Reaction mechanism (SN2 and SN1 reactions) How to determine whether a reaction is an SN1 or SN2 reaction?
I am unable to determine whether a reaction would form a transition state or not,
I referred the NCERT textbook but didn't found anything that could clear my doubt.

SN1 is characterized by the possibility of formation of a stable carbocation during the reaction, whereas in the case of SN2 there cannot be formed a stable carbocation, so the most stable "intermediate" structure would be a transitional state. Stable carbocations are formed via groups with strong inductive electronic effects (+I those that repell electrons) or they can be stabilized by conjugation (the case for this one is when you have a double, triple bond or an aromatic ring directly bound to the carbon atom where the substitution takes place)
so, for example, (CH3)3C-Cl would make for an SN1 when reactiong with alcohols. Also compounds like CH2=CH-CH2-Cl.
The remaining cases, for example CH3-Cl or CH3-CH2-Cl would work via SN2, because the carbocation cannot be stabilized.
Hope that's gonna help you :)


Condenser shooting out hexane in a soxhlet extraction Im running a soxhlet extraction using hexane as the solvent. Once the hexane has filled up the soxhlet, a considerable amount of hexane will come out on top of the condenser. Is this an issue with pressure or defects in the glasses?

Heating without boiling chips or stirring will cause vapor to burst. Your condenser is not the efficient type either. Suggestion: add boiling chips (after it is cooled down); change a more efficient condenser; heat up slowly.


Why a higher s character increases a carbon atom's electronegativity? To make us understand certain phenomenon like the higher acidity of alkynes than alkanes and alkenes, our teacher told us to learn this as a general rule:

  Higher s character leads to a more electronegative carbon atom.

Note: By s character I mean the percentage of s in the hybridisation of the carbon atom(eg $25\%$ s character in $\ce{sp^3}$ hybrid carbon)

Though this well explains the above mentioned phenomenon (and certain others like the decreased $\ce{C-Cl}$ bond length in chlorobenzene), but I was unable to figure out  why this thing(increased electronegativity) happens after all.

This did not provide me a satisfactory answer. Electronegativity is after all the ability of an atom to attract the shared pair of electrons towards itself. So, how would one explain this using the concept of orbitals or molecular orbital theory?

I would also like to know if the same argument is valid for other p-block elements?

Electronegativity is the power of an atom to attract bonding pairs of electrons to itself. It clearly depends on the nuclear charge: the larger it is, the more strongly the nucleus attracts electrons and the closer to the nucleus they stay, so the larger is the electronegativity. Besides, electronegativity also depends on the number of other electrons present in the atomic shells ahead of valence electrons participating in chemical bonding. These electrons shield the valence electrons from the positively charged nucleus decreasing its effective charge and consequently lowering the electronegativity of the atom.

Now, back to the question. $\mathrm{2s}$ orbitals are lower in energy than $\mathrm{2p}$ ones, so that electrons occupying the former are (at average) closer to the nucleus than that occupying the later. Thus, the more $\mathrm{s}$ character a hybrid orbital has, the lower its energy is and the closer (at average) electrons occupying it are to the nucleus. The lower the energy of a hybrid orbital of an $\ce{X}$ atom is, the lower is the energy of the molecular orbital in formation of which it participates, and consequently, the closer (at average) the bonding pair of electrons stay to that $\ce{X}$ atom.

So, with respect to a bonding pair of electrons that occupy a molecular orbital to which a hybrid orbital of an $\ce{X}$ atom contributes, the $\ce{X}$ atom indeed appears to be more electronegative as the $\mathrm{s}$ character increases. Specifically for $\ce{C}$ atoms, the larger the $\mathrm{s}$ character of a hybrid orbital is, the lower in energy is a bonding $\sigma_{\ce{C-C}}$ orbital formed from the hybrid ones: $E_{\sigma_\mathrm{sp}} < E_{\sigma_\mathrm{sp^2}} < E_{\sigma_\mathrm{sp^3}}$. So that electrons occupying the $\sigma_{\ce{C-C}}$ orbital stay closer to $\ce{C}$ atoms (at average) as the $\mathrm{s}$ character of hybrid orbitals increases.

With respect to the question OP asked in comments,

  But if we compare between $\ce{C}$ and $\ce{H}$, is there some other factor to be accounted for because clearly H atom due to its very lower potential energy electrons should attract the shared electrons more.

the factor OP is looking for is the nuclear charge, which is 6(!) times larger for $\ce{C}$ atom comparing to the $\ce{H}$ one.


Are there softeners in synthetic materials? Are there softeners used in polyester or polypropylene to make these materials more flexible? Or is this not needed for this kind of synthesis?

   Yes. All plastics have plasticizers in them. The particular plasticizers chosen depends on what properties are desired of the plastic. Some can make the plastic much stronger but also more brittle at the same time, so that it would be much more difficult to break but when/if it did snap, it would tend to shatter or snap violently. Others make the plastic softer and less brittle, so that the plastic form has less structural strength, but will bend, stretch and disform under heavy load instead of snapping.

   Plasticizers can also affect the  temperature resistance to deformation, volatility (temperature resistance to deflagragation, and properties of said polymer if ignited, such as, will it go out on its own or will it continue to burn the whole thing). Plasticizers can be added to affect the polymer's UV resistance, resistance to chemicals or resistance to abrasion.

   Once again, all plastics have plasticizers in them. If one were to magically remove all of the plasticizers from a plastic bottle (PETE or PE), the plastic would become brittle, dry, and probably reduce to tiny pieces (think shattered tempered glass, but smaller) or a powder upon touching the plastic form which no longer contained any plasticizers. In fact, you can observe this with some old plastics from the 1970s. Even though the plastic may have never been exposed to UV light, a plastic that is losing its plasticizer can take on the same cracked, opaque appearance. Most plasticizers are volatile at room temperature and will slowly evaporate away just by simply being exposed to air.

   Most plastics will leach their plasticizers more rapidly after being washed with soap and water, after being microwaved, while being heated, or while at a low temperature (such as in the freezer). Even still, some plasticizers will take 20 years or more to evaporate away enough to change the properties. Some plastics will never lose their plasticizer (that we know of).

According to Wikipedia:
Some of the most common plasticizers and their uses are:

Phthalate-based plasticizers - used in situations where good resistance to water and oils is required. Examples include Bis(2-ethylhexyl) phthalate (DEHP) and Butyl benzyl phthalate (BBzP).
Adipate-based plasticizers - used for low-temperature performance or resistance to ultraviolet light. Some examples are Bis(2-ethylhexyl)adipate (DEHA) and Dimethyl adipate (DMAD)
Trimellitates - used in automobile interiors and other applications where low volatility and resistance to high temperature is desired.
Other common plasticizers are Polybutene, Benzoates, Sulfonamides, Glycols/polyethers and poxidized vegetable oils


Phenyl group in proton NMR Suppose you have this spectrum and the molecular formula of $\ce{C10H12O}$

The structure is

My problem is with the peak at ~7 which is an H-Benzene signal, with quartet splitting and a relative area of 4.

Surely this means that:

The benzene would have only one Hydrogen environment
with 4 Hydrogens in that environment
and 3 hydrogens on the adjacent carbon atoms

But the answer has 2 H-Benzene environments.

How come this goes against normal? Please can answers be not too technical as I am a pre-university student

You asked to keep it simple, which is why I’m skipping explaining some details. Check the bottom of the answer to see the parts with superscript numbers explained.

The signal at $7\,\mathrm{ppm}$ is not a quartet. It would only be a (typical; there are exceptions, however not in typical organic chemistry) quartet if both of the following two conditions are met:

The signal ratio is $1\,:\,3\,:\,3\,:\,1$ and
The distance between all four signals is equal (they have the same coupling constant[1])

Both reasons stem from the way quartets are created due to coupling.

In your case, the signal is approximately $3.5\,:\,6.5\,:\,6.5\,:\,3.5$ (violating the first condition) and the distances are (enlarging the image on my screen and measuring with a ruler, so only a very rough approximation) $\mathrm{7\,mm, 12\,mm, 7\,mm}$ (violating the second condition).

What you are in fact observing is a pair of doublets. (It could be a doublet of doublets[2] as suggested by bon, but it is not.) You can think of it as two signals, each being a doublet, that are very close together. And that fits perfectly with the theory: Each signal would have an integral of $2$ and would correspond to one set of aromatic protons:

The two on the aldehyde’s side at lower field (slightly higher shift)
The two on the ethyl group’s side at higher field (slightly lower shift).

They are doublets because each one couples with one hydrogen of the other group. All that is left to explain is why one of the two signals of the doublet is higher than the other, creating a quartet-like effect. This is due to an effect called roofing or tenting. The bottom line about roofing is that it occurs if two coupling protons have close chemical shifts, and that once you have identified its existence, you can basically ignore it (it doesn’t give you any information that you couldn’t have derived otherwise).[3] You can observe the same effect in the triplet at $\mathrm{1\,ppm}$ and the quartet at $\mathrm{2.5\,ppm}$: The right-hand side signals of the quartet are slightly stronger than those on the left-hand side and vice-versa for the triplet.

[1] As you know, coupling stems from two protons being close, and connected via a chain of bonds — usually three for hydrogens. The reason that nuclei couple is because they have a spin. For hydrogen (and carbon), the spin can be ‘up’ or ‘down’. Consider two protons next to each other: Their spins can point in the same or in the opposite direction. Both cases have a slight difference in energy which gives rise to slightly different chemical shifts — the coupling you observe.

For reasons that I won’t go into detail into here, this difference does not depend on the strength of your NMR magnet (while the difference between the signals of different protons does). The ppm-scale has been chosen to remove the differences between different proton’s signals (so $2\,\mathrm{ppm}$ will always be $2\,\mathrm{ppm}$) and therefore the coupling splits will vary with the strength of the NMR magnet, which is why they aren’t given in ppm, but rather in Hz.

If you have multiple protons of the same type, the coupling value (one says: coupling constant) will always be the same, so $7\,\mathrm{Hz}$ three times in the event of a true quartet. The reason why the signals then need to be $1\,:\,3\,:\,3\,:\,1$ lies within Pascal’s triangle: Starting from a central signal split it into two halfes with a certain distance from the centre: You will get two $1\,:\,1$ signals. Do the same thing with each signal and the same distance: You will get a $1\,:\,2\,:\,1$ signal, because the two central ones will overlap. Do the same again: Voilà, $1\,:\,3\,:\,3\,:\,1$. So all true quartets that stem from this will obey the $1\,:\,3\,:\,3\,:\,1$ rule.

[2] If you have two different protons on a neighbouring atom instead of two identical ones, you will receive a doublet of doublets instead of a triplet. Think of it this way: One proton has a coupling constant of (say) $\mathrm{5\,Hz}$ with the proton of interest, and would give a doublet. The second one has a coupling constant of (say) $\mathrm{7\,Hz}$ with the same proton, and by itself would give a doublet again. Together, these two doublets would not meet in the middle as in the above $1\,:\,2\,:\,1$ case, because they have different distances. Instead, you will see a $1\,:\,1\,:\,1\,:\,1$ set of four signals, where the distance between the first and the second corresponds to $\mathrm{5\,Hz}$ while the distance between the first and the third corresponds to $\mathrm{7\,Hz}$.

But this doesn’t work here, because we are only coupling to one proton (and I would estimate a coupling constant of $\mathrm{19\,Hz}$ for the second coupling which is impossible in the structure given, and isn’t seen anywhere else in the spectra. Estimate was based on the aromatic coupling being $\mathrm{7\,Hz}$.)

