How is Bent's rule consistent with LCAO MO theory? Bent's Rule states that $s$ orbitals on a central atom participate contribute more to molecular orbitals directed towards more electropositive ligands. (For a longer explanation, see this question and answers: What is Bent's rule?). Lone pairs are considered to be the extreme limit of an electropositive "ligand".
Many sources claim that this Rule is consistent with LCAO MO theory representations of electron density. It seems, however, that this claim is easily disproved by considering the example of simple molecules of the type $\ce{AH2}$.
We can start with $\ce{H2O}$, whose LCAO MO diagram is well known:

Source: H2O-MO-Diagram.svg by Officer781 on Wikimedia Commons
The orbitals contributing to the O-H bonds are the lower two in energy. Ignoring orbital mixing for now, these are formed from linear combinations of the hydrogen $s$ orbitals with the oxygen $2s$ (giving $\mathrm{2a_1}$) and $2p_y$ orbitals (giving $\mathrm{1b_2}$). The lone pairs are a pure $p_x$ orbital ($\mathrm{1b_1}$) and a combination of $p_z$ with a small contribution from the hydrogen $s$ orbitals ($\mathrm{3a_1}$). Based on the contributing oxygen atomic orbitals, that gives 100% p lone pairs and $50\%$ $p$/$50\%$ $s$ for the bonds, the opposite of the result predicted by Bent's rule, which is that the lone pairs should have more $s$ character.
If we properly account for orbital mixing, there is no effect on the $\mathrm{1b_1}$ lone pair orbital, while the $\mathrm{3a_1}$ lone pair does in fact get more $s$ character, but this increased $s$ character is not expected to exceed $50\%$, since that would just mean a switch of label between the $\mathrm{2a_1}$ and $\mathrm{3a_1}$ orbitals. $50\%$ $s$ in the $\mathrm{3a_1}$ represents the maximum mixed case. That gives a final result of a maximum of $25\%$ $s$ in the lone pairs ($50\%$ in the $\mathrm{3a_1}$ and $0\%$ in $\mathrm{1b_2}$) and $25\%$ in the $\ce{O-H}$ bonds ($\mathrm{2a_1}$ and $\mathrm{1b_2}$), still contradicting Bent's Rule, which predicts that the lone pairs should have more $s$ than the bonds, not the same amount. [One a side note, we know that $\ce{H2O}$ does not represent the maximum mixing case, as $\ce{SH2}$ has a smaller bond angle consistent with even greater orbital mixing, but that's beside the point here.]
The paper by Clauss et. al.[1] reconciles this result by claiming that the 1b1 orbital is "always excluded from the Bent's rule competition for in-plane p-character", but I've never seen any formulation of Bent's Rule that indicates such an exception. If there is such a qualification to Bent's Rule, how do we know what lone pairs to include or exclude?
I conclude from this analysis that Bent's Rule is much like VSEPR - useful in some cases as a pedagogical tool, but ultimately inconsistent with LCAO MO theory. Or perhaps it's only consistent when applied to actual bonds rather than to bonds and lone pairs? I'm not concerned with why people still use it, but I am confused as to why so many people argue that it is consistent with LCAO MO theory. Am I missing something here?
References:

Clauss, A. D.; Nelsen, S. F.; Ayoub, M.; Moore, J. W.; Landis, C. R.; Weinhold, F. Rabbit-ears hybrids, VSEPR sterics, and other orbital anachronisms. Chem. Educ. Res. Pract. 2014, 15 (4), 417–434. DOI: 10.1039/C4RP00057A. Or as a free preprint

Beyond the educational value, Bent's rule has had significant impact on how we understand wave functions of molecules. It's simplicity is almost unique in the world of quantum chemistry; almost everybody can immediately understand it and apply it. Together with the VSEPR model it can lead to quite accurate predictions on the back of an envelope.
Molecular Orbital Theory vs. Valence Bond Theory (MO vs. VB)
In their respective infinite treatment these theories are congruent. For more on this topic please read What is actually the difference between valence bond theory and molecular orbital theory?
Important for this post is that both theories are approximations for the wave function of a molecule, hence they do not describe bonding beyond the stationary point they are applied to, and they operate within the clamped nuclei (Born-Oppenheimer) approximation.
In MO theory it is common to express the molecular orbitals as linear combinations of atomic orbitals. Those MO are usually chosen to be orthonormal. The MO themselves are delocalised, but there are algorithms to transform them into more intuitive VB type orbitals. This will provide us with a Lewis-like bonding picture. Famously used in this regard is Natural Bond Orbital (NBO) theory.
Bent's rule
Unfortunately I cannot find my copy of Bent's paper, so this answer will be based purely on the definition of Bent's rule.
The IUPAC gold book defines Bent's rule (DOI: 10.1351/goldbook.BT07000):

In a molecule, smaller bond angles are formed between electronegative ligands since the central atom, to which the ligands are attached, tends to direct bonding hybrid orbitals of greater p character towards its more electronegative substituents.

Water is a good example to demonstrate Bent's rule and its interplay with the VSEPR model. In $\ce{H2O}$ you expect oxygen to follow the octet rule, hence there will be four electron pairs around it, and it predicts a generally tetrahedral structure of the molecule. The angle for these shapes is about $\newcommand{\degree}{^\circ}109\degree$. Since hydrogen is more electronegative than no ligand, Bent's rule predicts a smaller angle than that.
Probably due to its simplicity, Bent's rule can be applied within the MO and VB bonding pictures.
Coulson's Theorem
Coulson's theorem is the formal theory which applies to Bent's rule (see Wikipedia).
We choose a set of orthonormal AO, as would be consistent with the LCAO approximation.
$$
\langle\chi_i|\chi_j\rangle = S_{ij} = \delta_{ij}
\begin{cases}
0, & i \neq j\\
1, & i =    j
\end{cases}\tag{1}\label{orthonormalAO}
$$
We can transform these orbitals into sets of orthonormal hybrid orbitals.
\begin{align}
\varphi_a &= \sum_i \lambda_i \chi_i &
\langle\varphi_a|\varphi_b\rangle = S'_{ab} &= \delta_{ab}
\begin{cases}
0, & a \neq b\\
1, & b =    b
\end{cases}\tag{2}\label{orthonormalHybridAO}
\end{align}
Let's look at a special case, where we choose atomic orbitals to be $\chi_i$ with $i \in \mathrm{s}, \mathrm{p}_x, \mathrm{p}_y, \mathrm{p}_z$. We can construct following hybrid orbitals (also compare Mathematical form of four hybrid orbitals):
\begin{align}\tag{3}\label{orthonormalSP3}
  \varphi_{xyz}
    &= \lambda_\mathrm{s}\chi_\mathrm{s}
     + \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x}
     + \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y}
     + \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}\\
  \varphi_x
    &= \lambda_\mathrm{s}\chi_\mathrm{s}
     + \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x}
     - \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y}
     - \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}\\
  \varphi_y
    &= \lambda_\mathrm{s}\chi_\mathrm{s}
     - \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x}
     + \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y}
     - \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}\\
  \varphi_z
    &= \lambda_\mathrm{s}\chi_\mathrm{s}
     - \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x}
     - \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y}
     + \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}
\end{align}
If all $\lambda_i$ are chosen to be $\frac{1}{2}$, then we would come to the famous $\mathrm{sp}^3$ hybrid orbitals.
Let's generalise this a bit and drop the normalisation:
\begin{align}\tag{4}\label{hybrid}
  \varphi_a
    &= N\cdot(\chi_\mathrm{s} + \lambda_{\mathrm{p}_k} \chi_{\mathrm{p}_k}) &
  \leadsto
  \varphi_a
    &= \chi_\mathrm{s} + \lambda_{\mathrm{p}_k} \chi'_{\mathrm{p}_k}
\end{align}
From \eqref{orthonormalHybridAO} and \eqref{hybrid}, and ignoring the trivial case:
\begin{align}\tag5
  \delta_{ab}
    &= \langle\varphi_a|\varphi_b\rangle\\
  0
    &= \langle
      \chi_\mathrm{s} + \lambda_{\mathrm{p}_k} \chi'_{\mathrm{p}_k} |
      \chi_\mathrm{s} + \lambda_{\mathrm{p}_l} \chi'_{\mathrm{p}_l}
    \rangle\\
    &= \langle\chi_\mathrm{s}|\chi_\mathrm{s}\rangle
    +  \lambda_{\mathrm{p}_k} \langle\chi_\mathrm{s}|\chi'_{\mathrm{p}_k}\rangle
    +  \lambda_{\mathrm{p}_l} \langle\chi_\mathrm{s}|\chi'_{\mathrm{p}_l}\rangle
    +  \lambda_{\mathrm{p}_k}\lambda_{\mathrm{p}_l}
       \langle\chi'_{\mathrm{p}_k}|\chi'_{\mathrm{p}_l}\rangle
\end{align}
Given \eqref{orthonormalAO} we can simplify this. Note that $\chi'$ is actually a linear combination of $\mathrm{p}$ orbitals, and therefore they don't necessarily have to be orthogonal. The angle between two such orbitals is given by the inner product, i.e. $\langle\chi'_{\mathrm{p}_k}|\chi'_{\mathrm{p}_l}\rangle = \cos \theta_{kl}$.
\begin{align}\tag6
  0 &= 1 + 0 + 0 + \lambda_{\mathrm{p}_k}\lambda_{\mathrm{p}_l} \cos \theta_{kl}
\end{align}
Therefore Coulson's Theorem is
\begin{align}\tag{7}\label{Coulson}
  \cos \theta_{kl} &= \frac{-1}{\lambda_{\mathrm{p}_k}\lambda_{\mathrm{p}_l}}
\end{align}
If you call $\lambda$ the hybridisation index, then you see that the angle between two hybrid orbitals is dependent on the this index. In other words, the angle between two ligands around a central atom determines the $\mathrm{p}$ character of the hybrid orbitals that ideally describe this bond.
What you can also derive from this formula is, given a constant angle $\theta_{kl}$, if you use a hybrid-orbital with increased $\mathrm{p}$ character, the other hybrid-orbital must have less $\mathrm{p}$ character, i.e. $\lambda_{\mathrm{p}_k} < \lambda_{\mathrm{p}_l}$.
This is in principle all that Bent's rule requires in the framework of LCAO-MO theory to be consistent; the argument about the electronegativity of a ligand is based on observation of molecular structure.

