's just focus on the 2 pi electrons in the acetylene that will be the pi electrons in benzene) we have 2 pi electrons in the HOMO and none in the LUMO.  As we bring these 3 acetylenes closer together to form benzene, the orbitals interact.  The 3 initially degenerate HOMOs in the 3 acetylene molecules split (they are no longer degenerate) as they move closer to becoming benzene and a net energetic stabilization results.

For high schoolers, this is how I explain it.  Remind them, that stability = less reactive or more set in place.... i.e. the electrons aren't going to go wondering off to another atom.

In our scenario... the carbons are moms/dads and the extra electrons are excitable little children who are prone to run off (think of little preschool brothers/sisters/cousins at a family reunion in a park).  Each individual mom/dad would be spending all their energy keeping their energized little kid from running off.  In the same way each carbon has trouble holding on to the extra energy electrons.... again, just as parents have a hard time holding their excitable children in place.

However, these carbons have linked up and are supporting each other and their extra energy babies (i.e. delocalized electrons) who have to move so they zip around in this ring of carbons.  Just like energized children zip around but stay in the safe part of the park when there are extra adults looking out for them and giving them physical safety boundaries.  By linking up, each of the carbons share the extra load of holding on to the extra electrons.  This stabilizes the whole group meaning the extra energy electrons won't go running off i.e. try to react with another molecule.  The electrons are happy and content staying within the ring of carbons.  AND (for my AP students, this keeps the carbons happy having the 4 bonds it needs without having to have a higher electronegativity number to keep them bonded; and how the stability of benzene makes it hydrophobic because it prevents the hydrogen bonding in water...)

I also remind them that if the double bonds in benzene were just double bonds, there would be a complete pi bond between two of the carbon atoms - i.e. these two carbons would own/hold the extra bonding energy and covalently share it.  But with a benzene ring, it is more like a co-op of carbons sharing the extra electron load and that is why we say that in reality it isn't 3 single and 3 double bonds in the ring, but more like (6) one and a half bonds.  Then we talk about bond lengths a little bit - comparing bond lengths of a single, double, and then a benzene ring bond.

AND (for my AP students, we talk about how conjugation stabilizes other organic molecules like antioxidants, etc.)


How to calculate the number of geometrical isomers in allenes? Can someone explain how to calculate the number of geometrical isomers in allenes like $\ce{CH3-(CH=CH)4-CH3}$ or $\ce{CH3-(CH=CH)5-CH3}$

If we have one double bond in a hydrocarbon compound we have an olefin or alkene.  Ethylene is the simplest example of this class of compounds.  The carbons in the double bond and the 4 atoms attached to them lie in the same plane.  One pair of cis-trans isomers is possible in compounds with a single double bond.

If we add a nother double bond directly on to the end of the double bond in ethylene such that an sp hybridized carbon is created in the process, then we have formed an allene.  The 4 substituents at the end of the double bonds in allene lie in planes that are oriented 90° to one another.  Look at the allene (bottom line) in the following drawing.  One set of substituents (R3, R4) are located in the plane of the screen; the other set of substituents (R1, R2) are located in a plane perpendicular to the screen.  Allenes cannot generate cis-trans isomers, but they can generate enantiomers.

Now let's add one more double bond to allene such that we now have 2 sp hybridized carbons (top line in the above drawing).  We see that, like in ethylene, the 4 atoms connected to the double bond lie in the same plane as the double bond and one pair of cis-trans isomers is possible.

Any compound with 3 or more cumulative double bonds is a member of the cumulene family. Any cumulene with an odd number of double bonds is geometrically structured like ethylene (the 4 atoms connected to the double bond lie in the same plane as the double bond) and is capable of having one pair of cis-trans isomers. Cumulenes with an even number of double bonds are structured like allene (the 4 atoms connected to the double bonds lie in perpendicular planes) and cannot display cis-trans isomerization, but can have enantiomers.

As a general rule, an even number of cumulative double bonds with different groups on terminal carbon atoms will be optically active and odd number of cumulative double bonds with two different groups on terminal carbon will be optically inactive.