[3] Two spins pointing in the same direction are different in energy from two pointing in different directions I just pointed out. However, if the chemical shifts of both protons that couple are similar, the difference is also noticeable without an external magnet, so that even without the sample being in the NMR magnet the two energy levels would differ — hence the protons are slightly more aligned per se than even distribution would suggest. This in turn means that one of the sides of the signal (corresponding to an energy difference) shows stronger absorption than the other, and because it is always the inner one, one can assume to see a roof — hence roofing.

The only information that roofing can give you — that two protons are close by in the molecule and their shifts are similar — can be derived from other information in the spectrum as well (information that corresponds to numbers! Real, mathematically physical hard numbers! Hence preferred by scientists): The proximity by the coupling constants and 2D-COSY and the similarity of the shifts … well, science doesn’t care too much.


How much additional hydrochloric acid should be added to arrive at ‘excess acid’? I am an IB student. And I am trying to find the mass percentage of $\ce{NaHCO3}$ in antacid. I am dissolving the antacid in $\pu{0.1 M}$ $\ce{HCl}$ solution. So, first of all I conducted some preliminary experiments with pure $\ce{NaHCO3}$ powder to optimize the experiment, get a hang of it etc.

For this preliminary experiment, where I used $\pu{0.5 g}$ $\ce{NaHCO3}$, I calculated the amount of $\ce{HCl}$ solution required as being approx. $\pu{60 ml}$. Hence I added $\pu{100 ml}$ in the flask so that I have an excess of the acid as required.

Is adding $\pu{40 ml}$ as the supposed excess amount reasonable? Should it be more or less? I am confused because my teacher suggested that by even putting $\pu{100}$ ml of $\ce{HCl}$ in $\pu{0.5 g}$ of antacid tablets, I still don't have the acid in excess? But do I not? Doesn't it only take $\pu{60 ml}$ to neutralize $\pu{0.5 g}$ of $\ce{NaHCO3}$? So $\pu{100 ml}$ is definitely in excess, isn't it?

The volume you've established seems correct, as well as your assumption that extra $\pu{40 mL}$ of $\ce{HCl}$ would be a good excess:

$$\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}$$

$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{C(\ce{HCl})} = \frac{m(\ce{NaHCO3})}{M(\ce{NaHCO3}) \cdot C(\ce{HCl})} = \frac{\pu{0.5 g}}{\pu{84.0 g mol-1} \cdot \pu{0.1 mol L-1}} = \pu{59.5 mL}$$

I suspect your teacher either prepared a series of various antacids, or this one contains additional components, such as $\ce{CaCO3}$, $\ce{Mg(OH)2}$ of $\ce{Al(OH)3}$. Note that for the neutralization of the same mass of, say, aluminium(III) hydroxide, over than 3 times more $\ce{HCl}$ is required:

$$\ce{Al(OH)3 + 3 HCl -> AlCl3 + 3 H2O}$$

$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{C(\ce{HCl})} = \frac{3m(\ce{Al(OH)3})}{M(\ce{Al(OH)3}) \cdot C(\ce{HCl})} = \frac{3 \cdot \pu{0.5 g}}{\pu{78.0 g mol-1} \cdot \pu{0.1 mol L-1}} = \pu{192 mL}$$

Either way, it looks like you understand how to properly find out the amount of acid for the neutralization. I think if you demonstrate your detailed attempt to the teacher, it should clarify the situation.


Most favorable position of attack of an electrophile on 4-trifluoromethyl-1,2-diphenylbenzene Problem
This problem is from a FIITJEE AITS paper

Which position is most favorable for the attack of an electrophile?

Answer

 Position 2

My attempt
The more favorable position for attack should be the on the para ring (considering ring with $\ce{-CF3}$ group as the reference ring), because of the lower inductive effect as compared to meta position.
If the electrophile attacks at position $1$ or $6$ (which means the carbocation will be formed at the carbon connected via a double bond), the carbocation would be equally stabilized by resonance, also the charge can move to the other 2 rings forming 9 Resonating structures.
If carbocation is formed at position $6$ by the electrophilic attack on 2nd position, the carbocation won't be able to move out to other rings for resonance implying that there would be only 3 resonating structures.
So my answer would be:

 Position 1 or 6

Why is my answer is wrong? What is the correct explanation?

TL;DR - The ring with positions 1, 2, and 6 on it is more deactivated than the ring with positions 3, 4, and 5, so the correct answer cannot be position 2 (as marked) or positions 1 and 6 (as you determined). The correct answer must be on the other ring. Since the central ring imposes a -I effect, the correct answer is position 4.

There are two factors to consider in this question. The first is that the trifluoromethyl group ($\ce{CF3}$) is electron-withdrawing by induction. Fluorine is the most electronegative element, and the $\ce{C-F}$ bond is polarized toward $\ce{F}$. The $\ce{CF3}$ group has a -I effect on the ring, and that effect is strongest at ortho and para.
The second is that the three rings in this structure are not coplanar. The steric interactions between hydrogen atoms on the two rings with labeled positions in your image causes the rings to twist out of coplanarity. This feature of the structure means that the central ring cannot exert a resonance effect on either of the other rings. The central ring will is electron poor thanks to the $\ce{CF3}$ group, and so it will provide a -I effect to the other two rings. Since the -I effect on the central ring is stronger at the position para to the $\ce{CF3}$ group than at the position meta, the ring with positions 1, 2, and 6 will be more deactivated than the ring with positions 3, 4, and 5.

Since the lower ring is experiencing a -I effect, electrophilic attack at meta is most likely, so position 4 is the true correct answer.


Do chemists use molecular orbital theory to explain physical characteristics of a compound (i.e. melting point, boiling point, etc) When you think of a molecule as simple as methane CH4, you can see that it has four identical C-H bonds. Carbon has 6 electrons, 2 in 1s, 2 in 2s and 1 in 2px, 1 in 2py and 0 in 2pz. Interestingly, one of the electrons in 2 s goes to 2pz. How does methane form 4 identical C-H bonds with just 1 s and 3 p orbitals? The answer is that carbon uses hybridized atomic orbitals called sp3 that has 25% s-character and 75% p-character.

Molecular orbital theory can explain why we have triple bonds. For example, ethyne. The sigma bond formed by the sp-sp overlap along with the two perpendicular p-orbitals that are parallel with each other that form 2 pi bonds explains the triple bond.

I've also noticed that unpaired sp3 bonds like those in ammonia (NH3) are used to form bonds and paired sp3 are the lone pairs.

I was wondering how any of this knowledge is useful for someone wanting to find out about the physical properties of compounds? Do chemists think about these things?

Certainly the electronic structure of compounds are of interest when predicting or interpreting the physical properties - but not necessarily the molecular orbital theory in and by itself. The symmetry of compounds (and by extension this also means;) the symmetry of the bonds in the compound are vital to many physical properties. However, MO theory is mainly used to describe bonding in more detail, not to infer "macro"-properties FROM bonding - so I guess the answer is... "no and yes". Yes, chemists think about these things, at least the ones that work close to the quantum realm. No, MO theory is not really useful in the way you are describing - with one giant exception:

When we are talking about "possible" physical properties of not-yet-synthesized compounds, the answer is resoundingly "yes". Graphene, for instance, has more theoretically predicted qualities than have been measured. Super conductance, tensile strengths, diffusion rates... there is just so much exiting being suggested all a result of computer modeling on the quantum level from MO theory and the likes.


Reaction-enthalpy for synthesis of 40 g hydrazine Calculate the reaction-enthalpy for the synthesis of 40 g hydrazine (rocketfuel): $$\ce{4NH3(g) + Cl2 (g) -> N2H4(l) + 2 NH4Cl(s)}$$

Attempt at solution:

First we have to account for the decomposition of four moles of ammonia into its elements.

Since $\Delta H_\mathrm f^o(\ce{NH3})$ = $-46.1 \mathrm{~kJ~mol^{-1}}$, we do:

$$\Delta H_1 = 4(+46.1) = 184.4 \mathrm{~kJ}$$

$\ce{Cl2}$ is in its standard state so we can ignore that.

We know that $\Delta H_f^o(\ce{N2H4})$ =  $50.6\mathrm{~kJ~mol^{-1}}$. Therefore, $\Delta H_2 = \pu{50.6 kJ}$.

Then, $\Delta H_\mathrm f^o(\ce{NH4Cl})$ =  $-314.2\mathrm{~kJ~mol^{-1}}$ to give

$$\Delta H_3 = 2(-314.4) = -628.8~\mathrm{kJ}$$

Finally,
\begin{align}
\Delta H_{\mathrm{rxn}} &= \Delta H_1 + \Delta H_2 + \Delta H_3\\
&= 184.4 + 50.6 + (-628.8) = -393.8~\mathrm{kJ}
\end{align}

However, my textbook says the correct answer is $-492.3~\mathrm{kJ}$.

Can someone tell me what I did wrong, please? Also, I didn't account for the $40\mathrm{~g}$. How does that affect the problem?

You are on the right track. Basically, you are just missing the last step. Your answer for the enthalpy of reaction should be $ΔH_{rxn}=−393.8\frac{\mathrm{kJ}}{\mathrm{mol}}$. So all you have to do is find the moles of hydrazine
$n=\frac{40}{32.0452}=1.248\space\mathrm{moles}$ and multiply it by your $ΔH_{rxn}$.


DIY Hand Sanitizer Clumping Hand sanitizer is impossible to find around me and I have 4 family members who are essential and not around a sink or soap during their work day (delivery drivers), and they are out.  I make a lot of soap, shampoo, lotions, things like that and I thought hand sanitizer would be easy - but I was WRONG.  I have 99% anhydrous isopropyl alcohol and aloe vera gel.  Tonight I weighed everything to make it 70% IPA and 30% aloe vera gel.  It formed these solid squishy jelly clumps and didn't form a gel at all.  What am I doing wrong?  I added the IPA in kind of slow, but was stirring the whole time. I saw all these YT videos of people doing this and their gel/IPA combination makes a perfect gel.  HOW?!  I'm thinking about adding some HEC to help thicken it but am not sure to how it will react.  Ideas?

HEC works ok, but it's not that easy to use. You have to hydrate in pure water for a few hours or more, ideally with stirring, and then incorporate the alcohol a little at a time with high shear mixing. You can use a blender, though it will incorporate a lot of air. If you use too much HEC, the "mother" gel (the HEC and water that will become 30% of your final product) will be very firm and it will be difficult to avoid lumps throughout, and the line is very thin between this and ending up with an unusually thin product. When I'm using this, I err on the thin side and use about 0.8% w/w HEC, so the "mother" is 2.67%.

Some other polymers are much easier to work with if you can get hold of them, notably Sepimax Zen aka. GelMaker PH (Polyacrylate Crosspolymer-6). These can be added directly to a 70% isopropanol or ethanol mixture and need no further ingredients to activate or correct the pH afterwards.

I have not tried aloe vera but I think it would be difficult to get a gel with 70% alcohol. I know lots of YouTubers and bloggers are recommending it. Success may depend on what other thickeners have been used in the aloe vera product. I've seen some videos where the result was the extremely thin liquid with a bit of stickiness that you would expect from the recipe.


PBP vs TBP geometry? Why are the axial bond lengths greater than those of the equatorial bonds in a trigonal bi-pyramid (TBP) geometry molecule; but the opposite is true for pentagonal bi-pyramid (PBP) geometry molecules? I think that it might have something to do with the angles between the bonds, so I thought of a possible explanation.