I hope the above answers the titular question sufficiently. In the second part I'll try to show where the original arguments fail.
First of all, you are basing your argumentation on even more handwavy mathematics then I did above. You are also not considering the actual molecular structure and the implications of that in terms of the LCAO-MO approximation of the wave function.

The orbitals contributing to the O-H bonds are the lower two in energy.

This is not quite correct. In LCAO-MO (or MO in general) theory all orbitals contribute to bonding.

Ignoring orbital mixing for now, these are formed from linear combinations of the hydrogen s orbitals with the oxygen 2s (giving 2a1) and 2py orbitals (giving 1b2). The lone pairs are a pure px orbital (1b1) and a combination of pz with a small contribution from the hydrogen s orbitals (3a1).

The shown MO scheme does not ignore orbital mixing, and neither do you in your argumentation. If the oxygen's $\mathrm{s}$ and $\mathrm{p}$ orbitals were far enough apart to not mix, then only one sort of these orbitals could interact with the hydrogen orbitals to lead to bonding. This would either lead to a situation where there is no oxygen $\mathrm{s}$-character in the bonds, or all of it.
This basically follows the argument that for the heavier homologues of water the oxygen valence $\mathrm{s}$-orbital becomes the lone pair.

Based on the contributing oxygen atomic orbitals, that gives 100% p lone pairs and 50% p/50% s for the bonds, the opposite of the result predicted by Bent's rule, which is that the lone pairs should have more s character.

I don't have an optimal LCAO-MO solution at hand, but I have the next best thing: a HF/STO-3G wave function with $C_\mathrm{2v}$ symmetry, $\angle(\ce{HOH}) = 100.0\degree$, and $d(\ce{OH}) = \pu{98.9 pm}$. That is off by plenty, but it's Hartree-Fock, so that was expected. The general tenor wouldn't change, if you were to run it on the experimental structure. This is the output of the wave function:
 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-20.252 is O1-s=1.0006
 Alpha occ 2 OE=-1.258 is O1-s=0.7899 H2-s=0.0928 H3-s=0.0928 O1-p=0.0245
 Alpha occ 3 OE=-0.594 is O1-p=0.5365 H3-s=0.2317 H2-s=0.2317
 Alpha occ 4 OE=-0.460 is O1-p=0.6808 O1-s=0.1328 H2-s=0.0932 H3-s=0.0932
 Alpha occ 5 OE=-0.393 is O1-p=1.0000
 Alpha vir 6 OE=0.582 is H3-s=0.3144 H2-s=0.3144 O1-p=0.2947 O1-s=0.0766
 Alpha vir 7 OE=0.693 is O1-p=0.4635 H2-s=0.2683 H3-s=0.2683

Unfortunately this doesn't tell us much, apart from that there is obviously one $\mathrm{p}$ lone pair, which was expected due to symmetry. (In MO theory though, this would still count as a bonding π orbital).
From the above table, one would derive that the majority of bonding comes from MO 3, which is a nearly pure oxygen $\mathrm{p}$ orbital.
I've run an NBO6 analysis on this calculation, which is basically a unitary transformation of the MO, and arrived at the following description:
     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   1. (2.00000) CR ( 1) O  1            s(100.00%)
                                         1.0000  0.0000  0.0000  0.0000  0.0000
   2. (2.00000) LP ( 1) O  1            s( 71.29%)p 0.40( 28.71%)
                                         0.0000  0.8443  0.0000  0.0000  0.5358
   3. (2.00000) LP ( 2) O  1            s(  0.00%)p 1.00(100.00%)
                                         0.0000  0.0000  1.0000  0.0000  0.0000
   4. (1.99926) BD ( 1) O  1- H  2
               ( 59.24%)   0.7697* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000  0.3789  0.0000 -0.7071 -0.5970
               ( 40.76%)   0.6384* H  2 s(100.00%)
                                         1.0000
   5. (1.99926) BD ( 1) O  1- H  3
               ( 59.24%)   0.7697* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000  0.3789  0.0000  0.7071 -0.5970
               ( 40.76%)   0.6384* H  3 s(100.00%)
                                         1.0000
 ---------------- non-Lewis ----------------------------------------------------
   6. (0.00074) BD*( 1) O  1- H  2
               ( 40.76%)   0.6384* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000 -0.3789  0.0000  0.7071  0.5970
               ( 59.24%)  -0.7697* H  2 s(100.00%)
                                        -1.0000
   7. (0.00074) BD*( 1) O  1- H  3
               ( 40.76%)   0.6384* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000 -0.3789  0.0000 -0.7071  0.5970
               ( 59.24%)  -0.7697* H  3 s(100.00%)
                                        -1.0000

This results in two $\mathrm{sp}^6$ bonding orbitals, one pure $\mathrm{p}$ lone pair and a $\mathrm{sp}^{0.4}$ lone pair. From this we would conclude that more $\mathrm{p}$ character is directed towards the hydrogen ligands.

If we properly account for orbital mixing, there is no effect on the 1b1 lone pair orbital, [...]

That is correct, this is due to symmetry.

[...] while the 3a1 lone pair does in fact get more s character, [...]

In your argumentation 3a1 would have been a pure $\mathrm{p}$ lone pair, with mixing it does gain $\mathrm{s}$ character, which is obviously more than none.

[...] but this increased s character is not expected to exceed 50%, since that would just mean a switch of label between the 2a1 and 3a1 orbitals.  50% s in the 3a1 represents the maximum mixed case.

The part with the switch is only true if the energy of these orbitals also switches and the contribution of the hydrogen were equal in both orbitals.
Does this argument also imply that an $\mathrm{sp}$ orbital was the maximum mixed orbital?
I'm very sorry, but I cannot follow the logic as I do not understand what is meant with maximum mixed case. In LCAO you are able to continuously mix $\mathrm{s}$ and $\mathrm{p}$ orbitals, if that actually produces a sensible approximation to the wave function is a completely different topic.

That gives a final result of a maximum of 25% s in the lone pairs (50% in the 3a1 and 0% in 2b1) and 25% in the O-H bonds (2a1 and 1b2), still contradicting Bent's Rule, which predicts that the lone pairs should have more s than the bonds, not the same amount.

This would be exactly the result which you were to expect from tetrahedral coordination: two $\mathrm{sp}^3$ bonds, one $\mathrm{sp}$ and one $\mathrm{p}$ lone pair.

[One a side note, we know that H2O does not represent the maximum mixing case, as SH2 has a smaller bond angle consistent with even greater orbital mixing, but that's beside the point here.]

I am really interested in that side note. I still want to understand what is meant by the maximum mixing case, and especially how a smaller bond angle is consistent with more mixing.


Calculating the percentage of (+)-2- butanol in the sample Specific rotation of (+)-2-butanol is 12.5$^{\circ}$ . A sample of 2-butanol containing both the
enantiomers was found to have specific rotation value of –2.8$^{\circ}$ . What is the percentage of (+)-2- butanol in the sample?

I attempted it by letting the percentage of (+)-2-butanol be $x \%$ and therefore percentage of enantiomers in the mixture would be $2x \%$ due to equal contribution by (-)-2-butanol. Since the net specific rotation of the mixture is given to be negative, we know that percentage of excess of (-)-2-butanol would be $(100-2x)\%$. It is also given that specific rotation of pure (+)-2-butanol is 12.5$^{\circ}$, but I am unable to understand how is that useful here.
In it's solution, they have calculated the percentage of (-)-2-butanol as $\dfrac{2.8}{12.5} \times 100=22.4 \%$ and thus percentage of (+)-2-butanol came out to be $\dfrac{100-22.4}{2}=38.8\%$.
I didn't understand this last step, why is the division of 2.8 and 12.5 giving the percentage of (-)-2-butanol?

A different, more logical solution gives the correct answer as well.
Since the optical rotation of (+)-2-butanol is 12.5 degrees, the optical rotation of (-)-2-butanol is -12.5 degrees. If the mole fraction of the (+) isomer is $x$ and the mole fraction of the (-) isomer is $1-x$, then
$$-2.8 = 12.5x -12.5(1-x)$$
Solving this simple linear equation gives $x = 0.388$, which when converted to mole percentage, is $38.8\%$.


Electrode potential for metal-metal ion and metal-metal insoluble salt-salt ion electrode So I read it in a book that

The potential of metal-metal ion electrode and metal-metal insoluble salt-salt anion electrode is same while their standard potentials are not same. Standard potential are related by the following equation:
$$E^\circ_{\ce{X-/MX/M}} = E^\circ_{\ce{M+/M}} + \frac{2.303RT}{F}\log_{10}K_\mathrm{sp}$$

My question is that $E^\circ$ is just the potential at some given standard condition, i.e. 1 bar and 25 °C, but when $E$ is same for both at all other conditions, why is it that it is not valid for the standard  particular case/condition?

My question is that $E^\circ$ is just the potential at some given standard condition, i.e. 1 bar and 25 °C....

No, $E^\circ$ is the potential at the given standard condition 25 °C, 1 ATM and activities of involved compounds equal to 1.

This may be the source of your confusion.

Standard potentials are for the unit activities. In my example, once for silver ions, once for chloride ions.

The activity of $\ce{Ag+}$ in 1M $\ce{AgNO3}$ Is very different to activity of $\ce{Ag+}$ in 1 M $\ce{KCl}$ with the $\ce{AgCl}$ precipitate.

You cannot find pressure in the Nernst equation. The pressure plays only very marginal role here, like being big enough to avoid water boiling at low pressure.

Usage of each of the standard potentials implies usage of activity of different ions to compute the final potential, or concentations as approximations.

One is using the metal cation activity, the other the anion activity.

One is using the metal cation activity directly, the other indirectly via the salt solubility product.

$$\begin{align}
E_\mathrm{Ag/Ag+}&=E^{\circ}_\mathrm{Ag/Ag+}+0.059\log{\left(a_\mathrm{Ag+}\right)}\\
&=E^{\circ}_\mathrm{Ag/Ag+}+0.059\left(\log{\left(K_\mathrm{s,AgCl}\right)}
-\log{\left(a_\mathrm{Cl^-}\right)}\right)\\
&=E^{\circ}_\mathrm{Ag/AgCl/Cl^-}
-0.059\log{\left(a_\mathrm{Cl^-}\right)}\\
\end{align}$$


What are point groups and the benefits of using them? I have heard vague information about point groups and symmetry before but I am unsure how they actually help me predicting physical properties of molecules. How do I assign point groups and what good does it do me?

Please answer at a low level.