The reason behind this is : For a even number of cumulative double bond compound(generally: Allene)  the plane of π bond of that Allene are perpendicular to each other. This geometry of this π bond causes the groups attached to terminal C atom to lie in perpendicular planes. Such Allene with different substitutents on end C are chiral  .Such Allene do  not show cis trans isomerism, but show optical activity.  (Due to the presence of stereo centres not chiral centres.)

Penta12diene do not show optical activity ,since terminal carbon has two same groups. (H atoms)


How to experimentally determine the elemental composition in fingernails? I am doing a project in my highschool about analysing C/N and C/S ratio of finger nails of various ages and sexes.
Please suggest me a way to do this. I don't know where to start, I know the various methods of estimating the amount of N or S present theoretically. Please provide info that how should I dissolve the nails etc.

I will be doing my experiment in a nice lab which has got all necessary equipment that I might need, so please answer without considering the limitations of my lab.

I got the idea by this website: https://pubmed.ncbi.nlm.nih.gov/18424895/

These analyses are difficult to carry out at a high school level. Nails are not soluble in any solvent. They must be burned and the gases coming out are analyzed and weighed. To burn them, they have to be mixed with a great excess of copper oxide, put in a long horizontal glass tube, heated from below. A current of oxygen gas should be sent through the glass tube, and the gas should be absorbed first by solid $\ce{CaCl2}$ which absorbs $\ce{H2O}$, then by $\ce{NaOH}$ which absorbs $\ce{CO2}$. By weighing the $\ce{CaCl2}$ and $\ce{NaOH}$ flasks before and after the operation, you can calculate the amount of $\ce{C}$ and of $\ce{H}$ contained in your original sample. To measure the amount of nitrogen of your nails, you have first to destroy them by sulphuric acid, then add them to $\ce{NaOH}$ and some Dewarda alloy (mixture $\ce{Al}$, $\ce{Zn}$) have it boiled to liberate $\ce{NH3}$ gas, and then absorbed this gas containing $\ce{NH3}$ into a solution of an acid like $\ce{HCl}$ or $\ce{H2SO4}$, and then titrate the remaining acid to deduce the percentage of $\ce{N}$ in your original sample. This can not be easily done in a high school lab, slowly.

Some suggestions:
First, collect the fingernail samples by applying a nail file to a subject's nail. This creates a fine powder.
Second, precisely weigh the sample and place the nail powder in distilled water.[EDIT] Here is a possible dissolution of nail path per a reference that should be verified in its accuracy, which claims:

COCO COLA IS CAPABLE OF DISSOLVING A NAIL
Water is great for your overall health including your nails, whereas fizzy drinks like Coke are not so good.  In fact Coco Cola is kind of scary.  If you place a piece of nail in a bottle of Coca Cola it will dissolve within 4 days, that’s how corrosiveness it is!

where the chemistry is likely conditional on the presence of the Aluminum (some electrochemistry at play). More interesting, do nails, in time, dissolve in an Aluminum can containing just Phosphoric acid, carbonated water, sugar and perhaps some NaCl, in a closed system (no gas escape)?
Third, I like a nascent hydrogen approach here to breakdown the nail powder (research the Marsh Test to detect small amounts of arsenic). So, further add Al/NaOH (or Zn/HCl) and collect the vapors in dilute H2O2 (which has been pH adjusted to a recorded value at over 7). [EDIT] This step may have to be re-evaluated based on revised Step 2, perhaps just neutralize with NaOH/H2O2.
Depending on the test equipment available, a simple comparative test would be to investigate the evidence/concentration of a visible sulfur suspension (a light-based test), as a pH check alone, for example, for any significant decrease in pH (a measure of increased acidity from sulfur-based acid formation) would not likely be sensitive enough for small differences in sulfur concentration.
One could refine the process starting with test samples containing but small differences in the amount of sulfur presence, as a feasibility check.
[EDIT] To be honest, I (and apparently at least one other) have doubts on the potential feasibility of this project. However, with the to-be-confirmed electrosynthesis dissolution of human nails, I would have to upgrade your chances especially based on your possible access to equipment to confirm relative elemental composition.


Examples of non-volatile, but flammable liquids? As far as I know, most flammable liquids are volatile. Are there flammable liquids that are non-volatile?