In a TBP molecule, the equatorial bonds are spaced apart from each other by 120 degrees, resulting in lesser repulsion; compared to the axial bonds at 90 degrees. The reduced repulsion leads to greater stability of the equatorial bonds.

In contrast, the equatorial bonds in a PBP molecule are separated by 72 degrees which results in them being less stable when compared to the axial bonds.

To what extent is this reasoning correct? Please provide a more detailed and accurate explanation, if there is one. Thanks!!

General Rule #1: Most elements use only s and p orbitals to form bonds, only transition elements and heavier elements use d, f, etc. orbitals in bonding.

General Rule #2: The more s-character in a bond the shorter the bond (reference).  For example

a $\ce{C(sp^3)-C(sp^3)}$ single bond length is ~ 1.54 $\mathrm{\mathring{A}}$
a $\ce{C(sp^2)-C(sp^3)}$ single bond length is ~ 1.50 $\mathrm{\mathring{A}}$
a $\ce{C(sp)-C(sp^3)}$ single bond length is ~ 1.46 $\mathrm{\mathring{A}}$

In an earlier SE Chem post, the structure and bonding in the trigonal bipyramid molecule $\ce{PCl_5}$ was discussed.  The molecular hybridization is pictured below.  The axial bonds are hypervalent (this concept is explained in the earlier post) and they are constructed from p orbitals.  The equatorial bonds are constructed from $\ce{sp^2}$ orbitals.  General Rule #2 suggests that the equatorial bonds will be shorter than the axial bonds since the equatorial bonds contain more s-character.

As the following structural diagram shows, this is indeed the case.

Now let's turn our attention to the pentagonal bipyramid structure as found in $\ce{IF_7}$ for example.  Here is the structure for this molecule

As you noted in your question, the equatorial bonds are now longer than the axial bonds.  Based on what we've discussed thus far, we might make an initial "guess" that in $\ce{IF_7}$ the axial bonds have more s-character than the equatorial bonds.  As it turns out, this is correct.

Here is a link to an article discussing the structure and bonding in $\ce{IF_7}$.  Only the abstract is freely available.  For completeness, I'll mention that there was an earlier SE Chem question related to $\ce{IF_7}$, but it doesn't seem to add much to the conversation.  Returning to the published abstract, the key part of the abstract states,

  "These features can be explained, however, by a bonding scheme
  involving a planar, delocalized pxy hybrid on the central atom for the
  formation of five equatorial, semi-ionic, 6-center 10-electron bonds
  and an sp hybrid for the formation of two mainly covalent axial
  bonds."

The abstract tells us that the equatorial bonds are formed from the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals; the axial bonds are formed from the $\ce{sp_{z}}$ hybrid orbitals (after using two p orbitals for equatorial bonding we are left with one $\ce{p_{z}}$ orbital and one s orbital on the central iodine, hence, when they combine two sp hybridized orbitals result).  That is the answer to the question.  As we guessed, there is more s-character in the axial bonds than the equatorial bonds.  Hence, the axial bonds are now shorter than the equatorial bonds.

For completeness

Someone, after reading that last paragraph, is going to notice that there are 5 equatorial fluorines, but only 4 bonding sites with the two p orbitals - what gives?  First off, note that the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals on iodine form hypervalent bonds with the p orbital on fluorine and so yes, there are only 4 equatorial points of attachment.  But note that the abstract also mentioned that the equatorial bonds are semi-ionic.  This suggests that we have a resonance structure where we have an [$\ce{IF_{6}^{+}}$][$\ce{F^{-}}$] ionic contributor.  We can draw 4 more resonance structures just like this for the other 4 equatorial $\ce{I-F}$ bonds.  This means that mixed in with our hypervalent p-p bonds is some ionic character - this allows us to make 5 equivalent equatorial bonds.  Said differently, our 5 equatorial bonds are all identical resonance hybrids obtained by mixing 2 hypervalent p bonds (each one providing bonding for 2 fluorines) with 1 ionic bond (providing bonding for 1 fluorine).


Which areas of chemistry require calculus? I didn't have much chemistry in school/college, and am now building up my knowledge again. I am studying maths in parallel, building up to calculus.

I noticed that the introductory chemistry books (even at the college level) don't use any calculus. However, more advanced books do tend to use calculus.

Can anyone tell me which areas of chemistry require (or make use of) calculus?

For first year college chemistry the only calculus, if there is any at a particular school, would be reaction rate equations.

$\mathrm{d}[A] = -k[A]\mathrm{d}t$ ;  $\mathrm{d}[A] =-k[A^2]\mathrm{d}t$ for example

Then second year you take organic chemistry; no calculus.

In the third year, thermodynamics has some multivariable calculus (partial derivatives) and quantum mechanics has partial differential equations.


Is alkene reaction with hydrogen to form an alkane an addition reaction? I have seen in some textbooks that such a reaction is listed under reduction of alkenes.

I understand that the double bond is getting reduced to single bond and one hydrogen atom is added to each of the carbons.

Should I classify this reaction as an addition reaction or a reduction reaction?

Yes, the hydrogenation reaction of alkenes is typically classified as an addition reaction. See here and there, for example.


Naming secondary amine IUPAC system $\ce{CH3NHCH(CH3)CH2CH3}$
Should this be named as 2-(N-methyl)butanamine or 2-(methylamino)butane?

In other words, do we name the compound as 'alkanamine' or treat amino group as a substituent of butane?

It should be N-methyl-2-butanamine. Amine is treated as a functional group attached. So we write it as alkanamine as it is the only functional group here.

Image source: ChemSpider


Significant figures in conversion factors? Example: I have $\pu{64 cm^{3}}$ of milk. How much is that in gallons?

When I use the conversion factors $\frac{\pu{1 mL}}{\pu{1 cm^3}}$
$\frac{\pu{1L}}{\pu{1000 mL}}$ $\frac{\pu{1.0567 qt}}{\pu{1 L}}$ $\frac{\pu{1 gal}}{\pu{4 qt}}$.

I know the final answer should only have two significant figures, but how many significant figures should the intermediate conversion factors have?

I have heard that:

conversion factors should have one more sig fig than the least precise measurement. (So three in this case)
But I also heard that we shouldn't round anything until the end.
(So use all 5, and round after conversions are complete)

The reason why I got confused is because of my textbook:

I'm not sure where your conversion factor came from as it is incorrect but regarding the correct factor:
Because the inch is defined using the metre, and the gallon defined using the inch, there is an exact conversion between them—1 US gallon is exactly 3.785411784 L.
Thus, like any exact quantity or conversion factor, we treat the gallon to litre conversion factor as having infinite significant figures.
Because we have defined it exactly, there is no implicit uncertainty in the number.
There are other conversion factors that are not exact and do actually have uncertainties, but for most practical purposes, these have far more significant figures than a normal measured quantity.
See this question for more details.


Negative charged intermediate in acidic conditions - what's the problem? Why positively charged intermediates are preferred over negative charged intermediates, when formed during a reaction taking place in acidic conditions?

For example:

Image source: Organic Chemistry by David R. Klein

Here, per Organic Chemistry by David R. Klein says,

  That is, the mechanism does not exhibit any negative charges, which is
  consistent with acidic conditions.

Why?

Acidic conditions — whether you are using the Brønsted or the Arrhenius definition of acidity — is characterised by readily available protons. Remember the general dissolution ionisation of a Brønsted acid in water  as shown below:

$$\ce{HA + H2O <=> H3O+ + A-}\tag{1}$$

The proton is readily liberated and transferred to species which are more happy to accept it — for strong acids, that means it is transferred to water, a really weak base. A proton typically gladly attaches to anything that has a negative charge, because those species are typically more basic than water in this example. Therefore, postulating an enolate or any other oxygen-centred anion in acidic solution is not considered correct since there will be enough protons to quench it.

However, this does not mean that no anions can exist in aquaeous solution. For example, a typical method to transform alcohols into iodides is refluxing in aquaeous hydrogen iodide. Here, $\ce{HI}$ is a very strong acid that very readily dissociates to give iodide anions, $\ce{I-}$. These, even though negatively charged, do exist in acidic conditions and are able to attack nucleophilicly. That’s all due to the high acidity of $\ce{HI}$.

$$\ce{R-OH + H3O+ + I- <<=> R-OH2+ + H2O + I- -> R-I + 2 H2O}\tag{2}$$


How to programmatically identify a chemical compound in my program? I have a small set (a hundred of items) of chemical compounds in XYZ formats in my DB.
How to quickly identify that a given chemical is or isn't in this DB?

I don't want to just scan through them, or do something else of an ad-hoc nature.

Is there an algorithm that allows to do this?

Either way, you need to define what it means for a chemical to be 'in' your database. Do you care about the 3D coordinates (given that you have XYZ) or just the atom types and connectivity?

Short answer. use a cheminformatics library that provides structure matching or can encode your molecules (query and targets) as canonical SMILES, InChI, etc.

See Is there a way to use free software to convert SMILES strings to structures?

Long answer. Use a series of filters (or questions):

Are there any molecues in my database with the same number of atoms?
Do any of these have the same atom types?
Finally, are these atoms connected to the same neighbours?

The third one is a little trickier, but is still a quick way to determine isomorphism. If you  need true structure matching, see the short answer! If you need 3D matching, then again it's going to be much quicker to use a library method.


Concentrations of mixed solutions If I mix a $\pu{2 M}$ solution with a different $\pu{1 M}$ solution, do the individual solutions reduce in molarity to $\pu{1.33 M}$ and $\ce{0.33 M}$ respectively?

You need to know how much of each solution you add and the final total volume. The molarity of each one would be

$$\frac{C_1 \cdot V_1}{V_1 + V_2} \quad \text{and} \quad \frac{C_2 \cdot V_2}{V_1 + V_2}$$

where $C_1 = \pu{2 M}$, $C_2 = \pu{1 M}$, $V_1$, $V_2$ - volumes of $\pu{2 M}$ and $\pu{1 M}$ solutions, respectively.


Which is the parent chain in Neopentyl chloride? and why? Which is the parent chain in this structure?
and since all have 3 carbons so should any of them be the parent chain, if not ..
then which should be and why? is it a rule in IUPAC naming?

(source: chemspider.com)

A chemist would say that all 3 methyl groups are equivalent (identical by symmetry).  Therefore they all produce the same chain, and that chain is the longest chain in the molecule.  Consequently, since the methyl groups are all the same, it doesn't matter which one you pick.


Product of lead and tin alloying It is common knowledge that tin melts at $449.5~\rm^{\circ}F$ and Lead at $621.5\rm~^\circ F$, and that when these are melted together you simply have an alloy.

What happens, or what product occurs, if you take pure lead to red heat ($1,400$ – $1,500~\rm^\circ F$) in a muffle furnace and cast in pure tin?

The mixture "Blooms" with somewhat small cubic crystals that, when scooped off the top into a receptacle, appears as a sulfur colored scoria that I was not able to re-melt into metal at even $2300\rm~^\circ F$.

I have a photo if I can figure out how to post it.

I think Ivan Neretin has given the right answer.

Any alloy between Sn and Pb melts at relatively low temperatures. You stated that the compound you fabricated did not even melt at 2300 °F. I have been looking at the stable lead oxides and tin oxides. The only compound of this set that has a melting temperature above 2300 °F is tin dioxide at 2970 °F.

Please note that I cannot be certain of this, as I only checked a small set of compounds I found to be the best candidates. It could be the case that you have created something more complex. (beautifully phrased by DrMoishe Pippik as a 'toxic miasma' :-) )


Do Grignard reagents deprotonate amines? Can I add Mg/dry ether to 2,4,6-tribromoaniline followed by $\ce{D2O}$ in order to replace the bromines with deuterium?