Point groups are a very valuable tool for analysing molecules without knowing much about them. They rely on analysing a molecule’s symmetry and deducing both physical parameters and the shape of molecular orbitals from the information gathered. The downside is that one must know a molecule’s geometry before one can apply group theory to it, but for most molecules the geometry is known anyway.
Picture a molecule, for example ammonia $\ce{NH3}$. Remember that it has a trigonal pyramidal shape with $\angle(\ce{H-N-H}) \approx 107^\circ$ Consider all atoms of the same type (only applicable to hydrogen here) identical. Look at this molecule naively and think what kind of symmetry it has.

Figure 1: Ammonia (image taken from Wikipedia).
One of the easiest symmetry elements that you can find is probably a mirror plane that transects the molecule along one $\ce{N-H}$ bond and places the image of the second hydrogen onto the third hydrogen. They are shown in figure two by the blue-ish ovals. A plane of symmetry is typically called $\sigma$.

Figure 2: Planes of symmetry in ammonia (image taken from Researchgate)
You may also see, that if you stick an axis into the nitrogen atom through its lone pair (the thick, blue axis in figure 2), you can rotate the molecule $120^\circ$ around this axis and it will again look the same. Three rotations along this axis make $360^\circ$, so due to some formalism (which is really helpful if you think about it), the rotation is called $C_3$.
All we have gotten so far is a simple set of symmetry operations but they don’t help much. We need to put them together. We can define the connection of two symmetry operations by saying ‘we’ll perform one first, the other after it’; since neither change the molecule the final molecule will remain unchanged. Now to prove that the symmetry elements present in a molecule constitute a mathematical group, we need to prove the group axioms:

Closure The connection $A \circ B$ of two elements of a group must always be an element of the group.
I just proved this before. A symmetry operation will transform all hydrogens onto a hydrogen position, no two hydrogens on the same and nitrogen will always remain unchanged. We will always gain the same molecule. We just need to make sure we find all possibilities correctly when searching.

Associativity $(A \circ B) \circ C = A \circ (B \circ C)$.
If you think about this practically, it makes sense. Why should it matter whether I rotate first and apply a mirror plane later or the other way around?

Existence of an identity element One element must exist so that $E \circ A = A$ and $A \circ E = A$.
Well, this is easily done; we just include the identity, i.e. no transformation as one of our symmetry operations. It is labelled $E$.

Inverse elements For each element $A$ there must exist an element $A^{-1}$ so that $A \circ A^{-1} = E$.
This requires a bit of working but in the end we can prove this. For $\sigma$, the same $\sigma$ is always inverse element. For $C_3$, $C_3^2$, the rotation around $240^\circ = -120^\circ$ is the inverse element.

The good thing is that if we didn’t identify all the possible operations initially, we can use the connection $\circ$ to indentify the ones we missed.
Now because there is only a certain number of inherently different groups around, they have been given names for easier identification. Our ammonia example would be the point group $\bf\it C_\mathrm{3v}$. $C$, because we have a rotation axis, $_3$ because rotation is $\frac{1}{3}$ of the full circle and $_\mathrm{v}$ because the planes of symmetry are vertical to the rotation axis (rather than horizontal).
(Naming point groups is rather simple. Find the most important axis — the rotation with the highest index. If that is the only rotation, it is a $C$ group. If there are rotations perpendicular to that one, the group is called $D$. If there is an improper rotation, it is labelled $S$. The highest index of rotation is also the group’s index. $_\mathrm{v}, _\mathrm{h}$ and $_\mathrm{d}$ denote planes of symmetry vertical, horizontal or diagonal to this axis. Finally, $C_s$ is the group that contains only one plane of symmetry, $C_i$ contins only the inversion and finally, tetrahedra ($T_\mathrm{d}$), octahedra ($O_\mathrm{h}$) and icosahedra ($I_\mathrm{h}$) have their own names due to their high symmetry.)
Part of the beauty of point groups is already apparant here. If we know what symmetry operations transform the atoms onto each other, this must also be true for some physical properties — those that can be expressed as vectors. Most notably, a dipole moment must transform onto itself during any symmetry operation, so ammonia’s dipole moment must be along the rotation axis. Molecules that display inversion symmetry cannot be a dipole. Molecules can only be chiral (but are always chiral) if their point groups do not contain inversion, planes of symmetry or improper rotation (a rotation followed by a mirror image perpendicular to the rotation axis). There are more examples, but these are probably the most important.
Finally, for each point group a character table exists. This groups sets of similar symmetry operations together according to a set of specific rules (e.g. the two rotations $C_3$ and $C_3^2$ will be grouped together) whose details I won’t go into, and also includes irreducible representations. Any spec of the molecule (and all its symmetry-equivalent specs) must transform as one of these irreducible representations. The $\mathrm{p}_z$ orbital of nitrogen in ammonia (the one along the $C_3$ axis) transforms according to the irreducible representation $\mathrm{a_1}$ — essentially, $\mathrm{a_1}$ means that something will always be the same. The $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals together will transform as $\mathrm{e}$. Knowing the irreducible representations of atomic orbitals helps to predict molecular orbitals, because only those of the same irreducible representation will mix and combine (for others, the constructive overlap is always $0$).
You can google character tables to find lists of them, and thereby also which point groups actually exist.

Point groups are even more powerful than shown here; but I tried to keep the application parts low-level. There are a number of questions here on Stack Exchange tagged group-theory that go more into depth.


Why is the pressure correction added in the Van der Waals equation? Since real gas particles experience attraction to each other, that means the pressure must be less from what we would expect. So why do we have to add a term to correct it?
For example, in the correction for volume, we subtract a term to find the effective volume or the "real" volume. So, it follows that the "real" pressure must be less than the given one and so, should we not subtract a term?

This is mostly with regard to solving problems based on this law, where the values of pressure, volume, $a$ and $b$ are given

You may consider the vdW equation as the deviation from the ideal gas state equation $pV_\mathrm{m} = RT$.

If you look at the vdW equation

$$\left(P+\frac a{V_m^2}\right)(V_m-b)=R T$$

you can easily see why the corrections have  their signs.

The pressure term correction has the positive sign,
as the real gas has  due molecule cohesion  smaller volume than  an ideal gas at the same pressure. Like if it was under bigger pressure than it really was.

Increasing $p$ term by a positive correction leads to lower predicted volume than for ideal gas, at the same pressure, because of cohesive force.

$$V_m=\frac {R T}{\left(P+\frac a{V_m^2}\right)}$$

The correction simulates this bigger pressure the ideal gas would need to get the same vol.

The volume term correction has the negative sign,
as the gas has at the given volume
higher pressure due molecule own volume.

Decreasing $V$ term by a negative correction leads to higher predicted pressure than for ideal gas, at the same pressure, because of own molecule volume.

$$P=\frac {R T}{(V_m-b)}$$


What is the most acidic proton of 2-methylcyclopent-2-en-1-one? In an acid-base reaction of $\ce{HO-}$ with 2-methylcyclopent-2-en-1-one, which is the most acidic proton?

I had assumed it was one of them coming off C-5 according to IUPAC numbering (the sp3 carbon α to the carbonyl group). However, the answer book claims it to be C-4 (γ when going in the direction of the double bond).
I assumed C-5, since it seemed it would distribute the charge due to resonance more with the $\ce{O}$ atom. What am I doing wrong?

There are two possible enolates that can be formed from your compound: the α deprotonation (left equilibrium arrow in the scheme below) and the γ deprotonation. Both protons have roughly the same $\mathrm{p}K_\mathrm{a}$ values, so deciding which one is more acidic is non-trivial.

However, experimental procedures give us a hint. Usually, the α deprotonation is achieved with strong bases at low temperatures — kinetic conditions. Typically conditions such as LDA in THF at $-78~\mathrm{^\circ C}$. This leads to the conclusion, that the α deprotonation is kinetically favoured. This is more evident in open chain compounds and especially if the α carbon is a methyl group.

The γ deprotonation is generally achieved under more thermodynamic conditions: Weaker bases, warmer temperatures. I didn’t do it myself, so I cannot state conditions off the top of my head, but I would guess conditions such as $\ce{$t$BuOK}$ and $0~\mathrm{^\circ C}$. Therefore, this system must be the thermodynamically favoured one and the γ proton must be slightly more acidic. (But only a tad, or the kinetic product would not be so easily accessable.)

I cannot provide a proper explanation, but it would seem to me like the non-branched π system achieved by γ deprotonation is more stable than the branched one following α deprotonation.


Number of possible isomers (including optical/geometric ones) with C4H8O molecular formula I found 32 isomers but I think I made a miscount.

Chain alcohols: 6, chain ethers: 5, and chain carbonyl compounds
(ketones/aldehydes): 3
Three-member ring alcohols: 7, three-member ethers: 6 (including ones
that have the ether-bond outside the ring)
Four-member ring compounds: 4, five-member compounds: 1

(Note: Enols are not counted because they're unstable)

It would have been nicer if you included the isomers what you have been counting. However, based on the isomer counts, I assumed you are in correct track but originally thought you might have counted meso-isomer twice (as two stereoisomers) for the total count. Thus I drew all possible isomers and listed accordingly:

Consequently, I also counted 32 isomers (same as yours). When compared to your count, I realized your count of chain alcohols (6), ethers (5), and carbonyl compounds (3) are all correct.

However, your count of three-membered ring alcohols (7) and three-membered ring ethers (6) are incorrect. There are only 6 possible three-membered ring alcohols (one less than your count) while 7 three-membered ring ethers are possible  (one more than your count; see the chart). Therefore, it might not be the miscount of meso-isomer (if this is the reason, you should count 8 possible three-membered ring ethers), but be an error on structure drawing.

Your final count of for-membered ring (4) and five-membered ring (1) ethers are also correct. Precisely, only 1 four-membered ring alcohol is possible.

Conclusively, you and I both have the same structure count but wrong counts on different kind of structures. You can compare your drawing with attached chart here and find out where you made the mistake.


Why do ionic compounds have higher melting and boiling points despite having smaller radii? Why do ionic compounds have higher melting and boiling points despite having smaller radii?

I thought that the closer the two ions are, the larger the repulsive force, and therefore it should rather require less energy to separate them (and at further distances apart, a lower melting point is observed because there is less attraction).

This is not a question of ionic compounds vs. covalent compounds, which is covered in this question. To be clear, I am interested in the comparison of ionic compounds with ionic compounds.

As pointed out in a comment, there is a problem with your second sentence:

  "I thought that the closer the two ions are, the larger the repulsive force, and therefore it should rather require less energy to separate them..."