The element caesium melts at 28.4°C which is slightly above "room temperature" but below human body temperature, so I think we can consider it a liquid in "normal" conditions. It is so flammable it readily self-ignites in the air, but if you put it in inert atmosphere you will see that it isn't particularly volatile.

Cooking oil is flammable, but doesn't evaporate or evaporates very slowly over months or years. http://www.oliveoilsource.com/asktheexpert/does-cooking-oil-evaporate It's flammable because if you heat it up enough, it will start to burn. Once it starts to burn, the heat will increase causing it to burn faster.


Intermolecular and Intramolecular Forces (Vapor pressure of n-butylamine and methanol) What are the different properties that intermolecular forces and Intramolecular forces attribute to? i.e. intermolecular forces contribute to BPs, vapor pressures, etc.

Specific problem: why does n-butylamine have lower vapor pressure at room temperature than methanol? Explain in regards to IMFs.

What are intermolecular molecular forces?
The forces(electrostatic force) between molecules of a compound. These forces are responsible for intermolecular distances and how tightly the molecules are held together. If the intermolecular forces(van der Waals force) are large, it means the molecules are tightly held together in a group and they are fixed in their position.

Now what is boiling point? It's a temperature at which liquid state of a compound changes into its gas state. What's the difference between gaseous and liquid phase? Gaseous phase has very large intermolecular distance when compared to that in liquid phase. If a large force exists between the molecules and they are quite fixed in their position, then large amount energy or heat will be required to move them away from each other. Large heat means greater temperature and hence greater boiling point. So the lesson is that, the stronger the intermolecular forces, the higher the boiling point.
The similar concept in applicable in melting point. the stronger the intermolecular forces, the higher will be the melting point.

You will recall that the vapor pressure is a direct measure of escaping tendency, so we can use these terms more or less interchangeably(In case you may be wondering, how are they related. Click here). If the intermolecular forces are large, then molecule in liquid phase are not able to leave to liquid easily, hence lowering the escaping tendency and therefore lower vapour pressure.

Note: This is a very basic explanation.

Butylamine does not exhibit H-bonding between each other, where as alcohols do. This added force from the hydrogen bond makes the boiling point of alcohols higher than those of most other ORGANIC molecules. Ionic bonds have the most strength.


Volume of composition of air My teacher stated:

  At standard conditions, 1 liter of air at 21% oxygen possesses $\pu{0.21 L}$ of oxygen. Since at STP, 1 mole of gas occupies $\pu{22.4 L}$, simply divide $0.21/22.4$, to arrive at 0.0094 moles of oxygen.

I have a question regarding this. What if I trap $\pu{100 ml}$ of air in a vessel and maintain $\pu{273 K}$ and pressure $\pu{1 atm}$? So, in that case, what does it mean that oxygen has volume of $\pu{21 ml}$? Won’t the volume of oxygen and nitrogen be same inside vessel? Or am I wrong somehow?

If you took the 100 mL sample of air, and took out everything but the oxygen, then the amount of oxygen trapped would fill 21 mL at STP.

One way to think of it is, even though the oxygen would expand to fill the container, its pressure would need to reduce according to Boyle's law: P1V1 = P2V2. This is where the concept of Partial Pressure comes from. But if you then compressed that oxygen so that P = 1 atm (since you specified initial pressure to be 1 atm), the oxygen would take up 21 mL of volume.

Air is 21% oxygen by moles, not by volume. Remember that gases occupy the whole of the container in which they are kept and not just a fraction of its volume. This is because the gas molecules are constantly moving about. Even if you tried to confine oxygen (along with air) in 21% volume of a container, the oxygen molecules would move about randomly and eventually occupy the whole container. The other gases can't stop this.

Clarification: Air is actually composed of approx. 21% dioxygen by moles, and not by mass or volume. A simple calculation of the molar mass of air assuming this composition and comparing it with the actual data will verify this. Sorry for the confusion.


How to calculate crystal field stabilisation energy? How to calculate it? I know that it depends on the coordination compound and the number of electrons present in t2g and eg orbitals and the final answer is expressed in terms of crystal field splitting parameter. But, how to know how many electrons are there in t2g and eg orbitals?

I think rather than a broad answer and example would help. So, consider the compound $\ce{[Ni(CO)4]}$ and how to calculate CFSE for it?