The reaction I'm proposing is part of a synthetic route as shown below.

If you really wanted to start with the substituted aniline, you could probably oxidize, deuterate, reduce to get the product if you indeed want the amine and not the amide, but the yield would be poor. Various peroxides and peroxybenzoic acids will do this for you. Lui, Li, et al. indicated that oxidation of substituted anilines would work best in dichloroethane or THF with 4-5 equivalents of m-CPBA, with little difference produced by electronic effects. Then you could use your metal reagent, deuterate, and reduce using a method of your choice.

Of course it's probably easier to just nitrate your 1,3,5-tribromobenzene directly with nitric acid.

If this is a question from a book, they're probably trying to make sure you notice that you can't use organometal chemistry when you have acidic protons, so yes, grignard reagents do deprotonate amines.


What are the prerequisites in hydrogen bonding? I know when a hydrogen bond is, but - can hydrogen bonds only occur between two molecules containing hydrogen? Or, only one of the molecules should contain hydrogen and the other one may not contain hydrogen, but should have a large electronegative difference between two of its sides?

Actually, theoretically speaking hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Usually the electronegative atom is oxygen, nitrogen, or fluorine, which has a partial negative charge. A hydrogen bond may occur between $\ce{H-O}$, $\ce{H-N}$, $\ce{H-F}$ (first two are more common). Apart from that, these hydrogen-bond attractions can occur between molecules (intermolecular) or within different parts of a single molecule (intramolecular)

See also the Wikipedia article about hydrogen bonding
A hydrogen atom attached to a relatively electronegative atom will play the role of the hydrogen bond donor An electronegative atom such as fluorine, oxygen, or nitrogen will be the hydrogen bond acceptor, irrespective of whether it is bonded to a hydrogen atom or not. An example of a hydrogen bond acceptor that does not have a hydrogen atom bonded to it is the oxygen atom in diethyl ether.

http://chemwiki.ucdavis.edu/@api/deki/files/4680/image115.png?revision=1


Why is there only one ionization isomer of [Cr(NH3)5(NO2)]Cl2? How many isomers are possible for $\ce{[Cr(NH3)5(NO2)]Cl2}$? Consider all possible types of isomerism and draw each isomer.

My texbooks presents only three possible isomers: $\ce{[Cr(NH3)5(NO2)]Cl2}$ (the original compound), $\ce{[Cr(NH3)5(ONO)]Cl2}$ (linkage isomerism through $\ce{NO2}$), $\ce{[Cr(NH3)5Cl]Cl(NO2)}$ (ionization isomerism).

However, why is $\ce{[Cr(NH3)5Cl]Cl(NO2)}$ the only possible ionization isomer? Why can't $\ce{[Cr(NH3)4Cl(NO2)]Cl(NH3)}$, $\ce{[Cr(NH3)3Cl2(NO2)](NH3)2}$, or $\ce{[Cr(NH3)4Cl2](NH3)(NO2)}$ form as well.

The proposed ionization isomers were constructed by substituting one, or both, of the $\ce{Cl-}$ ions in the complex ion structure with either $\ce{NH3}$ or $\ce{NO2-}$; the answer seems to indicate that $\ce{NH3}$ cannnot under ionization isomerism with $\ce{Cl-}$.

Because $\ce{NH3}$ is not a counter ion and thus would not be bound to the complex if not bonded to the center ion.  In $\ce{[Cr(NH3)5(NO2)]Cl2}$  the $\ce{Cl-}$ ions do not form dative bonds with the $\ce{Cr^{3+}}$ ion but are attracted by the positive charge of the $\ce{[Cr(NH3)5(NO2)]^2+}$ complex.  If you were to form$\ce{[Cr(NH3)4C(NO2)]Cl(NH3)}$, the $\ce{NH3}$ would not be datively bonded to the $\ce{Cr^3+}$ center it would have no bond or attraction to the complex and thus would be a free ion, not part of the complex anymore. Thus the compound would now be $\ce{[Cr(NH3)4(NO2)Cl]Cl}$ as opposed to $\ce{[Cr(NH3)5(NO2)]Cl2}$.


Stubborn CuAAC reaction as derivatization for ESI-MS Problem description
I have a substance that is extremely non-polar and therefore does not ionize in the ESI source, which makes it impossible to quantify small amounts. The substance contains only two functional groups, which are a point of attack for possible derivatization. One is an alkyne, and the other is a hydroxyl group. Both groups are attached to the same carbon. Since the substance is to be detected from blood plasma, I decided against derivatization at the hydroxyl group, since numerous side reactions with proteins, which also carry a lot of hydroxyl groups, are to be expected. The alkyne, however, is a good point of attack and is suitable for bioorthogonal reactions. A professor once said about the copper(I) catalyzed alkyne-azide cycloaddition (CuAAC) that this is a reaction that "any fool could manage". I have been trying to get the reaction going for a few days now, but so far, I have not been able to detect any product in LC-MS... Decide for yourself what this says about me as a chemist.
Since I have followed several approaches from papers, I slowly ask myself if my substance is not suitable for a CuAAC. Since the research project and the substance are new and unpublished, I cannot provide a complete structural formula here. But this much can be revealed: It is essentially a steroid, which carries a hydroxy group and an alkyne on carbon C17.

What was done
I performed the reaction with the pure substance (i.e. without blood plasma) according to Presolski et al., but I did not use THPTA nor aminoguanidine. All concentrations were used as described in the paper and the reaction was performed at 40°C according to the suggestion under "Troubleshooting". The azide was dissolved in DMF. Thus, there were 10 µL DMF (5%) in the buffer solution.
I also performed the reaction with the pure substance according to Hong et al.. Again, the proposed concentrations were used but neither THPA nor aminoguanidine were used.
I also tested different variations of those protocols with much longer reaction times (8 to 24 hours), more or less temperature, with and without pH 7 buffer.
As far as I know, the 4-substituted 1,2,3-triazoles formed by CuAAC should be extremely stable, even to metabolic degradation (Agalave et al.), so I do not believe that they decomposed so quickly that they were all destroyed by the measurement. To avoid ion suppression, I purified the reaction after the one-hour reaction time by RP-SPE, using the SigmaAldrich/Merck protocol.
Questions

Does anyone have experience with CuAAC reactions that do not want to run?
Which parameters should I at best adjust to make the reaction successful?
Is it possible that a hydroxide in alpha-position disturbs the reaction? I haven't found anything on this yet.
What advice is there to make the reaction successful even at very low (< 10 pg/mL) substance concentrations?

Update as of 24/07/2020
The reaction was, in fact, successful when applying the already cited protocol by Presolski et al. and using a DMF-water-ratio of 3:1 and incubating at 60 °C for 2 h. Unfortunately, though, the intensities are pretty weak which means that the reaction turnover is not great. The peak of my original substance is at 9.6 × 107 (50 µM concentration) and the intensity of the derivatized product, which should be way better ionized, is 4.2 × 107.
New question

How to get a better turnover while maintaining a reasonably short reaction time?

How to get a better turnover while maintaining a reasonably short reaction time?

To speed up the reaction, you should look at changing (1) the reagents, (2) the catalysts, or (3) the reaction conditions.  In particular you might be dealing with poor solubility for your alkyne substrate if you are doing the reaction in aqueous conditions.
1. Reagents
You could try the picolyl azides I mentioned in the comments in lieu of the 4-acetophenyl azide you originally tried.  These azide reagents chelate copper significantly, thus increasing reaction rates since proximity of two of the three components of the reaction (\ce{Cu+} and aizde) is enhanced by the chelation effect.
2. Catalyst
There are a variety of exotic $\ce{Cu+}$ organometallic complexes that work better than the "standard" copper salts.  An example publication is Sebest et al. 2018, which reports that the N-heterocyclic carbene complex $\ce{[CuI(Mes‐6)]}$ (Mes-6 being an NHC ligand pictured below) worked well at loadings down to 0.05 mol% in their trial reaction.

This might be overkill if you are not using the ligand THPTA used in your original protocol.  Maybe you could just use that?
3. Conditions
You an always heat more, but I think a more important variable is your solvent.  Is your reaction solvent one that dissolves all the species, including alkyne, azide, and copper(I) complex?  Instead of aqueous conditions, can you try acetonitrile or 100% DMF?  Or n-butanol?
Additionally, the Presolski reference you give mentions that high chloride concentrations (~0.2 mol/L) can compete with ligand for copper ion.  Plasma has about a 0.1 mol/L chloride concentration, so maybe it's an issue?  Probably speaks to the need to use a good copper ligand like THPTA.
Bottom line
If it were me, I'd try getting some THPTA or other ligand to improve $\ce{Cu+}$ solubility in organic solvents, and trying the reaction in non-aqueous solvents where your steroid is likely to be soluble.  I'd also get the picolyl azide if my lab could afford it.  If that doesn't work, and I still wanted to improve things, I guess I'd look at increasing temperature or using more exotic NHC ligands for the copper.


Wavenumbers of cyclic ketones Rank the compounds below in order of increasing C=O absorption frequency in IR spectrum (top - lowest wavenumber, bottom - highest wavenumber):

Question: Would I count compound #1's cyclopentane or all 11 carbons because wavenumber increases with 5C or smaller rings.

If I count #1 as a cyclopentane for this purpose then #3 has the lowest wavenumber and #2 has the highest.

Here is nice comparison of all 3 ring sizes. Octahydro-1H-inden-1-one (either cis- or trans-) carbonyl strechings fall between cyclohexanone and cyclobutanone, thus, fused ring is not a factor (it is to some extend because of sunstitution).


Is triclosan harmful? I keep hearing about this product known as triclosan, and that it is used in sanitizing soap. And I also heard that it is harmful, and causes some sort of muscle problems.

However some people aren't affected by it. Why is this, and should I
 be worried about washing my hands?

Triclosan (2,4,4′-trichloro-2′-hydroxydiphenyl ether) is a nonionic, broad spectrum, antimicrobial agent that, because of its favorable safety profile, has been incorporated into a variety of many personal care products, including deodorant soaps, underarm deodorants, shower gels, and health care personnel handwashes. Triclosan exhibits a moderate degree of substantivity to the skin, and, in many products, it imparts a remnant antimicrobial effect.Here is the link.

U.S. Food and Drug Administration (FDA) states that "Triclosan is not currently known to be hazardous to humans."

Despite having significant uses, it also posses little amount of adverse effects on prolong use.It impairs muscle function and skeletal muscle contractile property.This research has only been conducted in mice and results were not good; when mice were exposed to triclosan, heart muscle function was reduced by 25 percent and grip strength was reduced by 18 %.
So result has concluded that it might cause skeletal and heart muscle failure in humans as well.Here is the link.

Exposures to triclosan can also cause dermatitis, or skin irritation, and an increase in allergic reactions, especially in children.Due to its link with dioxin, it has been suspected to cause other health problems like weakening of the immune system, decreased fertility, miscarriage, birth defects, and cancer.Here is the link

But to eradicate these effects,FDA has been engaged in an ongoing scientific and regulatory review of this ingredient. FDA does not have sufficient safety evidence to recommend changing consumer use of products that contain triclosan at this time.

So we can conclude that, it cannot be considered entirely harmful for human body.


What is an NOx level What is an $\ce{NO_x}$ level? I have read some online - but I'm confused as what is used to calculate $\ce{NO_x}$. Reported $\ce{NO_x}$ levels by the government (where I live - Canada) have an $\ce{NO_x}$ level with the attached "On an $\ce{NO2}$ mass basis" What does that mean? I said $\ce{NO_x}$ doesn't exist.