This statement is technically correct, in that the repulsive forces between the two atoms do increase with decreasing bond length. And it is true that two atoms having a charge of the same polarity would exhibit increasing repulsion as the distance between them decreases.

However, in the case of the oppositely charged atoms found in ionic bonds, the attractive electrostatic forces also increase with decreasing bond length, which is the reason for the strong bonding and thus high melting points of ionic compounds.

A discussion of the bond strength of ionic compounds is given in this Wikipedia article, which summarizes that the ionic bond strength can be determined as the sum of the attractive electrostatic energy and short-range repulsive energy.


Concentrating mixtures of volatile compounds My interest is primarily food-related here but I think this is a fairly general chemistry question. Let's say I took some liquid like lemon juice and ran it though a rotary evaporator to get just the volatile compounds. In the case of lemon juice I'd assume I'd wind up with mostly water, and some other volatile compounds (that would smell exactly like lemon juice does). To increase the intensity of this we'd need to somehow remove water.

Similarly a common technique for making extracts is to dissolve volatile compounds of common aromatic foods into alcohol. In case this process left any non-volatile compounds you could certainly use a rotary evaporator. You wind up with a somewhat similar issue where alcohol is the primary liquid.

I'm wondering if it would be possible to concentrate such mixtures so that they contain higher ratios of smell-able volatile compounds to water/alcohol. One thing I've noticed is that most of the characteristic volatile compounds in say lemons have a much higher molecular weight than water or even alcohol. Is there someway to utilize this to like centrifuge out a more concentrated liquid? Are there good methods for either water or ethyl-alcohol specifically?

The simplest method would be fractional distillation: if you have a mixture of miscible liquids with different boiling points, as is the case in your question, then you can heat them slowly. When the boiling point of the most volatile compound in the mixture is reached, it begins to boil and the temperature does not increase until it has all boiled off. The vapour formed is cooled with a condenser and collected as a liquid. As soon as the temperature begins to increase again, all of this compound has evaporated, so it can be removed. Provided the boiling points aren't too close, you could use this to separate out all the compounds in the mixture. A diagram of the set-up (from Wikipedia) is below.

This would be easier with water than ethanol, since it has a higher boiling point so is less likely to boil at a similar temperature to your desired compound.
(The most common way to separate out mixtures in chemistry is chromatography - for example, HPLC or GC - but I'm not sure you'd want to put the results in food! HPLC, which stands for high performance liquid chromatography, involves the use of a solvent mixture which normally includes chemicals like acetonitrile or methanol, neither of which are safe to ingest.)


Planar transition-state of proline racemization facilitated by proline racemase The racemization of proline can be facilitated by the enzyme proline racemase.

Since the active sites of enzymes complement the conformation of the transition-states of the molecules they act upon, the most effective enzyme-inhibitors often resemble these transition states.

This enzyme can be inhibited by the transition-state (TS) analog pyrrole 2-carboxylic acid.

The strength of the bond between the TS-analog and the enzyme, is approximately 160 times stronger than the proline-enzyme bond.

When I was first reading about this, I did not believe it because of the negative charge on the alpha-carbon. I decided to draw the resonance structure of the transition state, to find a double bond between the alpha and carboxyl carbons.

My guess is that this resonance structure would make the transition state relatively planar (as is pyrrole 2-carboxylic acid).

But my question is, how "planar" is the transition state, and how might one determine the exact bond angles at the alpha-carbon?

TLDR: How "planar" is the transition state of proline isomerization, and how might one determine the exact bond angles at the alpha-carbon?

(P.S. Sorry for the bad scaling of my ChemDraw pictures, I don't know how to fix it.)

As others have written in the comments, there are large parts of your question that can't be answered by experiment. However, it is possible to come up with a mechanistic hypothesis (and even better, also an alternative hypothesis) and gather evidence that rejects one hypothesis and is consistent with another.

Structural data supports the mechanism you sketched for the PLP-independent racemases. The structure of proline racemase in complex with the transition-state analog (PDB ID 1W61) shows fairly symmetric binding interactions between the enzyme active site and the ligand:

In the figure (a screen grab from https://www.rcsb.org/3d-view/1W61?preset=ligandInteraction&sele=PYC), the inhibitor is labeled [PYC]700, and it has similar interactions on the left and the right (for example two main chain hydrogen bond interactions with the carboxylic acid group each, and one cysteine residue each in the vicinity of the inhibitor.

The binding pocket is fairly symmetric (view from the "backside" with respect to previous image):

In the paper reporting the structure, the authors point out "two Cys residues optimally located to perform acid/base catalysis through a carbanion stabilization mechanism". They also suggest that the sulfur atoms sandwich proline when it binds, "squeezing" it into a conformation closer to planar than it would be in solution.

However, as the proline-bound structure of a different, putative racemase shows, the active site does accommate proline with a non-planar alpha carbon:

http://www.rcsb.org/3d-view/6J7C?preset=ligandInteraction&sele=PRO

  When I was first reading about this, I did not believe it because of the negative charge on the alpha-carbon. I decided to draw the resonance structure of the transition state, to find a double bond between the alpha and carboxyl carbons.

As the enzyme is stepping through catalysis, the enzyme active site binds to the molecule in different charge states, and itself changes charges as the cysteine residues accept and donate protons. In the cited paper, it is left open whether the reaction proceed in a concerted manner (one proton abstracted on one side while one is donated on the other) or via a carbanion intermediate. They do point out the the pKa of the alpha carbon is  "in the range 21–32", arguing against a long-lived intermediate.

  But my question is, how "planar" is the transition state, and how might one determine the exact bond angles at the alpha-carbon?

As you can read from the excellent comments, there is no good experiment to get at the exact structure of the transition state, and it is not a given that the transition state is planar. However, if the carboxylate group is on one side of the ring before the reaction and on the other side of the ring after the reaction (and stays attached to the alpha carbon throughout), at one point on the reaction coordinate the structure around the alpha carbon has to be planar (or some version of trigonal bipyramidal for a concerted mechanism). Therefore, the active site has to be able to accommodate that structure.

It would be possible to study this system using computational methods, to at least get a sense of the energy landscape in the absence of enzyme. The calculation with enzyme is complicated by the fact that the enzyme undergoes fairly large conformational change upon ligand binding.


Why concentrated sulfuric acid for dehydration and dilute for hydration? To dehydrate ethanol one uses concentrated sulfuric acid:

$$\ce{C2H5OH ->[\text{conc.} H2SO4] C2H4 + H2O}$$

but to go in the reverse direction, dilute sulfuric acid is used:

$$\ce{C2H4 + H2O ->[\text{dil.} H2SO4] C2H5OH}$$

Why is one concentrated, and one dilute? For that matter, how does the sulfuric acid even help in the first place?

Think of the reaction as an equilibrium equation:

$$\ce{CH3CH2OH <=>[H2SO4] CH2=CH2 + H2O}$$

Now, using Le Châtelier's Principle, we can deduce Dissenter's comments.

In concentrated sulfuric acid there is little to no water (note that we also start off with little to no ethylene). The equilibrium shifts to produce products.

In dilute sulfuric acid there is lots of water (and little to no ethanol). The equilibrium shifts to produce reactants.

As for the role of sulfuric acid, it acts as a proton source to enable the loss of leaving group in the forward direction and the electrophilic addition in the reverse direction.


The impossibility of 100% ionic bond Recently, I read the definition of  oxidation state on Wikipedia. It read that a 100% ionic bond is impossible. So what does a 75% ionic and 25% covalent bond mean at all?

A "100% ionic" bond would be a bond whose bonding electron(s) were never in the vicinity of the cation, but rather always in unperturbed valence orbitals of the anion.  That's not far from the truth in a paradigmatic ionic compound like $\ce{NaCl}$, but no matter how electronegative the anion is, the bonding electrons will still experience some attraction toward the positive charge on the cation, and so a detailed model of their wave functions will show a small but nonzero amplitude near the cation.  That's just basic electrostatics; you have a positive charge and a negative charge not that far away from each other, they will attract.

Despite that, we can still talk about 100% ionic bonds as a theoretical limit and an acceptable approximation for modeling some compounds.  It's kind of like how we can use absolute zero as the theoretical zero point of a temperature scale even though nothing in the universe can ever be that cold.


How to create a buffer solution from NaOH and CH₃COOH? In my chemistry book, they say that an approach to making a buffer solution of $\mathrm{pH}=5.09$ is by adding an appropriate amount of strong base $(0.052\ \mathrm{mol}\ \ce{NaOH})$ to $0.300\ \mathrm{l}$ of $0.025\ \mathrm{mol/l}\ \ce{CH3COOH}$. But, I haven’t been able to figure out how they got those numbers. If I am not mistaken the relevant equation is,

$$\ce{CH3COOH +OH- <=>CH3COO^- + H_2O}$$

where the weak acid and conjugate base are respectively $\ce{CH3COOH}$ and $\ce{CH3COO^-}$. My concern is that this equation above goes nearly to completion $(K=1.8\times10^9)$ so I don’t see how this can even be a buffer solution. If not, how do they get the values? I would appreciate any help.

Try considering reaction stoichiometry. You know the extent of the reaction. Now consider stoichiometry.

EDIT: If these are really the numbers your textbook gave you, then the textbook is wrong; you cannot create a buffer system with these numbers specifically. I suspect that you didn't transcribe the numbers from your textbook correctly. In any case the general idea, however, is correct - you can indeed form a buffer solution using sodium hydroxide and a solution of acetic acid.

All you have to do to realize this is to know what a buffer is - a buffer has significant concentrations of both an acid and its conjugate base (or base and its conjugate acid) - and consider reaction stoichiometry.


Confusion about the relation between gibbs free energy and electromotive force $$\Delta G = -nFE$$

My textbook says that we can measure that this $\Delta G$  is measured when the work is done reversibly. But my confusion is how will we work reversibly when we have an electrochemical cell? And what guarantee do we have that while measuring $\Delta G$, E will remain constant? Isn't it possible that after 0.25 moles of electrons have reacted, E would change because it depends on concentration.

What that means is that $\Delta G$ is the maximum work available and therefore is only a thermodynamic limit. That's why $E$ is measured with a voltmeter where no current is flowing (actually extremely low current, i.e. the voltmeter has very high impedance). An analogy would be measuring gas pressure in a pressurized vessel, only in e-chem this is "electron pressure". Once you do work with an e-chem cell, the current will encounter many resistances which translate into voltage drop and in turn are energy losses and E will change.