Step 1: Look up Nickel Carbonyl and find out what geometry it has.

We need the geometry to know how the $d$ orbitals will split in the ligand field. The geometry can also be predicted: late transition metals with strong field ligands tend to be tetrahedral.

  Step 2: Find the appropriate crystal field splitting diagram for this geometry.

  Step 3: Figure out how many $d$ electrons there are.

Nickel is in Group 10, with a configuration of $\ce{[Ar]}4s^2 3d^8$ or $\ce{[Ar]}4s^1 3d^9$. In coordination compounds we will consider this configuration to become $\ce{[Ar]}4s^2 3d^8 \implies \ce{[Ar]}3d^{10}$. The $s$ electrons are considered to move to the $d$ sub-shell in bonding - bonding need not represent ground state atomic electron configurations just group state molecular electron configurations. For more information on $d$ electron count, go to this Wikipedia page.

  Step 4: Fill in the $d$ electrons.

You can do this part yourself. How do you put 10 $d$ electrons in the orbital diagram?

  Step 5: Determine $\Delta E$.

$\Delta$ is the energy different between the $e_g$ and $t_{2g}$ sets of orbitals. This Wikipedia page walks through an octahedral complex. In tetrahedral complexes, the energy of the $e_g$ orbitals is lower and the energy of the $t_{2g}$ orbitals is higher. The energies are:

$e_g \ (d_{x^2-y^2},\ d_{z^2}) \ E=-\frac{3}{5}\Delta_\textrm{tet}$ (stabilized)
$t_{2g} \ (d_{xy},\ d_{xz},\ d_{yz}) \ E=\frac{2}{5}\Delta_\textrm{tet}$ (destabilized)

Then multiply the numbers of electrons in the orbitals by the stabilization/destabilization values, and sum.

$$\Delta E = 4\times\left(-\frac{3}{5}\Delta_\textrm{tet}\right)+6\times\left(\frac{2}{5}\Delta_\textrm{tet}\right)=?$$

The process is quite simple. .. to determine CFSE  we need to know that the diffidence of energy levels of two sets have been arbiter taken as 10 dq. . Where t2g set is more stable then eg set octahedral and vice versa in tetrahedral complex.
-0.4x + 0.6y will help to calculate it.
X indicates electrons in t2g and y in eg.
Sum up all and find the diffidence ...
That gives cfsc
Usually tetrahedral splitting is shown in terms of octahedral as
4/9 OS


Why are peroxides unstable but disulfide bridges considered stable? Why are esters stable but thiolesters are unstable? I can not understand why a peroxide $\ce{R-O-O-R}$ is considered reactive and unstable.

Going down one row on the periodic table, a disulfide bridge ($\ce{R-S-S-R}$) is apparently super stable and super important to proteins 3d structure.

At the same time, a thioester is considered to be as unstable/reactive as ATP and the ester is the stable one?

What am I conceptually missing here to explain this contradiction?

Some hard data: bond enthalpies (in $\pu{kJ mol-1}$)

$$\begin{array}{c|c|c|c} \text{Bond} & \text{Enthalpy} &
     \text{Bond} & \text{Enthalpy}  \\ \hline
\ce{C-C} & 350 & \ce{Si-Si} & 226\\
\ce{N-N} & 163 & \ce{P-P} & 201\\
\ce{O-O} & 146 & \ce{S-S} & 226\\
\ce{F-F} & 155 & \ce{Cl-Cl} & 240 \\
\end{array}$$

We observe a decrease in bond energy from carbon to nitrogen and from silicon to phosphorus. It is clear and probably have same reason: appearance of lone electron pair of the atom. There is a repulsion between lone pairs, that is in partially compensated by stronger bonds formed by smaller atoms (the atomic radii falls down to the end of the row). When moving from second to third row, this repulsion is reduced thanks to larger size of atoms of the elements of the third row.

The general decrease in energy from carbon to silicon is thanks to increased size of the orbitals of valence level, making them more diffuse with less efficient overlap and longer bond length.

The stability of esters vs thioesters is ruled by different reasons. Sulfur is not as electronegative as oxygen, but when you consider an anion, sulfur anions have larger radius and are less prone to grabbing first positive charge they find. So sulfur anions are more stable intermediates (kinetically), and thus polar dissociation of thioester bonds is easier.