The atmospheric $\ce{NO_x}$ (usually pronounced as "nox" or "knocks") concentration is the sum of the $\ce{NO}$ and $\ce{NO2}$ species.  These get combined into a single term as they each play a roll in photochemical ozone formation. $\ce{NO_x}$ can form as a byproduct of the combustion of fuels containing nitrogen, like coal, or from atmospheric nitrogen in the high-temperature environment of an internal combustion engine.

Because photochemical ozone production requires $\ce{NO_x}$, volatile organic carbon (VOC) and sunlight, the $\ce{NO_x}$ concentration is really most relevant for environments already rich in VOC that get plenty sunlight.  Modern automobiles have emission systems that have significantly reduced VOC emissions.

Depending on just where you live in Canada, $\ce{NO_x}$ levels are likely very important for photochemical ozone production during the summer, when you have plenty of natural VOC's available from local vegetation (be it forest, tundra, or any other of your rich and diverse ecosystems).

During the wintertime in Canada, or summer in places like the desert Southwest US, the atmospheric VOC concentrations are small.  In these cases it is the VOC concentration rather than that of $\ce{NO_x}$ that limits ozone production.  Of course the deserts also get plenty of the other ingredient, sunlight.

Depending on your latitude, there may be enough sunlight to produce some ozone, and due to temperature inversions the lower rate of ozone production can still lead to the buildup of ozone; so long as you have some $\ce{NO_x}$ and VOC's present.


Water exchange at Al (III) Why is the rate of water exchange at Al(III) centres so slow ?

According to this medical book (p.5) it is $10^{5}$ times faster at Mg(II).

I actually tend to agree with the answer from @JanJensen based on electrostatic effects, rather than with the "covalency" argument.

In particular, it has been mentioned in the comments that the $\ce{Al^{III}-OH2}$ is a "far more covalent" bond than one $\ce{Mg^{II}-OH2}$ one. This remark, however, does not actually explain anything, but rather it shifts the question to a new one, i.e. why is it, then, that the $\ce{Al-O}$ bond is more covalent?

Water exchange reactions at the Mg and Al centers have actually been studied extensively, but unfortunately not to the same extent as those for transition metal ions, or for selected lanthanides. The latter in particular are a subject of continuous and very intense interest, because water exchange at e.g. $\ce{Gd^{III}}$ or $\ce{Eu^{II}}$ ions play a pivotal role in the effectiveness of complexes of these ions as MRI contrast agents.

Nonetheless, here are a few very relevant references that will certainly offer some more insight, but probably not the final answer that you are looking for.

An $\ce{^{17}O}$-NMR Study of the Exchange of Water on $\ce{AlOH(H2O)5^{2+}(aq)}$, Inorganic Chemistry, 1998, 37 (19), 4760–4763;
Water exchange on metal ions: experiments and simulations, Coordination Chemistry Reviews, 1999, 187 (1), 151–181.

In fact, when studying exchange reactions at metal ions, the authors of these and other studies tend to consider ions with equal charge as homogeneous series, because the effect of the charge will swamp out any other effect, some times even the relatively large effects of electronic configuration in d-transition metals!

That notwithstanding, the papers above agree that the mechanism for the water exchange reactions for both Al and Mg metal complexes is dissociative in nature, which means that the rate determining step involves the loss of a $\ce{H2O}$ ligand from the inner sphere of the complex.

We can combine this piece of information with the fact that, as mentioned in the second paper above, there is a significant increase in charge density when going from $\ce{Mg^2+}$ to $\ce{Al^3+}$. First, the aluminum cation is smaller (reported values from that reference are: 72 pm for $\ce{Mg^2+}$ and 53 pm for $\ce{Al^3+}$); secondly, it has higher charge. This rather staggering difference can again be explained using the usual electrostatic arguments (i.e. the two ions are isoelectronic, but the Al nucleus contains more protons, so electrons in the latter are held more closely, leading to smaller cations).

Trying to put this all together:

removing a water molecule determines the rate of the overall exchange process according to the experimentally determined dissociative mechanism;
the smaller, more highly charged $\ce{Al^3+}$ exerts a much stronger attraction on each of its bound $\ce{H2O}$ ligands, making said removal much more costly;
in macroscopic terms, the $\ce{Al^3+}$ has a much higher activation energy, which ultimately manifests itself in the much lower rates of exchange that the OP referred to.


Uncertainty in measuring the mass of the palm oil and the base in saponification I am a student coming from the domain of mathematics, and I am working on error analysis and uncertainty, but I have a question  in the domain of chemistry.
I am  studying the effect of the measurement error in saponification.  According to different research references (this is one) we need $\pu{142 g}$ of $\ce{NaOH}$  to react with  $\pu{1000 g}$ of palm oil to produce soap.

What if we use  $\pu{140 g}$ of  $\ce{NaOH}$  with  $\pu{1000 g}$ of palm oil? Would this change the mass of the obtained soap? Or would it just change the quality of the soap?

It would effect primarily the mass obtained (yield) and probably the quality, too, but that depends on how you measure this. Given that you use some procedure to purify your crude product that is equivalent in all cases independent on the mass of sodium hydroxide used, you simply will get less purified product.

Palm oil is a natural product and composed of a mixture of compounds. These compounds are know as fats, which are esters of propane-1,2,3-triol (glycerine) and fatty acids. Saponification is the hydrolysis of these esters. In that case sodium hydroxide is the strong base used, and the obtained soap is a mixture of glycerine, and the sodium salts of the fatty acids.

In principle for complete saponification you need one equivalent of base for each ester group.
$$\ce{RCOOR' + NaOH -> RCOONa + ROH}$$

Extending this scheme to palm oil, for the fat or triglyceride you need three equivalents of base:
$$
\ce{\underset{triglyceride}{H5C3-(OOCR^{1/2/3})3}
   + 3 \underset{base}{NaOH}
   -> \underset{glycerine}{H5C3(OH)3}
   + 3 \underset{salts of fatty acids}{R^{1/2/3}COONa}}
\tag{1}\label{eq:fat}$$

Since it is a natural product you probably have a number of environmental factors influencing the composition of your specific sample of palm oil. Wikipedia cites the following approximate composition:
\begin{array}{lr}
\text{Type of fatty acid}    & \text{Abundance}/\%\\\hline
\text{Myristic saturated C14}       &  1.0  \\
\text{Palmitic saturated C16}       & 43.5  \\
\text{Stearic saturated C18}        &  4.3  \\
\text{Oleic monounsaturated C18}    & 36.6  \\
\text{Linoleic polyunsaturated C18} &  9.1  \\\hline
\text{Other/Unknown}                &  5.5  \\\hline
\end{array}

As you can imagine this composition of the palm oil has an uncertainty. You probably want to look up more accurate values and the errors of the employed techniques. However, the mass of base needed can therefore be only a crude recommendation. In one case be too much, resulting in leftover (unreacted) base, while in another case it might be too little, resulting in only partially or not at all hydrolysed fat.

A crudely (over)nsimplified calculation:
Assuming palm oil is made out of 100% triglyceride of palmitic acid, we know the molecular mass is $M(\text{palm oil})\approx\pu{807 g/mol}$, we also know $m(\text{palm oil})=\pu{1 kg}$, $m(\ce{NaOH})=\pu{142 g}$.
\begin{align}
  M &= \frac{m}{n} &
  \Leftrightarrow&&
  n &= \frac{m}{M} &
  \implies&&
  n(\text{palm oil}) &= \frac{\pu{1kg}}{\pu{807 g/mol}} \approx \pu{1.24 mol}
\end{align}

From $\eqref{eq:fat}$ we know we need three equivalents of the base to produce three equivalents of soap. Ergo $n(\text{soap}) = n(\ce{NaOH}\text{ needed}) = \pu{3.72 mol}$. With $M(\ce{NaOH}) = \pu{40 g/mol}$ we know we need $m(\ce{NaOH},\text{ needed}) = \pu{149 g}$. We have $m(\ce{NaOH})=\pu{142 g}$, or $n(\ce{NaOH})=\pu{3.55 mol}$, ergo we only produce that much of soap. The molecular mass of our soap is $M(\text{soap})=\pu{278 g/mol}$. Therefore we will obtain $m(\text{soap}) = \pu{987 g}$.
I'll leave it to you to do the more precise calculations, since you are the mathematician of the both of us.


Why does the Oxymercuration Demercuration mechanism lead to stereorandomization? I'm having a doubt on the mechanism of oxymercuration Demercuration reaction.
How does the reaction, initially heading towards anti addition(after cyclic transition state formation) ,get spoiled by $\ce{NaBH4}$ and ends up being stereorandom? Can anyone point out the importance of using mercury as well? The explanation would be greatly helpful.

Thank you

The stereorandomization occurs in the formation of the cyclic mercurinium ion intermediate.

When the mercury acetate approaches the alkene, it can do so from either the top face:

or the bottom face:

Since the initial alkene is achiral and the steric factors are the same for both approaches, we get both ions, which are enantiomers.

Each enantiomer reacts stereospecifically with water to produce a chiral product. Note that the position where the mercury sits is still a chirality center in this example, but in many simpler cases it is not.

Finally, $\ce{NaBH4}$ is added to this compound as a second step to reduce the $\ce{C-Hg}$ bond to a $\ce{C-H}$ bond. In most cases, if the mercury was at a chirality center, that position no longer is one because we have two $\ce{C-H}$ bonds present.

  Why do we use mercury anyway?
  It works. It works to put the OH on the more substituted position without rearrangements typical of carbocations.

However, there is ongoing research to replace toxic mercury.


The formation of ice crystals on/in frozen food that has been defrosted and then frozen again I made a mistake. I put my ice cream in the refrigerator instead of the freezer and left it there for a night. I realized my folly the next day.
I put it in the freezer. When I took it out, I noticed it had a bunch of ice crystals inside it and on its surface. This intrigued me (not that it is a new discovery of mine or anything). I was left with the question: Are these ice crystals the product of added water, or is it just the phase change of water already present in the ice cream.
Okay, so I have to hypotheses:
Hypothesis 1
When the ice cream was defrosted, it became more liquid and as such, the bonds of the chemicals were weakened. This allowed the water that was mixed with the milk to break loose a little. When the ice cream was put back inside the freezer, it cooled down at too quick of a pace, making it so that some of the water didn't have time to fall back in line before it underwent a phase change.
Hypothesis 2
While the ice cream was hotter and more liquid, its permittivity increased. Atmospheric water, of which there is probably much of inside a refrigerator, entered the semi-melted ice cream and over-saturated it with water. When it was put back into the freezer, this over-saturation led to the crystallization of excess water.

So, which one of these hypotheses are most likely, if not surely, correct or most correct? Or is none of them correct? Or is the phenomenon caused by a combination of the two hypotheses?

Industrial ice cream makers make it frozen much faster than household freezers. They freeze faster and ice cream is disturbed to take out heat and to keep the emulsion well dispersed. As result, frozen water based emulsion droplets are very small.
When melted and standing, emulsion slowly separates either and oily phases.
She put in household freezer, it is getting frozen very slowly. That helps further emulsion separation and ice have enough time to form big crystals.
Note that it happens for frozen vegetables as well, if melted and refrozen again. Without any added water.