Hydrolysis and Hydration Does Ammonium Acetate, a salt of a weak base and weak acid, undergo both hydration and hydrolysis in aqeous solution? What is meant by the statement that " Some hydration reactions are hydrolysis"

Hydrolysis is reaction with water, forming usually 2 different reaction products. It can be also named decomposition by water.
Hydration is (usually) reversible chemical bonding of water. It can be also named water addition.
One action of above can preceed or follow the other. If the hydration step preceeds hydrolysis in the reaction kinetics, the distinction may look vaque from some points of view.
Ammonium acetate undergo hydrolysis.

$$
\begin{align}
\ce{CH3COONH4 &-> CH3COO- + NH4+} \\
\ce{CH3COO- + H3O+ &<=> CH3COOH + H2O } \\
\ce{CH3COO- + H2O &<=> CH3COOH + OH- } \\
\ce{NH4+ + H2O &<=> NH3 + H3O+} \\
\ce{NH4+ + OH- &<=> NH3 + H2O} \\
\ce{2 H2O &<=> H3O+ + OH-}
\end{align}
$$


Decarboxylation of Salicylic acid I would like to know how and under what conditions can decarboxylation of salicylic acid (o-hydroxybenzoic acid) occur. I couldn't find any sources for it online. Here are some reasons I think it would happen:

The $\ce{COOH}$ group is out of plane of the benzene ring due to the ortho $\ce{OH}$ group, making it easier to decarboxylate.
The keto form is a beta keto-acid, which makes it easier to undergo decarboxylation.

From (1):

  Salicylic acid decarboxylated to give phenol and carbon dioxide, at a convenient rate for kinetic measurements, at $200-230^\circ$, in a homogeneous solution of benzoic acid and certain other high-boiling solvents.

  ... there was a considerable amount of interest in the possibility of a bimolecular decarboxylation mechanism involving an attack by a proton on the free acid or its anions. A variety of compounds such as anthracene-9-carboxylic acid, substituted cinnamic mesitoic acid, aminosalicylic acid, and salicylic acid, appeared to represent examples of this type of decarboxylation mechanism. In most cases, pseudo-first-order rate constants were
  observed experimentally.

  ... mechanism is also proposed for the decarboxylation of salicylic acid in benzoic acid solution. A competition between an oxygen atom and the terminal ring carbon atom for the proton to give either the undissociated acid or a $\sigma$-complex leading to products is shown by eq. 3.

  The effect of substituents on the rate of decarboxylation of salicylic acid has been used to support a bimolecular mechanism. Electron-donating groups in the ortho or para positions have led to an increase in rate. The increased electron density on the ring carbon atom bonded to the carboxyl group would presumably facilitate an attack by a proton at this position.

  ... An attack by a proton on the terminal ring carbon atom of the salicylate ion would appear, at first glance, to be extremely unfavorable because of the close proximity of a formal negative charge. Any interaction should merely lead to undissociated acid. The hydroxyl group in the ortho position appears to influence the distribution of the charge because of the steric configuration.
  Resonance and tautomeric forms, eq. 5, which do not involve an extensive shift of atoms, can be drawn which would distribute the charge in a manner favorable
  to the proposed mechanism, eq. 3.

  This unique steric arrangement is not possible with m- or p-hydroxybenzoic acid, the effect of which is borne out by the fact that these two acids did not decarboxylate at a measurable rate in benzoic acid solution at $212^\circ$. Since inductive and resonance effects would be expected to be similar for the ortho and para isomers, the steric effect appears to exert a critical influence.

(1) Kaeding, W. W. Oxidation of Aromatic Acids. IV. Decarboxylation of Salicylic Acids J. Org. Chem. 1964, 29 (9), 2556–2559. DOI: 10.1021/jo01032a016


How would substitution reaction results differ between reacting halides with a tertiary bromoalkane vs reacting it with a primary bromoalkane? Say, you conducted two separate reactions; in one, you combined $\mathrm{50~mmol}$ saturated $\ce{KCl}$ solution with $\mathrm{10~mmol}$ 1-bromooctane (assume heating), and in the other you combined $\mathrm{10~mmol}$ saturated $\ce{KI}$ solution with $\mathrm{10~mmol}$ 1-bromoctane. In both cases $\mathrm{0.5~mmol}$ hexadecyltributylphosphonium bromide was used as a catalyst.

So, due to relative charge densities and the fact that the electrophillic carbon is primary, you would have a low yield of 1-chloroocatane vs starting material in reaction 1 and you would have a high yield of 1-iodooctane vs starting material in reaction 2, both cases obtained via SN2 reaction.

Now, suppose you would have repeated the same two reactions, only this time you would have reacted the $\ce{KCl}$ and $\ce{KI}$ with 2-bromo, 2-methyloctane, how exactly would the results differ from the results of the previous two reactions, or would they have been the same?

I know that the reaction would have no "choice" but to react via the SN1 (or perhaps E1) route, as the electrophillic carbon is tertiary and thus very sterically hindered. Also, I know that $\ce{I-}$ ions are good nucleophiles, but that doesn't matter in this case, does it?

Perhaps this reaction would go slower than the previous, owing to the fact that it requires the $\ce{Br-}$ to dissociate, and therefore there would be lower yields of both haloalkane products. But then again, wouldn't it go faster owing to the fact that it is in a polar protic solvent, and that the formed carbocation would be tertiary?

There are (at least) four major factors to consider in predicting the results of these reactions: (a) the nature of the substrate, (b) the strength of the nucleophile, (c) the quality of the leaving group, and (d) solvent effects. Halides are able to function, to varying degrees, as both good leaving groups and strong nucleophiles.

In protic solvents, negative charges are stabilized by the $\ce{\delta^{+}}$ of the solvent hydrogens bonded to some electronegative atom (conventionally limited to $\ce{N, O, F}$). Since nucleophiles carry either a full or partial negative charge, protic solvents tend to reduce the reactivity of nucleophiles. Because smaller nucleophiles are more easily and thoroughly solvated than large and polarizable nucleophiles, the effect on them is more drastic. Consequently, in protic solvents, the relative nucleophilicity of the halides is generally given as $\ce{I- > Br- > Cl- > F-}$. In aprotic solvents, the reactivity is precisely the opposite. Hence, the proper selection of solvent is crucial. In your first example, there is no particular reason I can think of to suppose that the yield of 1-chlorooctane would inevitably be low. Indeed, in an aprotic solvent, the usual choice for SN2 reactions, $\ce{Cl-}$ can be expected to be a stronger nucleophile and poorer leaving group than $\ce{Br-}$. I would expect the yield to be good, especially with an excess of $\ce{KCl}$.

On the matter of the substrate, with a tertiary carbon, SN2 is all but completely impossible. In that situation, elimination reactions are typically favored if some base is present. The halides, however, are extremely weak bases (being the conjugate bases of strong acids), so if a non-basic solvent is selected, substitution by an SN1 mechanism seems most probable. Selection of a relatively non-basic, non-nucleophilic protic solvent would be ideal. (Off the top of my head, DMSO comes to mind, which is only weakly nucleophilic, especially by comparison to the halides.) You're right that the strength of the nucleophile is not the deciding factor in an SN1, since the rate-determining step is the loss of the leaving group and the resultant carbocation formation. In a protic solvent, $\ce{I-}$ should theoretically be more effective as a nucleophile than $\ce{Cl-}$. However, I would still expect the latter reaction to work, particularly if the concentration of $\ce{Cl-}$ is high. I'm fairly confident that the difference in reactivity between, e.g., $\ce{Cl-}$ and $\ce{Br-}$ would be largely negated if the former is available in vastly larger concentration in the solution.

Edit: On a second reading, I just noticed that you labeled the halide solutions as aqueous. I don't know if that's deliberate, but I don't think that would make sense. For one thing, the reagents (namely, the haloalkanes) would be largely insoluble in water. Second, because water is an amphoteric species, it could be potentially reactive.


Which is the most reactive metal: Caesium or Francium? I searched on Wikipedia for reactivity series but it showed me Caesium as the most reactive metal. I know and have studied that reactivity increases as we go down the first group. Then shouldn’t it be Francium more reactive than Caesium than the other way around.

PS: I study in class 10 so please explain in that way only.

The reactivity series consists of elements whose properties are FULLY known and understood. It is on the basis of whether an element can displace another from its salt solution (this is what you've been told at grade 10 Indian schooling level I guess?).
Francium, on the other hand is a radioactive element, with a half life of only 22 minutes for its most stable isotope
(Francium-223). Therefore, we can only make predictions, and not observations, about the nature of its salts and whether they would displace other elements from their salt solution or not.


Antibonding, Nonbonding and Bonding orbitals Recently I came across nonbonding orbitals. I know something about bonding and antibonding orbitals and their formation, but I have no idea about nonbonding orbitals. Could someone explain their formation and the difference when compared with the other two?

In basic MO theory, when two atomic orbitals mix, they form an in-phase (bonding) and out-of-phase (antibonding) molecular orbital. The primary criteria for this mixing are having the same symmetry and being near in energy. If you mix your atomic orbitals together, but find that some of the atomic orbitals have no symmetry match with a similar enough energy, then these unmatched atomic orbitals do not mix and become non-bonding MOs.

Consider hydrogen fluoride. The 1s orbital on hydrogen has the same symmetry as the $p_z$ orbital on fluorine. These mix into a bonding and antibonding orbital. Hydrogen however does not offer any other occupied atomic orbitals with matching symmetry for the $p_x$ and $p_y$ atomic orbitals on fluorine. Therefore, these orbitals remain as non-bonding. Furthermore, the 2s fluorine atomic orbitals are much lower in energy than hydrogen's 1s orbital, so this does not mix appreciably either, and can therefore also be treated as non-bonding.

For elementary concepts, I think you just need to know about how these contribute to bond order. The occupancy of non-bonding orbitals does not have any effect on the bond order and are typically considered lone-pairs.


Is liquid water in equilibrium with water vapor at room temperature? This question was inspired by Why does liquid water form when we exhale on a mirror?.

This question is different from Why does water evaporate at room temperature? because it asks about whether an equilibrium is established. At room temperature under certain circumstances, water molecules leave the liquid phase and end up in the gas phase, as explained in the answers to the linked question. Whether an equilibrium is established is a related but distinct question.

Consider the reversible process of water moving from the liquid phase to the gas phase:

$$\ce{H2O(l) <=> H2O(g)}$$

At room temperature and atmospheric pressure, for which of the following scenarios does this process reach equilibrium?

A glass filled with water
A bottle of water after you poured half of the water out and recapped it.
A zip lock bag filled with water (no air).
Humid air in a zip-lock bag.