It has to do with electronegativity (EN). Remember the concept of "formal charge"?
In a bond between molecule 1 and 2, if molecule 1 is more electronegative than molecule 2, then the electrons will be drawn closer to molecule one.

That is general, but to get to your specific question, Oxygen's EN is 3.5. Sulfur's EN is aprox 2.5. Carbon's EN is also 2.5.

So in a Carbon oxygen bond, the electrons will shift to oxygen.
In a carbon-sulfur bond, the electrons will be relatively in the middle, thus one can expect a carbon-sulfur, and sulfur-sulfur bonds to be relatively equal in strenght.

On the other hand, the peroxides (when bonded to carbons) will always pull the electrons towards them more than the carbons wil, and at some point, the bond will eventually snap.


pH of an acetate buffer after dilution 100 mL of a 0.1 mM buffer solution made from acetic acid and sodium acetate with pH 5.0 is diluted to 1 L. What is the pH of the diluted solution?

The pKa value for acetic acid is given as 4.75.

The answer provided in my textbook is pH = 5.53. The closest that I can get to that is pH = 5.507. I know that's an almost insignificant difference but I'm just confused how they got to their answer, so if anyone can give me some suggestions that would be awesome. I'll outline my approach below:

Using the Henderson-Hasselbalch equation to find the ratio of [A-] to [HA] in the 100ml:
$$10^{5-4.75} = \frac{[A^-]}{[HA]}$$

$$[A^-] = \left(\frac{10^{0.25}}{10^{0.25}+1}\right)\times\left(\frac{0.1mM}{1000}\right) = 6.4{\times}10^{-5}M$$

$$[HA] = \left(\frac{1}{10^{0.25}+1}\right)\times\left(\frac{0.1mM}{1000}\right) = 3.6{\times}10^{-5}M$$

The addition of 900ml of water to 100ml of buffer is a 10x dilution:

$$[A^-] = 6.4{\times}10^{-5} \times 0.1 = 6.4{\times}10^{-6}M$$

$$[HA] = 3.6{\times}10^{-5} \times 0.1 = 3.6{\times}10^{-6}M$$

$$\ce{[H+]} = 10^{-5} \times 0.1 = 10^{-6}M$$

ICE table to determine equilibrium concentrations after dilution:

$$
\begin{array}{c|c@{hello}c@{}c@{}c@{}c@{}c@{}c}
   &   [CH_3 COOH] & + &[H_2 O] & {}\rightleftharpoons{} & [H_3 O^+] & + &[CH_3 COO^-] \\
\hline
I & 3.6 \times 10^{-6} && - && 10^{-6} && 6.4 \times 10^{-6} \\
C & -x && - && +x && +x \\
E & 3.6 \times 10^{-6} -x && - && 10^{-6} +x && 6.4 \times 10^{-6} +x \\
\hline
\end{array}
$$

$$K_a = \frac{\ce{[H3O+][CH3COO-]}}{\ce{[CH3COOH]}}$$
Then substituting in Ka and the equilibrium values from the ICE table: $$(10^{-4.75}) = \frac{(10^{-6}+x)(6.4 \times 10^{-6} + x)}{(3.6 \times 10^{-6} -x)}$$

$$x^2 + 2.518 \times 10^{-5}x -5.762 \times 10^{-11} = 0$$

$$x = 2.111 \times 10^{-6}$$ $$ [H^+] = 10^{-6} + 2.111 \times 10^{-6} = 3.111 \times 10^{-6}$$

$$ pH = -log_{10}(3.111 \times 10^{-6}) = 5.507 $$

Depending on your first calculations before dilution:
\begin{align}
[\ce{CH3COO-}]&=[\ce{CH3COONa}]=\pu{6.4\times{10^{-5} M}}=[\ce{Na+}]\\
[\ce{Na+}]_\text{(After dilution)}&=\pu{6.4\times{10^{-6} M}}\\
[\ce{CH3COO-}]+[\ce{CH3COOH}]&=\pu{1\times{10^{-4} M}}~\text{ (Before dilution)}\\
[\ce{CH3COO-}]+[\ce{CH3COOH}]&=\pu{1\times{10^{-5} M}}~\text {(After dilution)}
\end{align}
So,you have the the following $5$ independent equations in order to specify the unknown concentrations of $5$ species as present in an aqueous solution consisting of $\pu{1\times{10^{-5} M}}$  buffer solution made from acetic acid and sodium acetate :