Why is the boiling point of stibane higher than that of ammonia? I recently came across the fact that the boiling point of $\ce{SbH3}$ (stibane) is greater than that of $\ce{NH3}$ (ammonia). I was expecting $\ce{NH3}$ to have a greater boiling point as a consequence of intermolecular hydrogen bonding. I understand that as molecular mass increases, van der Waals interactions between molecules also increase, but believing that hydrogen bonding will be a more dominant factor than van der Waals interactions, I expected $\ce{NH3}$ to have a greater boiling point.

Now, I already know that:

$\ce{NH3}$ and $\ce{SbH3}$ are gases at room temperature.
Gases including $\ce{NH3}$ do not showing hydrogen bonding in gaseous phase, which I also don't understand why.

But I want to know:

Why does $\ce{NH3}$ have a lower boiling point than $\ce{SbH3}$ even though it shows hydrogen bonding?
Why does $\ce{NH3}$ not show hydrogen bonding in gaseous phase?

Though it does go against your intuition, you've actually mentioned the answer in your question. Stibane has a higher boiling point than ammonia/azane on account of van der Waals interactions (owing to the larger size of the antimony atom).

Our teacher had actually posed this question to us during my first year of high-school. All of us were incredulous when we were told that stibane had a higher boiling point than ammonia, and like you, we were fixated on H-bonding and chose to dismiss van der Waals' interactions as insignificant in this regard.

First, a look at the H-bonding argument:

At school we're taught that H-bonding only occurs in molecules where you find hydrogen atom(s) bound to either nitrogen, oxygen or fluorine ($\ce{N,O,F}$). This is because $\ce{N, O}$ and $\ce{F}$ are sufficiently electronegative (about 3.0, 3.5 and 4.0 on the Pauling scale respectively) to polarize the $\ce{N/O/F-H}$ bond. Do note, that nitrogen is only barely able to establish H-bonding, while it's much easier for oxygen and fluorine to polarize the $\ce{O/F-H}$ bond on account of 1) their higher electronegativites and 2) their smaller sizes.

As antimony is much larger than nitrogen, and since the hydrogen bonding due to the nitrogen in ammonia is pretty weak, the van der Waals interactions in stibane 'outdoes' the H-bonding in ammonia.

However the same argument does not follow for water and hydrogen fluoride, which have higher boiling points than tellane and hydrogen iodide respectively.

$$\textbf{Melting } (\vartheta_\mathrm{m})\textbf{ and boiling } (\vartheta_\mathrm{b}) \textbf{ points of}\\\textbf{Group 15, 16, and 17 hydrides  in K}$$
\begin{array}{lrrlrrlrr}\hline
&\hspace{-2em}\text{Group 15}\hspace{-1em} &&& \hspace{-2em}\text{Group 16}\hspace{-1em} &&& \hspace{-2em}\text{Group 17}\hspace{-1em}&\\ \hline
\text{Hydride} & \vartheta_\mathrm{m} & \vartheta_\mathrm{b} & \text{Hydride} & \vartheta_\mathrm{m} & \vartheta_\mathrm{b} & \text{Hydride} & \vartheta_\mathrm{m} & \vartheta_\mathrm{b}\\ \hline
\ce{NH3} & 195.5 & 239.6 & \ce{H2O} & 273.0 & 373.0 & \ce{HF} & 180.7 & 292.4\\
\ce{PH3} & 138.0 & 185.0 & \ce{H2S} & 190.3 & 211.0 & \ce{HCl} & 161.0 & 189.4\\
\ce{AsH3} & 159.0 & 218.0 & \ce{H2Se} & 209.0 & 231.0 & \ce{HBr} & 184.5 & 206.0\\
\ce{SbH3} & 184.0 & 256.0 & \ce{H2Te} & 222.0 & 271.0 & \ce{HI} & 222.2 & 237.0 \\ \hline\end{array}

Now here's where your logic comes into play; the H-bonding's pretty strong in case of water and hydrogen fluoride (after all, oxygen and fluorine are the two most electronegative elements we know) and this time 'round the H-bonding outdoes the van der Waals interactions.

So you see, you should've made a comparision between the higher and lower hydrides (for groups 15, 16 and 17, to really be able to spot the exception).

As for your spin-off question:

  Why does $\ce{NH3}$ not show hydrogen bonding in gaseous phase?

Well, the H-bonding here is strong enough to keep the ammonia molecules together at normal temperatures. But you must never forget that H-bonding is not a true chemical bond, but an intermolecular interaction. If you provide sufficient energy to a sample of ammonia (i.e. by heating it) the ammonia molecules wiggle and wiggle and wiggle, until at some point they wiggle so rapidly, that they overcome the attractive forces brought about by hydrogen bonding and become a gas. This is why you don't observe H-bonding in gaseous ammonia: because the particles are too energetic to be held down.


According to what criteria was the periodic table of elements ordered? I was wondering why the elements in the periodic table were disposed the way they are. I understand, of course, that they are put in increasing atomic number fashion, but I'd like to know more about the topic. Could you link some reference that examine in detail the structure of the periodic table?

I do not address the question of history, of Mendeleyev's investigations, that is answered in How was Mendeleev able to develop his table?

Modern understanding of basic principles follows. Let us have a piece of periodic table

  ... A0 B0 ...
  ... A1 B1 ...
  .............

Element "B0" must have the next atomic number after "A0", i.e. one more electron. It is different chemically, although may have (albeit not necessarily) similar physical properties.

Element "A1" must be chemically similar to "A0", although physical properties may differ significantly (as a general rule, ".1" elements are denser than ".0"). It happens that for each element there exist its "analog" in the next period, that has similar chemical properties (based on similar outer-shell electron configuration). This pattern, once noticed, permitted for good predictions (search for Mendeleyev's "eka" elements in Internet for more information).

Periods have different lengths. Analog's atomic number may be 8, 18, or 32 ahead off the prototype's atomic number (the number of electrons expended to "raise" all the configuration one shell up). This difference is due to quantummechanical effects.


Mistake in Wikipedia article on fugacity? The Wikipedia Article on Fugacity contains statement

  The contribution of nonideality to the chemical potential of a real gas is equal to RT ln $\phi$

Here $\phi$ is fugacity coefficient.I think it is wrong it should be fugacity coefficient multiplied by partial pressure of gas divided by standard pressure.So for the example of nitrogen gas at 100 atm it's chemical potential should be μ = μ$_{id}$ + RT ln 97.03 and not μ = μ$_{id}$ + RT ln 0.9703 as given in Wikipedia Article.

I am confused, who is wrong me or wikipedia ?

I think you have misinterpreted the passage on Wikipedia a little. Of course, you are right that the chemical potential is given by the formula

\begin{align}
 \mu_{i} (p_{i}, T) &= \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \log a_{i} \\
 &= \mu_{i}^{0} + RT \ln \left(\phi_{i} \frac{p_{i}}{p^0} \right)
\end{align}

where $R$ is the universal gas constant, $T$ is the temperature, and $\phi_{i}$, $p_{i}$, $a_{i}$, and $\mu_{i}$ are the fugacity coefficient, the partial pressure, the activity and the chemical potential of the $i$th component in the system, respectively.
But the passage you are quoting says that the contribution of nonideality to the chemical potential of a real gas is equal to $RT \ln \phi$ and in the equation

\begin{align}
 \mu = \mu_{\text{id}} + RT \ln \left(\phi \right)
\end{align}

$\mu_{\text{id}}$ is not the same as $\mu^{0}$. Rather it is meant in the following way: For an ideal gas the fugacity coefficient is $1$ by definition and so the chemical potential of an ideal gas, $\mu_{\text{id}}$, is given by

\begin{align}
 \mu_{\text{id}} = \mu^{\text{id}}_{i} (p_{i}, T) = \mu_{i}^{0} + RT \ln \left(\frac{p_{i}}{p^0} \right)
\end{align}

If you use the mathematical property of logarithms that $\log(a \cdot b) = \log(a) + \log(b)$ you can rewrite the equation for the chemical potential of real gases to

\begin{align}
 \mu_{i} (p_{i}, T) &= \underbrace{\mu_{i}^{0} + RT \ln \left(\frac{p_{i}}{p^0} \right)}_{= \, \mu_{\text{id}}} + RT \ln \left(\phi_{i} \right) \\
 &= \mu_{\text{id}} + RT \ln \left(\phi_{i} \right)
\end{align}

and so you see, that the contribution of nonideality to the chemical potential of a real gas is indeed equal to $RT \ln \phi$ and you get to the equation from the Wikipedia article.


Breaking degeneracy in phosphorus pentachloride Recently, in a class discussion, the following point was brought up:

  Although the axial and equatorial bond lengths in $\ce {PCl5}$ are inequivalent, when we consider the time-averaged $\ce {P-Cl}$ bond length, there should only be one value sincethe molecule is symmetric and thus, all the chlorine atoms interact with the phosphorus atom to the same extent.

The idea of "time-averaged" means that we are considering the $\ce {P-Cl}$ bond over a long duration. This bond length varies over time since the atoms are constantly in motion. When this bond length is "time-averaged", it means that we are taking the average of the bond lengths measured at infinitely many instants in time.

The argument is that although at any particular instant, the molecule would adopt a configuration with axial bonds being longer than equatorial bonds, when the bonds are time-averaged, they should all be the same length.

This has caused much dispute among members of the class and it has also led me to consult a professor who is well-versed with quantum chemistry. The professor mentioned that the argument put forward is not valid because in $\ce {PCl5}$, there is a breaking of degeneracy due to perturbation. He cited the Jahn-Teller distortion as a situation similar to that in $\ce {PCl5}$, where degeneracy is broken.

I am very confused by this. What exactly is this perturbation? Why does it exist for $\ce {PCl5}$ and not for similar molecules, such as $\ce {SF6}$ (as seen in the comments for this question)?

I would guess he means a spontaneous symmetry breaking, in which case the "perturbation" is any old force from the environment.  This can happen in physical systems: it's possible for a system of high symmetry to spontaneously fall to a lower-energy state of lower symmetry.  Only the tiniest of perturbations is required, if you wait long enough.  A familiar classical example is crystallization, in which below a certain temperature a system of atoms spontaneously loses complete translational symmetry even in the absence of an external potential.  A quantum example is magnetization when a ferromagnet cools below the Curie temperature -- the spins lose rotational symmetry and pick some essentially random preferred direction, in response to the tiniest of perturbations from the environment.  In the case of PCl5, one would assume, very roughly (and it would take detailed calculation to verify) that if 3 of the Cls pull in closer to the nucleus and 2 move further away, you get a better balance of nuclear-electron attraction and nuclear-nuclear and electron-electron repulsion than if all 5 remain at the same distance.  So the system spontaneously breaks rotational symmetry, in response to any old tiny perturbation in the environment.

That said, in the absence of an external potential, the rotational quantum state of the molecule will be entirely symmetric, and you will have an equal chance of finding the long Cl-P-Cl axis pointing in any direction.


Change in gas phase reaction equilibrium when not all product and reactant are increased $$\ce{2NOCl <=> 2NO + Cl2}$$
What will be the effect on equilibrium concentration of $\ce{NOCl}$ when equal moles of $\ce{NOCl}$ and $\ce{NO}$ are introduced in the mixture at constant temperature?
Now this is what I think
Increasing the moles will result in increased concentration of both $\ce{NOCl}$ and $\ce{NO}$. But there also lies a second product $\ce{Cl2}$. Since there's an increase in $\ce{NO}$ and $\ce{NOCl}$, there should be an equal increase in the concentration of $\ce{Cl2}$ to keep the equilibrium-constant constant. But since there's no addition of $\ce{Cl2}$ molecules there should be a decrease in $\ce{NOCl}$ concentration which will decompose further to increase the molar ratio of $\ce{Cl2}$ and hence achieve equilibrium. Please enlighten me where my logic is flawed.