According to Wikipedia,

  [...]chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system. Usually, this state results when the forward reaction proceeds at the same rate as the reverse reaction. The reaction rates of the forward and backward reactions are generally not zero, but equal.

I am pretty confident in my answers for 1. and 2., but I am having trouble with 3. and 4.

Scenario 1. is an open system, and unless the air is at 100% humidity, the water will evaporate. This means the forward rate is larger than the reverse rate, and it is not at equilibrium.

Scenario 2. is a closed system, and if the temperature is constant, the water will evaporate until the air is at 100% humidity and the equilibrium is reached.

Scenarios 3. and 4 are not clear to me. Would you say these are at equilibrium because there are no changes over time? Or would you say these are not at equilibrium because either product (3.) or reactant (4.) is absent? What about if there is a tiny air bubble in scenario 3., or the bag is wet on the inside in scenario 4., would that make a difference?

It depends on the circumstances, but mostly not

Under most normal atmospheric circumstances water in an open vessel is not in equilibrium with water vapour in the atmosphere. We know this because water usually evaporates, albeit slowly.

One way to measure whether water is in equilibrium is to measure the humidity which is a measure of the maximum concentration of water vapour under the prevailing temperature and pressure. Humidity is mostly less than 100% which means that the air could have more water vapour in it before the vapour spontaneously condenses. This isn't always true, which is why we get fog and why some countries are unbearable to live in.

At normal temperatures the equilibrium between water and water vapour is established slowly. Any vessel containing water that is open to the atmosphere will try to establish that equilibrium, but that won't be complete until the air is 100% saturated with vapour (humidity=100%). And you'd have to saturate the entire atmosphere in the room or, if the room is open, the entire planet's atmosphere to achieve equilibrium.

That this doesn't happen quickly is obvious. But it can happen locally (we call that weather). It it obviously not true across the whole planet at the same time or the planet would be called "sauna" and not earth.

The point is that the atmosphere is rarely 100% saturated with water vapour, so nothing open to the atmosphere will normally be in equilibrium with liquid water. A close bottle left long enough might establish equilibrium (with 100% humidity) inside the closed bottle, but any exchange with the surrounding air will lower the humidity and the equilibrium might take some time to re-establish.

The key to understanding this is that the kinetics of establishing the equilibrium are slow and the normal atmosphere is not saturated with water vapour. This means that the air above a glass containing open water will have a slightly higher concentration of vapour than the air surrounding the glass but with not be fully in equilibrium because the room surrounding the glass is big. And the atmosphere around the room is even bigger and clearly isn't in equilibrium (we have weather and wind to demonstrate that). The only systems where equilibrium will be established are closed ones which have been left a reasonably long time. atmospheric exchange causes the first; slow kinetics the second of these constraints.


Calculate the partial pressure of NO2 Calculate the partial pressure of $\ce{NO2}$.

$$p_{\ce{N2}}/p_{\ce{O2}}=5$$

$$\ce{N2(g) + O2(g) <=> 2NO(g)}$$

Let’s assume $K=0.1$

$$K= \frac{p_{\ce{NO}}^2}{p_{\ce{O2}}\cdot p_{\ce{N2}}}$$

$$K= \frac{p_{\ce{NO}}^2}{p_{\ce{O2}}\cdot p_{\ce{O2}}\cdot5}$$

But how do I proceed?

Ok.
Now you have 3 variables and 2 equations.
To solve the problem you need 3 equations.

Third equation you'll get from the partial pressure law:

  The total pressure of an ideal gas mixture is the sum of the partial pressures of each individual gas in the mixture.

Tip: suppose total pressure is 1.


How do epoxys cross-link? What's the mechanism of cross-linkage for a typical household 2-party epoxy mix? I was working with some epoxy the other day and I was instructed to mix equal parts of epoxy and epoxy hardener. Stoichiometry at its finest.

The hardener contained "polymercaptans amines" according to the label. Google told me that a mercaptan is a broad name for a class of thiols. Makes sense; the $\ce{-SH}$ group is likely strongly nucleophilic (to better pop open the epoxide rings). Amines contain the also nucleophilic $\ce{-NH2}$ group.

However, what are polymercaptan amines? Do these contain both thiol and amine functional groups? And again, how does cross-linkage work? I presume that this involves opening the epoxide ring. But then does the resulting oxide anion attack another epoxide ring? Wouldn't we then need a 2:1 ratio of epoxide to hardener?

"Epoxy" here refers to an epoxy resin, which is comprised of oligomers that contain at least two epoxide functional groups. Hardening agents, like polyamines and polysufides, react with the resin via nucleophilic attack on the epoxide (as you desribed). Because the hardening agent has multiple nucleophilic groups, it will react with multiple epoxy oligomers. Likewise, the epoxy oligomer reacts with multiple hardening agents. This allows a complex crosslinked network to form.

Looking at the MSDS for a typical epoxy hardener, "polymercaptan amines" is listed as a "blend." Likely this is just a mix of polysulfides and polyamines that was given an ambiguous name by the manufacturer. These molecules likely contain a varying number of nucleophilic functional groups.

  Stoichiometry at its finest.

The stoichiometry is actually not straightforward here at all. It is possible that the total number of epoxide groups in the resin equals the total number of nucleophilic groups in the hardener, but that isn't necessarily the case. The reason you get to mix the parts 1:1 is because the manufacturer provides an approprite blend of hardeners to provide the desired stoichiometry, whatever it may be.

  But then does the resulting oxide anion attack another epoxide ring?

This reaction certainly competes with attack by the hardening agent. Whether it is competitive enough to be relevant depends on the identity and available amount of the hardening agent. It may even intentionally utilized in the crosslinking process.

As a side note, it is unlikely that an alkoxide anion is involved here. Often times a Lewis acid catalyst is added to facilitate crosslinking, which would provide an alkoxide-Lewis acid adduct. Alternatively, there are plenty of hydrogen bond donors available in these mixtures, so a proton would likely be associated with the epoxide during the ring opening, providing a neutral alcohol upon completion of the reaction.


Molecular mechanics force fields for metals? I know of several general molecular mechanics methods suitable for metals, including Dreiding and UFF.

I know it's difficult to get many things right for such classical treatment of inorganic / organometallic molecules, because oxidation state, spin state and other factors can greatly influence geometries and energetics.

But I'm surprised that so few attempts have been made to improve on these methods. Are there no other methods published since 1992? I'm looking for force fields that cover a substantial part of the periodic table (or at least most metal-ligand interactions) and are completely published - proprietary methods don't count.

The modelling of metals using a 'simple' potential is relatively difficult. I have a suggestion for a force field, though it does not fit exactly all your criteria, but I would like to offer it anyway.

Adri van Duin has developed a force field called ReaxFF. This force field is non-proprietary and is readily available in the open-source package LAMMPS. This force field gives pretty good results and can be applied to any kind of elements, it can describe chemical reactions and is also valid for hetero-atomic interactions. That being said, the biggest challenge when using the ReaxFF force field is that you have to fit / train it to your system. This requires, for instance, a database of ab initio calculations that are representative for your system.

If you cannot find any force field, than this might be a interesting alternative to developing a force field completely from scratch.


Chemical Thermodynamic Functions I'm a high school student with basic knowledge about chemical thermodynamics. I have three questions regarding the thermodynamic state functions $H, S, G$. I have searched a lot in textbooks but they don't give the complete information as to whether the properties taken are that of the system or surroundings.

Q1):-Enthalpy is given by the formula

$H = U +PV$ .
 What I want to know is that whether $P$ mentioned above is the internal pressure of the system or the externally applied pressure?

Q2):- The change in entropy is given by the formula
$\Delta S= $$\int_1^2$${\frac{dq_{reversible}}{T}}$. Is the temperature mentioned of the system or the surroundings.

Q3)-On the same lines what temperature is taken to define the Gibbs free energy function:-$G=H-TS$.

Thanks in advance.

The three functions you mentioned are applicable only to thermodynamic equilibrium states of the system, for which the pressures and temperatures of the system and its surroundings match one another to within insignificantly small differences.  In the integral to get the entropy change (Q2), the same matching applies; that is, a reversible path consists of a continuous sequence of thermodynamic equilibrium states.

Note that, at the interface between the system and the surroundings, the temperature and pressure of the system always matches the temperature and pressure of the surroundings (irrespective of whether the system is at equilibrium).  It's just that, along an irreversible path of the system (for which the system passes through a sequence of non-equilibrium states), the pressure and temperature within the system vary with spatial position, and their average values (averaged over the volume of the system) do not match those of the surroundings at the interface.


Do disaccharides undergo anomerization? I am interested in disaccarides of D-glucose linked 1-4 style. Anomerization is the interconversion of alpha-D-glucose into beta-D-glucose. I realize that this could only be possible on one of the glucose sub-units: The other glucose is obviously unable to inter-convert due to the glycosidic bond preventing ring-opening.

I do not see what would prevent it:

Not only is there no reason to think anomerization does not occur in the disaccharide you mention, it most definitely does occur.

As long as water has access to carbon-1 of a glucose unit, there will be an open-chain form at that site, and then when it re-closes, the possibility of alpha- and beta-anomers.  This will be the case for di-, tri- tetra-, oligosaccharides, polysaccharides, etc... and applies to monosaccharides of any carbon length or type (aldoses or ketoses).

The reason is that hemi-acetals or hemi-ketals are intrinsically unstable in water. By contrast, when that carbon is a full acetal or full ketal (a so-called non-reducing sugar) then it is stable in water, and anomerization does not occur.  Examples include trehalose and sucrose.


Is there a potentiometric method for monitoring salt formation in organic solvent? I work in a pharmaceutical R+D lab, and currently aim to develop a validatable potentiometric method for determining the optimal amount of acid to add to a basified small organic molecule for conversion to an ionic species (i.e. sulfate salt, hydrochloride salt). Our API (active pharmaceutical ingredient) has two protonation sites, so I also need to determine the stoichiometric ratio of API to counterion, and therefore which specific salt is present at specific points of the titration curve. Ideally, this will be carried out in methanol or ethanol.

Is it possible to accomplish these goals by titrating with acid and monitoring with a conductivity meter/probe? A consultant has advised me to do so, and I have a meter with two conductivity probes. I've already tried using a pH meter, but neglected to factor in that pH is only relevant to aqueous systems. The pH method did produce decent looking titration curves with obvious inflection points. However, as major amounts of salt 'crash out' of solution, meter drift becomes an issue, the reference junction gets clogged, etc. It gets messy and I have to continually clean the probe. I've read that they make pH meters that can withstand as much as 80% organic in solution, but this is not representative of our 100% organic protocol. Furthermore, I'm having trouble finding non-aqueous calibration solutions for the pH (and conductivity) meter.