\begin{align}
6.4 \times 10^{−6} &= [\ce{Na+}] \tag{1}\\
K_\mathrm{a} &= \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}
     = 1 \times 10^{−4.75} \tag{2}\\
K_\mathrm{w}
    &= [\ce{H+}][\ce{OH-}]
     = 1 \times 10^{−14} \tag{3}\\
1\times{10^{-5}}
    &=[\ce{CH3COOH}] + [\ce{CH3COO-}]\tag{4}\\
[\ce{H+}]+[\ce{Na+}]
    &= [\ce{OH-}] + [\ce{CH3COO-}] \tag{5}
\end{align}


Why can't alcohols turn blue litmus red? I came across the fact that alcohols cannot turn blue litmus red. The fact confuses me as they are acidic and acids do turn blue litmus solution red. What prevents alcohols from turning blue litmus solution red?

Litmus is a weak acid, with a $\mathrm{p}K_\mathrm{a}$ of about 6.5.

Alcohols are extremely weak acids, with $\mathrm{p}K_\mathrm{a}$'s (typically) around 17.

The litmus itself is a much stronger acid than the alcohol. Dissociation of the alcohol won't produce enough protons to shift the equilibrium between the two colored forms of the indicator, so you'll see no color change.

The pH of ethanol ($\ce{C2H5OH}$) or pure alcohol is $7.33$. This means that ethanol is slightly basic. And the colour of litmus paper change at $\mathrm{pH} =7$.
Consider the ionization of indicator  in solution represented by this equation:

$$\ce{HIn  <-->  H+    +   In-}$$

And the acidic  colour  of  $\ce{HIn}$ is red, the basic colour of $\ce{In-}$ is blue.
According to Le Chatelier’s principle, if we put indicator in basic solution, $\ce{H+}$ reacts with base, more $\ce{HIn}$ ionized, more $\ce{In-}$ formed, so the colour of blue litmus unchanged.


Why is the atomic radius of krypton smaller than that of potassium? A potassium atom comprises 19 protons and 18 core electrons. Those electrons screen 18 protons worth of Coulomb force. Therefore, there is one proton worth of Coulomb force attracting the valence electron of potassium. Whereas, krypton comprises 36 protons and 28 core electrons. Its electrons screen 28 protons wroth of Coulomb force. Therefore, there are eight protons worth of Coulomb force attracting the valence electrons of krypton.

Consequently, there is more Coulomb force attracting the valence electrons of krypton than there is attracting the valence electron of potassium; accordingly, the atomic radius of krypton is smaller than the atomic radius of potassium.

Is that correct?

Your concept is close to the real answer. You should think it as no. of protons per electron over all. This is the way things have been done in literature and if you are not satisfied then consider the fact that not all electrons are shielding the nucleus to same extent. There are certain rules [slaters rule] made and which works in most cases by which you can actually predict how much of nuclear charge is acting on the electron. Even with that you will see valence electron for K is more shielded than for Kr. But the numbers you came up with don't make that much sense. It does a bit but better if you go by the way it has been discussed.

Due to the poor shielding effect, of "d" orbital of" Kr" so the "z"effect is increase and radius is decrease but in "K" no poor shielding effect so it's radius is greater than "Kr"


Change in pH when boiling off acidic water Assume 1 L of deionized water at pH 2 at ambient pressure open to atmosphere. I boil off 500 mL. I believe the pH will not change because to the same extend as water boils off it will also boil off H+ ions. Correct?

$\rm H^+$ ions (as well as any other ions, for that matter) do not boil off. Acid does. Depending on the nature of the acid, it might boil off faster or slower than water, and hence your pH might shift up or down.

$\ce{HCl}$, in particular, forms a high-boiling azeotrope with water at 22%, so the solution of pH 2 (which is pretty dilute) would form a vapor enriched with water, and the liquid phase would get enriched with acid. So despite the seemingly volatile nature of $\ce{HCl}$, it will mostly stay in the solution and make it progressively more acidic as it boils.