From comments (edit by BuckThorn): The reported answer is that the position of equilibrium remains unchanged and hence the concentration of $\ce{NOCl}$ remains constant.

Given the reaction $$\ce{2NOCl <=> 2NO + Cl2}$$  we can formulate the equilibrium as $$\mathrm{K} = \dfrac{\ce{[NO]^2[Cl2]}}{\ce{[NOCl]^2}}$$

The problem then asks "What will be the effect on equilibrium concentration of NOCl when equal moles of NOCl and NO are introduced in the mixture at constant temperature?"

Now if we add equal amounts of $\ce{NOCl}$ and $\ce{NO}$ we don't have to do it at the same time to end up with the final equilibrium.

So let's add the $\ce{NOCl}$ first. We now have two possibilities (assuming that it is very unlikely that K=1 exactly).

If K>1 then more $\ce{NO}$ and $\ce{Cl2}$ form.
If K<1 then less $\ce{NO}$ and $\ce{Cl2}$ form.

Now let's add the $\ce{NO}$. This has to push the reaction to the left.

So the net result is it depends. You have too have enough data to plug into the equilibrium expression and other equations to work the net result out.


Calculating the biological standard potential I have an electrochemical cell with following overall reaction:
$$\ce{5 Sn^4+ + 2 Mn^2+ + 8 H2O <=> 5 Sn^2+ + 2 MnO4- + 16 H+}$$

I have defined the half reactions as:

$$
\begin{align}
\ce{Sn^2+ &-> Sn^4+ + 2 e-} \\
\ce{MnO4- + 8 H+ + 5 e- & -> Mn^2+ + 4 H2O}
\end{align}
$$

I have trouble calculating the biological standard electrochemical potential. This is what I've tried:

The standard potential can be found from tables:

$$E^\circ = \pu{1.51 V} - \pu{0.15 V} = \pu{1.36 V}$$

Then I use Nernst equation in which all species except $\ce{H+}$ are in their standard states, so their activities are all equal to $1:$

$$
\begin{align}
E &= E^\circ - \frac{RT}{vF}\ln(10)\cdot \mathrm{pH} \\
  &= \pu{1.36 V} - \frac{\pu{8.314 J mol-1 K-1} × \pu{298.15 K}}{10 × \pu{96485 C mol-1}}\ln(10) × 7 \\
  &= \pu{1.31 V},
\end{align}
$$

which is wrong. Is there something wrong with the calculations or the understanding?

Unfortunately, the OP's Nernst equation is incorrect for several reasons. The balanced net redox equation is

$$\ce{5 Sn^{4+} (aq) + 2 Mn^{2+} (aq) + 8 H2O <=> 5 Sn^{2+} (aq) + 2 MnO4^- (aq) + 16 H+ (aq) \tag 1}$$
and n = 10 electrons for the reaction as written. The reaction quotient, Q, is

$$\ce{Q = \frac{[Sn^{2+}]^5 [MnO4^-]^2 [H+]^16}{[Sn^{4+}]^5 [Mn^{2+}]^2}              \tag 2}$$

and the Nernst equation is

$$\displaystyle{E_{cell} = E_{cell}^\circ - \frac{RT}{nF} \ln(Q) = E_{cell}^\circ - \frac{RTln(10)}{nF} \log(Q)         \tag 3}    $$

The OP's question states that "all species except $\ce{H+}$ are in their standard states, so their activities are all equal to 1". Hence, in this very contrived example, Q reduces dramatically:

$$\displaystyle{Q = [H^+]^{16}             \tag 4}$$

The OP notes that $\displaystyle{E_{cell}^\circ }$ = 1.36 V (from tables). With T = 298.15 K, $R = \pu{8.31446261815324 J mol-1 K-1}$ and $F = \pu{96485.3321233100184 C mol-1}$, then

$$\ce{\displaystyle{ \frac{RTln(10)}{nF} \approx \frac{0.05916 \ V}{n} }        \tag 5}$$

So the Nernst equation is simply

$$\displaystyle{E_{cell} \approx 1.36 \ V - \frac{0.05916 \ V}{n} \log(Q) = 1.36 \ V - \frac{0.05916 \ V}{10} \log([H^+]^{16})      \tag 6}    $$

Hence, using $\ce{pH = -log[H^+] = +log(1/[H^+])}$, the Nernst equation can be expressed as

$$\displaystyle{E_{cell} \approx 1.36 \ V - 16 \times \frac{0.05916 \ V}{10} \log[H^+] =  1.36 \ V + 16 \times \frac{0.05916 \ V}{10} pH      \tag 7}    $$

If pH = 7, then $\ce{E_{cell} \approx 2.02 \ V}.$

It is better to use expression (3) in order to avoid getting too tricky. This is where the OP went wrong: forgetting the "16" exponent on the hydrogen ion concentration and also missing the fact that the negative sign before the pre-log correction term must be subsumed into the pH definition. In a realistic example, it will not happen that everything is in standard state except the hydrogen ion.

What about dividing the net ionic equation, i.e., equation (1), by 16? Then n = 10/16 = 5/8 and all the exponents in the Q expression also get divided by 16. This is no problem if everything is in standard state except the hydrogen ion. Otherwise, it is just facilitating making a silly mistake.


Why aren't there 8 double bonds in octasulfur (S8)? I know that the $\ce{S8}$ molecule contains 16 pairs of lone electrons. My teacher always says that we must try to reduce the no of lone pairs, so, had there been double bonds between 8 atoms of sulfur then the no of lone pairs would be reduced. Then why aren't there any double bonds?

An $\ce{S}$ atom in a molecule must have 2 bonds to be neutral and in the $\ce{S8}$ ring it needs both of these bonds to continue the chain/ring. If you introduce some double bonded S atoms into the ring each of these will have 3 bonds and will thus be positively charged. So, while reducing the number of free electron pairs you increase the charge of the compound which would be wrong because $\ce{S8}$ isn't charged. You can only use free electron pairs to form double bonds if you have some electron deficient atoms in the neighborhood which can "receive" them.


galvanic corrosion in the wild: where is the reduction? Galvanic corrosion occurs of an active metal (i.e. iron or zinc) in contact with a passive metal (i.e. copper) in a conductive solution. A famous example is the corrosion of iron girders in contact with copper in the statue of liberty.

Although the difference in potential can account for the protection of copper, it does not explain the girders in had a greatly accelerated corrosion rate. In a zinc/copper battery there is a solution of copper sulfate that gets reduced at the copper electrode. However, if all the copper is in metallic form in the first place, no galvanic reaction can occur.

It seems galvanic corrosion in real-life, does not increase total corrosion, but rather concentrates it (at lest for pairs of metals with similar non-galvanic corrosion rates). Copper sheets have a large surface area that initially reacted non-galvanically to form the green color we see. Some of these ions are then reduced at the expense of the oxidizing the girders. The statue of liberty and other infamous examples of corrosion involve large amounts of inactive metals connected to small amounts of active (iron/steel) girders/fasteners/etc.

This logic predicts that the attaching a copper bar to an iron bar of equal surface area will only double it's corrosion rate. That being said, predicting the corrosion rate is not easy so there could be some other effect I am overlooking. Is the surface area logic correct or would there a large increase in the total corrosion?

Comparing galvanic corrosion to the zinc/copper battery is a bit unfair. A fast corrosion rate is still much much slower kinetics compared to the expected rates of the electrochemical reaction in the battery.

The counter reaction in the galvanic corrosion case is almost always reduction of water. This is why it is much worse in conditions near the sea.

Galvanic corrosion rate can be measured and there is an ASTM standard(G71) to make the measurement.

To answer your comment below, I am trying to say that it is impossible to say if it will double or triple or just increase by 10%. The factors are complicated and hard to predict a priori. That is why it is a good idea to measure using the above mentioned standard.


Why is edta in its calcium complex still a tetraanion at pH 10? $\ce{edta}$ has different forms according to the pH. For example if the pH is $7{-}10$, $\ce{HY^3-}$, whereas if the pH is $11{-}14$ its form is $\ce{Y^4-}$.

But I've come across an exercise where it says that we titrate $\ce{CaCl2}$ with $\ce{edta}$ and the pH is $13$, and then we titrate it again with $\ce{edta}$ but the pH now is $10$.

The reactions that are given are the following:

$$\ce{CaCl2 + Na2H2Y <=>  Na2[CaY] + 2HCl}$$

Why is the $\ce{edta}$ given as $\ce{Na2H2Y}$ both times? Shouldn't its form be different at different pHs?

The $\mathrm{pH}$ discussions only make sense in the way you are using them for free $\ce{edta}$. Free $\ce{edta^4-}$ can indeed accept up to six protons depending on the acidity of the solvent; the most common forms are probably the neutral compound $\ce{H4edta}$ and the disodium salt $\ce{Na2H2edta}$. For these species, a $\mathrm{pH}$ equilibrium exists as you know it for polyprotic acids and only for these species is the discussion you’re beginning your post with correct.

Once $\ce{edta}$ forms a coordination compound, e.g. with calcium, things change. Six atoms of $\ce{edta}$ — four oxygens and to nitrogens — are now octahedrally surrounding the central metal ion. The complex will look something like this:

The calcium ion is of course calcium(II), the $\ce{edta}$ ligand carries four negative charges resulting in the complex anion being a dianion $\ce{[Ca(edta)]^2-}$. Technically, this species can still accept up to four protons, but only the carbonyl oxygens can now be protonated to some extent. If you do, you will encounter a very acidic oxygen-centred cation which will want to get rid of the extraneous proton again. Note that the two sites which I intuitively expect to be the most basic, the two nitrogen atoms, cannot accept any more protons since they are already tetracoordinated.

Thus, once you form a complex from free $\ce{edta}$, the equilibrium situation changes completely. We now no longer may consider $\ce{edta}$ by itself but must consider the complex as a whole; the remaining acid-base equilibrium is

$$\ce{[Ca(edta)]^2- + H+ <<=> [Ca(Hedta)]-}\tag{1}$$

As I mentioned, protonating the ligand will lead to a very acidic species (akin to protonating a protonated carboxylic acid) and therefore the equilibrium is shifted to the left.

So even though the $\ce{pH}$ values may tell you that you are dealing with a tri- or tetraanion of free $\ce{edta}$ in solution, the complex has quite different acidity constants.


Why isn't it commercially suitable to dissolve sulfur trioxide in water to produce sulfuric acid? Choose the two correct options from A-F, why isn't it commercially suitable to dissolve $\ce{SO3}$ directly in water to give concentrated sulfuric acid instead of using the contact process? The major reaction of the contact process are

  $$\ce{S + O2 -> SO2}$$
  $$\ce{2 SO2 + O2 <=>2 SO3}$$
  $$\ce{SO3 + H2SO4 -> H2S2O7}$$
  $$\ce{H2S2O7 + H2O -> 2H2SO4}$$

  A. To reduce the hazards of spillage
  B. Because the density of the product is too high
  C. To minimise transport costs of large volumes
  D. Because the last step is too expensive.
  E. Because the last step of the process is too exothermic.
  F. An aerosol of the sulfuric acid rapidly fills the container.

The solution is C and F, but I think it should be A and D. Is it because to make gas react with $\ce{H2O}$, too much pressure is needed and it will be expensive? Can someone explain why each choice is false and true?