My idea: Make a set of organic conductivity calibration solutions using methanol (and/or ethanol) and neat sulfuric acid at concentrations of, say, 0.010 M, 0.10 M, and 1.0 M (arbitrarily chosen), and then carry out the titration. Ignoring solutes, this is the most representative solvent system
I can think of since I would be adding acid to methanol during the titration.

Or, am I completely going about this the wrong way? My boss and I have been researching this for over a month and have found nothing specific to the application. I suspect that I'm over complicating the matter. This type of process is surely carried out daily in drug manufacturing facilities (tons of API's are in salt form, and GMP facilities require method validation/quality assurance/etc of all analytical methods), so I assume that there is a known and developed method for this out there somewhere.

Add methanol with excess $\ce{H2SO4}$ to your solution. In acid solution to can use HPLC to determine the concentration of your mystery goo compound.
Add water to ppt whatever organic goo you're making.
Mechanically separate your organic goo (filter, centrifuge.). Instead of 2-3 maybe solvent extraction?
Back-titrate excess $\ce{H2SO4}$ in water/methanol solution with $\ce{NaOH}$ and phenolphthalein.

There has to be a wealth of information on the titration of "dicarboxylic acids" in methanol.


Is it likely that increased understanding of quantum physics will change our understanding of chemistry? Reading that the large hadron collider will be up and running with twice as much energy in March 2015, I was curious whether our understanding of subatomic particles has changed our understanding of chemistry.

Edit: To attempt to narrow down the question to something with a definitive answer: has the discovery of subatomic particles changed experimental chemistry?  If so, what is an example?

Well, you asked for an example, I'll give you one.
"Quantum" is undeniably the most effective keyword our science has seen, it's changed our 'understanding of chemistry and physics' drastically over the past century.
For example, between 1900 to 1925 (years vary from source to source, but this era exists, that's what that is important), in which Bohr's model was the main focus of study (as wikipedia puts it), is known to be the era of "old quantum theory". There were series of theories, which started by Max Plank, and partially came to results with the final modifications of Sommerfeld; and ended in Erwin Schrodinger's famous model. New efforts led to new results and therefore the "old quantum theory" was never consistent.
What was all of this about? Scientists are always eager to find out the differential behavior of matter and energy. Sub-atomic particles (in first level, n, p & e) were discovered in order to explain the phenomenons that Roentgen saw (radioactivity of Uranium and other heavy elements in 'pitchblende'); Faraday experimented (the $\ce{SnCl2}$ solution experiment); Greek Leucippus and Democritos tried to explain; and so on. After every discovery, since the 'rules' that were discovered usually were up to a point in explaining what's around us, we change them and make new discoveries. And that was the reason of what Einstein, Bell and people like those did. They corrected the quantum for it to be what we know today.
So you see, the science was always about understanding; so why wouldn't a big discovery like quantization that led to quantum mechanics be able to change our direction of thought, either we are chemists, physicists or simply humans?
Links you might find helpful: About the old quantum theory in wikipedia; About the old quantum theory 2; Bell's theorem(which linked classical quantum to quantum mechanics) and finally quantum mechanics according to wikipedia.


What is an effective fabric-friendly solvent for aluminium chlorohydrate? I have recently found out that the white/yellowish "sweat" stains one  occasionally finds on t-shirts might be aluminium chlorohydrate polymers.

While talk about aluminium and antiperspirants is all over the press and there are lots of DYI-instructions for removing them, they all sound quite pseudoscientific.  I find it strange, for instance, that some guides recommend both vinegar and/or baking soda as a remedy...

Any ideas what else I could use to dissolve such stains without also dissolving much of the pigments?

My undershirts have turned yellow in the armpit area without the benefit of aluminum antiperspirants. I assumed the stains were from carbonaceous exudants from my pores - or the biogenic products of bacterial life on my skin.
So I bleached the undershirts, soaked them in a diluted bleach (sodium hypochlorite) solution (about 1 in 10, I think). BIG improvement. Not perfect, but very good. One of the side effects was that the undershirts turned blue in the bleach because of the whiteners in the surfactant used to wash them. But that faded as the bleach continued to oxidize the stains.
You can go overboard and use more concentrated bleach: this will perk up the entire undershirt! Interestingly, it seemed to not only clean the underarm stains, but also attack/weaken the fabric there, so the stains went away, but small holes started to appear there, which grew larger and larger until the undershirts became nice dust cloths.
Next time, I think I will try an oxygen laundry bleach, even if it takes longer. Or H2O2.
Addendum: the supposition that the brown deposits in the armpit area are due to aluminum compounds is widespread. Even undershirt makers suggest it - but of course, take no blame for providing the substrate that shows the stain. One manufacturer (https://thompsontee.com/blog/the-science-behind-your-pit-stains-and-what-to-do-about-it/) even has a link to fixing it. What do they suggest? Change your antiperspirant; take proper care of your clothing; wear a sweatproof undershirt. (They offer the best sweatproof undershirt: 20% off!) But how to get rid of armpit stains?
#1. Vinegar, baking soda and HYDROGEN PEROXIDE.
#2. OxiClean or Raise Stain Remover (mixed reviews)
Nivea makes deodorants and has a site that recommends ways to remove armpit stains: try an acid. For example: #1: HYDROGEN PEROXIDE #2. Baking soda (?) #3.White vinegar #4.Lemon juice. https://www.nivea.com.au/advice/skin/removing-deodorant-sweat-stains
The desire to use environmentally safe washing agents in certainly admirable, but when you have to do a job, it is quicker to use the powerful tool. Of course, there is a learning curve. I love bleach because its activity is so adjustable and controllable - from an additive at low concentration to eradicate gingivitis by waterpicking to clothes cleaning to mildew removal from house siding, and probably a million more uses.
But I suppose hydrogen peroxide will be adequate. Or OxiClean.


Standard electrode potential for electrolytic vs galvanic cells Whys is the standard electrode potential (-) for electrolytic cells but positive for (+)  galvanic cells.

  If E°cell > 0, then the process is spontaneous (galvanic cell)
  If E°cell < 0, then the process is non-spontaneous (electrolytic cell)

  Thus in order to have a spontaneous reaction, E°cell must be positive,
  where:

  E°cell = E°cathode − E°anode

^ above taken form wiki.

Also can someone please explain the follow:

In both cells is the electron flow from

anode -> cathode ??

For electrolytic cells isn't the electron flow from the battery (-) anode to cathode (-) of the cell.

Do we then show this by electron flow as cathode (-) ------> anode (+) [in the actual cell itself]??

Potential refers to the amount of energy produced or consumed by the reaction. If the sign for potential is (+) than the reaction is spontaneous (moves on its own). If the sign is (-) then the reaction requires energy to proceed.

In the case of electrolytic cells, the cell is used to decompose a compound. this can require energy depending on the reaction. Think of water and how much energy it takes to produce hydrogen gas and oxygen, that's an example of electrolytic cell doing work.

The definition of an anode is an electrode in which a positive charge flows into an electrical device, this is the direction of the electrons.

For both types of cells direction of electron flow would remain the same, but with electrolytic cells, there would be an additional object, the electrolyte, in the diagram.

Here is a decent picture i found that also does some explanation.


Why Ionic compounds dissolve into water First, I would like to say that I understand the basic idea of disassociation and solubility. However, as I study intermolecular forces I feel like this basic Idea is being challenged. My understanding from basic chemistry is that ionic bonds are much stronger than covalent bonds, due to the actual exchange of electrons and the formation of cations and anions, causing an attractive force between the opposite charges to "hold" the compound together. When explaining solubility with IMF theory, however, I fail to understand how the IM forces present in water--dipole-dipole, H-bonds, and dispersion--are able to overcome that attraction between the cation and anion present in the ionic bond; because, I was under the impression that actual bonds are always stronger than IMF bonds. So, for instance, does it not infer that if, say, Sodium Chloride is placed into water, and the attraction of the h-bonds in water cause the NaCl to essentially break apart into ions, that the inherent attractions between cations and anions in an ionic compound are weaker than the attraction force of the h-bond water molecule?

It's not that ionic bonds are weaker than hydrogen bonds, but that the energy cost of breaking the ionic bonds is well compensated by the gain in solvation energy when you put NaCl in solution.  Solvation energy drives the spontaneity of this dissolution process, because of entropic considerations.


Which Chemical Database Software Are Available? By chemical database software I mean software with similar functionalities to Instant JChem in that they have a large database of different compounds and have the basic capabilities you, or at least I expect from a chemistry software. Namely:

Can approximate the chemical and physical characteristics of the compounds in question.
Can approximate products of reactants and their corresponding proportions.
Can show the 2d and 3d structure of compounds in the database.
Can connect to online chemical databases like PubChem, ChemSpider, NCBI, etc.
Has a molecule editor.

I would also prefer it if it is free of charge.

A widely used database in academia and industry is ChemACX.The only one of your requirements that I don't believe it does is the product prediction.

http://www.cambridgesoft.com/Ensemble_for_Chemistry/ChemACX/Default.aspx


How to make p-chloroanisole from methanol and benzene? I am tasked to synthesise para-chloroanisole from methanol and benzene. Other reagents are permitted.

Would it be correct to have the benzene attack the methanol $\ce{H}$ and then attack the carbocation that forms with the methoxy ion? From there use $\ce{Cl2}$ and $\ce{AlCl3}$ which would install para and ortho or a combination?

Also if you write one equivalence, does that suffice to say that you are only having it attach once, so that you would only have a max of two groups at the end for each possible product e.g. addition at the 1,2 positions and 1,4 positions?

Benzene is not basic enough to be protonated by methanol, which is precisely the first step you are proposing.