Chloride is well known (by chemists) as a volatile element. Chloride salts are easily lost upon heating. At STP, HCl is a gas. Consider these factors and you should not be surprised to learn that in a solution of H2O and HCl, both will be lost upon heating. Since HCl is a strong acid, a pH of 2 indicates that there is 0.01 M/L of HCl present. At 100°C, the solubility of HCl in water is about half what it is at 0°C, but it is still much higher than 0.01 moles/L. I would expect just a little HCl loss of chloride during boiling. Therefore a reasonable first approximation would be to assume no loss of HCl, and hence the concentration will go from 0.01 to ~0.02 moles/L. As a second approximation, you could determine the vapor pressure of HCl at STP and at 100°C, and somehow account for the difference. Since vapor pressure assumes equilibrium, rather than a dynamic process, this would only indicate the approximate magnitude of the loss. In other words, since the distillation process would depend on the heating rates, the surface area of the liquid, the air pressure, the air flow speed and volume, there's no accurate way -other than actual measurement - to calculate the loss. I suspect it would be small, but I'd expect the distilled water to have a pH slightly lower than predicted for pure deionized & distilled water (assuming it's saturated with the other gasses found in the air stream).    The idea that H+ will vaporize off along with the H2O is wrong. You must always assume charge balance. Which means that both the H+ and the Cl- would have to be volatile and exactly as volatile as the water. Like balancing a pencil on its point, that would be possible, but highly unlikely.


Isn't the mechanism of aqueous sodium hydroxide and 1-bromobutane involves both heterolytic bond fission and attacking by a nucleophile? What is involved in the mechanism of the reaction between aqueous sodium hydroxide and 1-bromobutane?

A. attack by a nucleophile on a carbon atom with a partial positive charge

B. heterolytic bond fission and attack by a nucleophile on a carbocation

C. homolytic bond fission and attack by an electrophile on a carbanion

D. homolytic bond fission and attack by a nucleophile on a carbocation

I think that A and B would be correct answers, but the correct answer is only A. Bromine accepts both two electrons from the carbon atom and the nucleophile attacks on that carbocation of the carbon atom. Why choice (B) is incorrect? I also searched the internet and found that this reaction is SN1, but why it occurs with the primary carbon, which is usually a reactant in SN2 reaction.

Well your thinking is absolutely right but only for tertiary halogenoalkane as this is a prinary halogenoalkane it will follow SN2 mechamism. In which necleophile attacks on a partial positive carbon


Calculating pH of reaction mixture of silver nitrate and ascorbic acid I am designing an experiment but before I do, I wanted to understand if the following method is right to calculate the expected $\mathrm{pH}$ when I mix $\pu{0.736 M}$ $\ce{AgNO3}$ and $\ce{C6H8O6}$.

I want the $\ce{AgNO3}$ to be the limiting reactant for this reaction. I think chemical equation will be:
$$\ce{2AgNO3 + C6H8O6 <=> 2Ag (s) + C6H6O6 + 2HNO3},$$

which would mean that there would be 2 hydrogen ions.

$$\ce{2Ag+ + C6H8O6 <=> 2Ag (s) + C6H6O6 + 2H+}$$

If I start with $\pu{0.736M}$ $\ce{AgNO3}$ and I plan to have that in $\pu{1L}$ water solution (also intend the $\ce{C6H8O6}$ to be in $\pu{1L}$ water solution), then the moles of hydrogen generated will be:
$$\pu{0.736 mol} \ \ce{H+}$$

This would mean the concentration of $\ce{H+}$ will be
$$\frac{\pu{0.736 mol}}{\pu{2L}} = \pu{0.368M}$$

So the $\mathrm{pH}$ of the solution would be $-\log 0.368  = 0.434$? This number does not seem right since ascorbic is a weak acid, is my logic correct?

For solutions of weak acids and with simplifying assumptions [H+]=[A-], [A-] << [HA ], there is formula:

$$\mathrm{pH}=\frac12(\mathrm{p}K_\mathrm{a} - \log {c})\tag{1}$$

For our case, if the stoichiometric ascorbic acid concentration $0.368/2=\pu{0.184 mol/L}$ is used:

$$\mathrm{pH}=\frac12(4.1 - \log {0.184}) \simeq 2.42\tag{2}$$

If reaction is completed, $\mathrm{pH}$ would be given by $\ce{HNO3}$ only.