Part of the difficultly with this question is that it is asking about what is commercially viable as opposed to what is merely possible. It is, of course, quite possible to dissolve SO$_3$ in water, but doing so creates a runaway chemical reaction. So let's go through each of the points and explain why they don't make commercial sense.

(A) The contact process produces SO$_3$, as well as several other noxious gasses anyway, so they is no reduced risk as regards spillage.

(B) The density of the product is totally irrelevant, as you get the same product in both cases.

(C) Sulphur is a solid at standard temperature and pressure (STP), so one can transport a large amount using a standard container and transport vehicle. As a gas at STP, one can transport far less (many millions of moles less) SO$_3$ at STP for the same size container, so an unfeasibly large container would be needed unless a highly pressurized or very low temperature container is used. Either method is considerably more expensive, so less commercially viable.

(D) The last step of the contact process is to react oleum (H$_2$S$_2$O$_7$) with water. This is in fact relatively safe to do if done slowly and water is extremely cheap, so the process is not expensive

(E) The reaction of oleum and water is exothermic. However, in practice water is added relatively slowly to the oleum (both of which are controllable liquids), which controls the process.

(F) As said earlier, the reaction of water with SO$_3$ gas cannot be easily controlled. It runs away with itself, and the water starts to boil. This throws up lots of small droplets (an aerosol) of conc. sulfuric acid. This is highly dangerous, can increase the pressure in the reaction chamber, and also makes it difficult to cheaply extract the product from the reaction chamber.

Therefore, on economic grounds, the answer is indeed C and F.


Effect of enatiomers in pharmaceuticals Often (or always?) one enantiomer of a certain drug will be effective and the other ineffective or harmful. The famous example is thalidomide, where one enantiomer caused mutilation of the unborn child and the other relieved morning sickness.

Are there any other examples where the other enantiomer of a drug that is used today or in history causes some adverse side effect, but has obviously been separated from the product to leave the effective drug, in addition what is(are) that side effect(s)?

Does absolute configuration at chiral/stereogenic centres affect the effectiveness of a drug molecule or perhaps cause unwanted side effects?

In pharmaceutical industries, $56\%$ of the drugs currently in use are chiral molecules and $88\%$ of the last ones are marketed as racemates (or racemic mixtures), consisting of an equimolar mixture of two enantiomers.

More recently, drugs originally marketed as racemic mixtures are reintroduced using the active isomer. Examples include racemic citalopram (Brand Name: Celexa among others) and its S-enantiomer, escitalopram (Brand Name: Cipralex and Lexapro among others); racemic omeprazole and its S-enantiomer esomeprazole; and racemic modafinil and its R-enantiomer armodafinil.
Sometimes, the difference in pharmacological activity between stereoisomers is dramatic. The dextrorotatory isomers in the morphine series are cough suppressants with less risk of substance abuse, whereas the levorotatory isomers contain the analgesic activity and significant risk of substance abuse. Although the direction of optical rotation is opposite to that of the morphine series, dextropropoxyphene contains the analgesic activity, and the levo-isomer contains antitussive activity.

Some were originally approved as racemic mixtures, and later a specific isomer was marketed with claims of having fewer adverse reactions in patients.
An example of the latter is the local anesthetic levobupivacaine, which is the S-isomer of bupivacaine. Both the R- and S-isomers have good local anesthetic activity, but the R-isomer may cause depression of the myocardium leading to decreased cardiac output, heart block hypotension, bradycardia, and ventricular arrhythmias. In contrast, the S-isomer shows less cardiotoxic responses but still good local anesthetic activity. Escitalopram is the S-isomer of the antidepressant citalopram. There is some evidence that the R-isomer, which contains little of the desired selective serotonin reuptake inhibition, contributes more to the adverse reactions than does the S-isomer.

The reintroduced S-enantionmer of citalopram with less pronounced side effects
The second part is a bit tricky, but there are different possibilities.
i. Absolute configuration
The two enantiomers of a chiral drug are best identified on the basis of their absolute configuration or their optical rotation. The absolute configuration at a chiral center is designated as R- or S- to unambiguously describe the 3-dimensional structure of the molecule. A chiral drug may have more than one chiral center, and in such cases it is necessary to assign an absolute configuration to each chiral center within the molecule.
Although the enantiomers of chiral drugs have the same chemical connectivity of atoms, they exhibit marked differences in their pharmacology, toxicology, pharmacokinetics, metabolism etc. Therefore, when chiral drugs are synthesized, as much effort goes towards the rigorous separation of the two enantiomers. This ensures that only the biologically active enantiomer is present in the final drug preparation.

The difference between two enantiomers of a drug is illustrated below using a hypothetical interaction between a chiral drug and its chiral binding site. In this case, one enantiomer is biologically active while the other enantiomer is not. The portions of the drug labeled A, B, and C must interact with the corresponding regions of the binding site labeled a, b, and c for the drug to have its pharmacologic effect. The active enantiomer of the drug has a 3-dimensional structure that can be aligned with the binding site to allow A to interact with a, B to interact with b, and C to interact with c. In contrast, the inactive enantiomer cannot bind in the same way no matter how it is rotated in space. Although the inactive enantiomer possesses all of the same groups A, B, C, and D as the active enantiomer, they cannot all be simultaneously aligned with the corresponding regions of the binding site:

Here is another example of epinephrine:

A postulated fit to epinephrine’s receptor can explain why (-)-epinephrine exhibits 12 to 15 times more vasoconstrictor activity than (+)-epinephrine.
This is the classical three-point attachment model. For epinephrine, the benzene ring, benzylic hydroxyl, and protonated amine must have the stereochemistry seen with the (-) isomer to match up with the hydrophobic or aromatic region, anionic site, and a hydrogen-bonding center on the receptor. The (+) isomer (the mirror image) will not align properly on the receptor.
ii. Another possibility
It is difficult to conclude that one isomer is superior to the other. For instance, S-verapamil is a more active pharmacological stereoisomer than R-verapamil, but the former is more rapidly metabolized by the first-pass effect. First-pass refers to orally administered drugs that are extensively metabolized as they pass through the liver.

This illustrates that sometimes absolute configuration has little impact on pharmacological activity and sometimes makes little difference whether a racemic mixture or one isomer is administered.
The popular nonsteroidal anti-inflammatory drug (NSAID) ibuprofen is sold as the racemic mixture. The S-enantiomer contains the anti-inflammatory activity by inhibiting cyclooxygenase. The R-isomer does have centrally acting analgesic activity, but it is converted to the S-form in vivo.

In some cases, the portion of a molecule containing the chiral center(s) may be in a region that does not play a role in the molecule's ability to interact with its target. In these instances, the individual enantiomers may display very similar or even equivalent pharmacology at their target site. Even in these cases, the enantiomers may differ in their metabolic profiles as well as their affinities for other receptors, transporters, or enzymes.

The body with its numerous homochiral compounds being chiral selective, will interact with each racemic drug differently and metabolize each enantiomer by a separate pathway to generate different pharmacological activity. Thus, one isomer may produce the desired therapeutic activities, while the other may be inactive or, in worst cases, produce undesired or toxic effects.
Although slightly off topic, it is important to note that in pharmacology area, activity of racemic drugs can be divided into three main groups:

Group 1. Racemic drugs with one major bioactive enantiomer

Group 2. Racemic drugs with equally bioactive enantiomers

Group 3. Racemic drugs with chiral inversion

Overall, it is critical to distinguish the single enantiomer from the racemic form because they may differ in their dosages, efficacies, side effect profiles, or even indicated use.
The decision to use a single enantiomer versus a mixture of enantiomers of a particular drug should be made on the basis of the data from clinical trials and clinical experience.
The two enantiomers of a chiral drug may differ significantly in their bioavailability, rate of metabolism, metabolites, excretion, potency and selectivity for receptors, transporters and/or enzymes, and toxicity. The use of single-enantiomer drugs can potentially lead to simpler and more selective pharmacologic profiles, improved therapeutic indices, simpler pharmacokinetics due to different rates of metabolism of the different enantiomers, and decreased drug interactions.

References

Chiral drugs

Pharmaceutical Analysis: David C. Lee, Michael L. Webb, Eds,, Pharmaceutical Analysis; 1st Edn., CRC Press LLC (Behalf of Blackwell Publishing Ltd.): Boca Raton, FL, 2003.

Chemistry for the Biosciences: Jonathan Crowe and Tony Bradshaw, In Chemistry for the Biosciences: The Essential Concepts; 3rd Edn., Oxford University Press: Oxford, United Kingdom, 2014.

Chiral Drugs: Lien Ai Nguyen1, Hua He, Chuong Pham-Huy, "Chiral Drugs: An Overview," International Journal of Biomedical Science 2006, 2(2), 85-100 (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3614593/); Also availble at: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3614593/).

Stereochemistry in Drug Action: Jonathan McConathy, Michael J. Owens, "Stereochemistry in Drug Action," Primary Care Companion to The Journal of Clinical Psychiatry 2003, 5(2), 70–73 (doi:10.4088/pcc.v05n0202); Also available at: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC353039/.

Organic Medicinal Chemistry and Pharmaceutical Chemistry: John M. Beale, Jr., John H. Block (Eds.), Wilson and Gisvold's textbook of Organic Medicinal Chemistry and Pharmaceutical Chemistry; 12th Edn., Lippincott Williams & Wilkins (a Wolters Kluwer business): Baltimore, MD, 2011.


Density of water increases from 273 to 277 K due to increase in entropy? In a set of questions that I was solving, the answer to one of the questions asking why the density of liquid water increases from 0 to 4 degrees C had the above answer: due to an increase in entropy.

This website does explain this a bit with two thermodynamic equations, but I am unable to make sense of how entropy comes into the picture here. Please explain how the increase in density of water is due to an increase in entropy.

At $0°$C, the water is not yet made of independent $H_2O$ molecules. It is still a mixture  of hexagones containing 6 molecules $H_2O$. And these hexagones are separated by independent $H_2O$ molecules. There is still a certain order in the water at 0°C. These hexagones are destroyed between $0°$C and $4$°C. These hexagones are more cumbersome than 6 molecules $H_2O$, because there is a hole in the center of the hexagones. That is why the density of water at $0$°C is lower than at $4°$C.


Addition of HBr to Vinyl bromide The addition of $\ce{HBr}$ to $\ce{CH2=CHBr}$ produces $\ce{CH3-CHBr2}$. Why is this so?

According to me, halogens have a stronger $-I$ effect than a $+M$ effect (for example a halogen substituent on benzene, deactivates the ring), and hence the $\ce{CH3-CHBr+}$ carbocation should be more unstable as compared to $\ce{CH2+-CH2Br}$. Where am I going wrong?

Edit : I have seen this question but it doesn't answer my query.

The reaction of $\ce{HBr}$ with olefins is a relatively fast reaction, in other words, the activation energy required is small.  Further, formation of the carbocation intermediate is the rate determining (slow) step in the overall process.  Considering both of these facts and applying the Hammond postulate we can say that the transition state leading from the reactants to the carbocation intermediate will be an early transition state, it will look more like the reactants than the carbocation.

This suggests that resonance and inductive effects need to be examined with regard to the reactants, rather than the transition state, in order to determine how the molecule will react with an electrophile (the proton).

As the above diagram illustrates, both resonance and inductive effects (in an early transition state that resembles the reactant) suggest that the beta carbon will be more nucleophilic than the alpha carbon.  Therefore addition of the proton (electrophile) will occur at the beta carbon.  This will be followed by rapid addition of $\ce{Br^{-}}$ at the carbon already bearing a bromine atom (the alpha carbon) to yield a geminal dibromide.