Situation would get complicated, if nitric acid starts to react with metallic silver or residual ascorbic acid. These side reactions would decrease mineral and eventually also organic acidity.

$$\ce{3 Ag + 4 HNO3 -> 3 AgNO3 + 2 H2O + NO}$$
$$\ce{3 C6H8O6 + 2 HNO3 -> 3 C6H6O6 + 4 H2O + 2 NO}$$

If more of the ascorbic acid is used, initial $\mathrm{pH}$ is again given by the above equation.

The ending $\mathrm{pH}$ is given by equation:

$$[\ce{H+}]= c_1 + [\ce{A-}]\simeq K_\mathrm{a}\frac{c_2}{[\ce{A-}]}\tag{3}$$

where $c_1$ is the nitric acid concentration, $c_2$ is the residual ascorbic acid concentration. and the right equation side is taken from $(1)$

Concentration of $\ce{H+}$ is equal to sum of $\ce{HNO3}$ concentration $c_1$ + $\ce{H+} from weak acid, that is equal to A- concentration ( HA -> H+ + A-). The latter is then approx equal to simplified extression derived from the acid dissociation constant.

$$c_1 \cdot  [\ce{A-}]  + {[\ce{A-}]}^2 - K_\mathrm{a}\cdot c_2  \simeq 0 \tag{4} $$

$$[\ce{A-}]=\frac{-c_1 + \sqrt{{c_1}^2+4\cdot c_1\cdot c_2 \cdot K_\mathrm{a}}}{2} \tag{5}   $$

From (3):

$$[\ce{H+}]=\frac{c_1 + \sqrt{{c_1}^2+4\cdot c_1\cdot c_2 \cdot K_\mathrm{a}}}{2} \tag{6}  $$

$$[\ce{H+}]=c_1 \cdot \frac{1 + \sqrt{1+ \frac{4\cdot c_2 \cdot K_\mathrm{a}}{c_1}}}{2} \tag{7}  $$

In the case by we use acid concentration twice as much as stoichiometric one, then $c_1 = 2c_2$. As $K_\mathrm{a}=\pu{7.94e-5}$:

$$[\ce{H+}]=c_1 \cdot \frac{1 + \sqrt{1+ 2 \cdot K_\mathrm{a}}}{2} \simeq c_1 \cdot  1.00004 \tag{8}  $$

You can see the dissociation of the weak acid is almost suppressed by $\ce{H+}$ from nitric acid, so we can neglect the weak acid contribution to $\mathrm{pH}$.


What is "gold number"? I was reading about gold number and a book defined gold number as following:

  It is the minimum mass in milligrams of a stabilizing agent which is added to the 10 mL of red gold sol to protect it against coagulation caused by 1 mL of 10% (by mass) NaCl.

Is this the definition? Because I am not able to understand above definition.
Is NaCl used as the standard?

A red gold sol is a colloidal suspension of gold nanoparticles with average size of particles less than 100 nm. The suspension is quite stable — the nanoparticles don't aggregate because of the existence of an electrical double layer on the surface of particles which causes electrostatic repulsion between the particles.

If you add a sufficient amount of an electrolyte (e.g., NaCl) to the colloid, the ions will disrupt the electric double layer which will cause aggregation (coagulation) of gold particles. A stabilizing agent (e.g., citric acid) is a compound that is able to adsorb on the surface of gold nanoparticles and protect them from aggregation (it acts like a shield), even in a high electrolyte concentration.

Practically it means:

Dissolve different amounts of your reagent in 10 mL water. Make a solution of colloidal gold. Specifically this colloidal should be "red gold" (there are different ways to prepare colloidal gold, but "red gold" is a common standard. Add 1 mL of solution of NaCl. It will cause precipitation in a sample with no stabilizing reagent, but if the amount of stabilizing reagent is sufficient then the gold will remain in "red gold" colloidal state. The amount of required stabilizing agent (in grams) is the gold number of this reagent.

The reason they do it is because it provides an standard, easy and reproducible way to compare different stabilizing agents. Part about 10 (by mass) is unclear.