The use of free methanol to generate an aromatic methyl ether immediately inspired me to take a look at the Buchwald-Hartwig coupling. While the website only mentions the applicability to phenol couplings, the bible of organic chemistry, Strategic Applications of Named Reactions in Organic Synthesis also mentions the possibility of using alcoholates as a coupling partner.[1]

This leaves us the question of how to get to a compound suitable to be subjected to a Buchwald-Hartwig coupling. We would need an aryl halide into which the palladium catalyst may insert. Fortunately, we need halogenation anyway, so we might just make it a two-step procedure. Chlorination of benzene is well-known and well-taught, so I decided to check whether bromination of chlorobenzene is also known. Turns out that Ferguson et al. published $89~\%$ yield of the bromination of chlorobenzene using bromine and $\ce{AlBr3}$ in $\ce{CS2}$ — fully para-selectively.[2] This gives us the following sequence:

$$\ce{C6H6 ->[Cl2/AlCl3] Cl-C6H5 ->[Br2/AlBr3][CS2, $89~\%$] Cl-C6H4-Br ->[Pd(OAc)2, NaOMe][base] Cl-C6H4-OMe}$$

Wherein both the latter two are para-substituted.

The required sodium methanolate is easily synthesised from methanol using sodium hydride:

$$\ce{MeOH + NaH -> NaOMe + H2 ^}$$

References:

[1]: L. Kürti and B. Czako: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 70.

[2]: L. N. Ferguson, A. Y. Garner, J. L. Mack, J. Am. Chem. Soc. 1954, 76, 1250–1251. DOI: 10.1021/ja01634a013.


How do I interpret characters that are not 1 or -1 in a point group table? When the character in a point group table is listed as 1 or -1, I understand how an object/orbital etc. transforms to give either itself or the negative of itself.

When the character in a point group table is 2 or -2, what does this mean in terms of transformation? To understand this would be very helpful in terms of assigning the correct irreducible representations to atomic orbitals/LGOs to then determine which pairs mix.

Similarly, I do not understand how to interpret a character of sin or cos of an angle, nor a character of 0.

I am hoping there is a conceptual/visual way of explaining this as I am not up to speed on the matrix math, and the class that this is relevant too does not include that math either.

When you observe that in the characer table of a symmetry group there are irreducible representations where the character of the identity operation (E) is 2 or higher, you may find that the character of some symmetry operations is 0, +2, -2, etc. This is due to the fact that this IR are degenerate.

To illustrate this idea, consider that you trying to determine the symmetry of the MO of bezene (D6h). You will find that some of these orbitals have the same energy (are degenerate) for example

When you apply the symmetry operations of the D6h point group you can check that what you obtain is a linear combination of these orbitals. If you express this linear combinations in matrix form, the character of the matrix will give you the value that appears in the character table for the symmetry of that orbital. In this case these orbitals have a E1g symmetry.


Acidic strength of diethyl malonate vs ethyl acetoacetate Which would be more acidic, diethyl malonate $\left(\ce{CH2(CO2Et)2}\right)$ or ethyl acetoacetate $(\ce{CH3COCH2OOC2H5})$

In the above compounds, I have marked the hydrogen which I think will be the most acidic. Now removing that hydrogen would yield a carbocation. Both the carbocations would be equally stabilized due to resonance effect. In fact diethyl malonate acid would be more stabilized due to the extra inductive effect of the extra oxygen. Yet Wikipedia shows that ethyl acetoacetate has a p$K_\text{a}$ less than that of diethylmalonate acid.

What is making it more acidic?

The hydrogens which you have chosen are correct but both the carbanions are NOT eqaully stablised via the resonance. In case of malonic ester, the elctrophilicity of the carbonyl(c=o) is reduced due to the resonance from the oxygen beside it. So the carbanion will not be pulled very effectively into resonance as the carbonyl becomes electron rich due to the oxygen of the ester.

However, in the ethyl acetoacetae only one of the carbonyl groups is affected by this reasonace. The other one is still highly elctrophillic. So it stablises the carbanion even more. So it has a lower pka(more acidic) than the malonic ester.


Is this the NMR spectrum of 3-methyl-1-butanol? It is a $\ce{^1H}$-NMR spectrum and I think the signal near 3,61 ppm is the CH2 group next to the hydroxy group. But which signal is the hydroxy group and which protons are in the multiplet?

Welcome to ChemistrySE.
In short: is the spectrum consistent with 3-methyl-1-butanol? No. The triplet splitting and integral of the signal at 0.86ppm is inconsistent with an isopropyl sidechain. This suggests a methyl group attached to a methylene, $\ce{-CH2-CH3}$.
The signal at $\pu{3.6 ppm}$ has a chemical shift symptomatic of an alcohol, as you suggest. It has diagnostic splitting of $\pu{4.2 Hz}$, also consistent with coupling to the alcohol peak. So, this peak is a doublet of triplets..... which makes it look a bit like a quartet. To see this coupling, the sample needs to very dry to eliminate exchange decoupling. So where is the alcohol peak? In the absence of any exchange, and in chloroform (most likely solvent used given the spectrum shown), alcohol peaks usually come in the region of about $1.2$-$\pu{1.5 ppm}$. So possibly under the multiplet. Which would then leave you with 6 protons to account for......or $3 \times \ce{CH2}$ groups. The peak at $\pu{1.5 ppm}$ has coupling that suggests it is attached to the $\ce{-CH2OH}$, giving a $\ce{-CH2CH2OH}$ group. There are 14 protons in all. You've got a $\ce{-CH2-CH3}$, a $\ce{-CH2CH2OH}$ and the rest are probably $\ce{-CH2\!-}$. How might you put them together?
In the real world, identification of complete unknowns are rarely done using $\mathrm{^1H \ NMR}$ alone. Mass Spec, $\mathrm{^{13}C}$, and a range of $\mathrm{2D}$ options are very useful.
And to answer a follow-up question below, the splitting at 1.54 is from the protons on the beta carbon, and is a triplet of triplets. When coupling constants are close (6.5 and 6.8) this will appear as a pseudo-pentet, as shown below:


Rate constant and Michaelis Menten Please consider the following figure of a concentration over time (msec) diagram with variables for an enzyme, a substrate bound enzyme and a product.

Recall that $\ce{E + S \leftrightharpoons ES \rightharpoonup E + P}$ and also recall that the velocity constant of the respective reactions are noted $k_1 , k_2, k_3$.

What I would like to know is:

On the plot provided, is the slope of $[ES]$ equal to $k_2$ and is the slope of $[E]$ equal to $k_1 + k_3$ thereby giving some insight/explanation for $k_m$ which is used to determine the affinity of an enzyme?

Any more details on Michaelis Menten is greatly appreciated but please note that my level of understanding is limited on the topic.

Given $\ce{E + S <=>ES -> E + P}$

$\frac{d[\ce{ES}]}{dt}=-\frac{d[\ce{E}]}{dt}=  k_1[\ce{E}][\ce{S}]-k_2[\ce{ES}] -k_3[\ce{ES}]  $

  is the slope of $[ES]$ equal to $k_2$ ?

no

  is the slope of $[E]$ equal to $k_1 + k_3$

no


Do chemical compounds have spectral lines, or only elements? I know that elements (ex. Ag) have spectral/emission/absorption lines, but do chemical compounds (ex. NaCl) have spectral lines as well? Ones that when seen you can tell that that compound (NaCl) is made up of certain elements (Na and Cl).

Yes, but for "spectral lines" the gaseous state would generally be the implication. In solids and liquids photon interactions are generally over "bands" of energy not a very discrete wavelength as in a "spectral line."


How do chemists verify certain phenomena like 1,2-hydride shifts, 1,2-methyl shifts and ring expansions? I've been learning about organic chemistry independently on a ship without an instructor or university resources like a lab setting for the past year. I was wondering how do chemists verify that their mechanisms are correct?

For example, take 3-methylbutene and add the reagent HCl. From the 1,2-hydride shift we get 3-chloro-3-methylbutane. For a 1,2-methyl shift take 3,3-dimethylbutene with the same reagent HCl. We know that from the mechanism we get 2-chloro-2,3-dimethylbutane. But how do you verify that these are the major products?

I wanted to put this as a comment but I don't have enough rep :p

It happens the other way round, first the products are obtained, then the mechanism is proposed.

The point is, according to me that once it has been verified which product is major and which minor, chemists may try to come up with a suitable explanation for the product obtained, on paper. They may try some things, come up with possible mechanisms and sometimes even phenomena like hydride shift, methyl shift etc. Once they come up with such an explanation, they try to apply it to other related problems and upon reaching the right solution every time, they can say this is what might have happened to a high degree of certainty.

So you see, chemistry is also sometimes glorified problem solving :p


Why is the acylium ion stable? Why is the acylium ion stable? Doesn't the inductive effect of oxygen destabilize it? Compared to an alkyl cation, is an acylium ion more stable or less stable?

The acylium ion is stabilized by the positive mesomeric effect of the lone pair of electrons on the oxygen atom onto the empty p-orbital of the $\ce{C+}$. This helps achieve octet completion of the carbon atom, and thus imparts stability to the carbocation.

(Image by Jü CC BY-SA 4.0, from Wikimedia Commons)

Resonance effects mostly dominate the weak inductive effect, so here the destabilizing negative inductive effect of oxygen atom has only a negligible role.

An alkyl carbocation has no such positive stabilizing mesomeric effect, rather only hyperconjugative effects, which are not able to provide as much stability as is possible in the acylium ion.


Buffer made of salts of polyprotic acid: reactions I was trying to make a buffer with $\ce{K2HPO4}$ to use it in a bacterial culture.
If I use $\ce{K2HPO4}$ I think I have to add $\ce{KH2PO4}$. I am trying to think it writing the equations of the process but I have troubles.
I think the buffer's reactions will be:

Salt dissociation in water:
$$\ce{KH2PO4_{(aq)} <=> K+_{(aq)} + {H_2PO4^-}_{(aq)}} $$
$$\ce{K2HPO4_{(aq)} <=> 2K+_{(aq)} + {HPO4^{-2}}_{(aq)}}$$

Anion hydrolysis:
$$\ce{HPO4^2-_{(aq)} + H2O_{(l)} <=> H2PO4^-_{(aq)} + OH-_{(aq)}}$$

$$ \ce{H2PO4-_{(aq)} + H2O_{(l)} <=> HPO4^2-_{(aq)} + H3O+_{(aq)}}$$

If I add acid:
$$\ce{K2HPO4_{(aq)} + H3O+_{(aq)} <=> H2O_{(l)} + H2PO4-_{(aq)}} $$

If I add base:
$$\ce{H2PO4-_{(aq)} + OH-_{(aq)} <=> H2O_{(l)} + HPO4^2-_{(aq)}}$$

Is it correct? And if so, is it possible to add only $\ce{K2HPO4}$ and water to make the buffer?

For use in a bacterial culture, you no doubt want a buffer solution with a $\mathrm{pH = 7.20}$, which corresponds to the $pK_{a2}$ for phosphoric acid. At that pH $\ce{[H2PO4-] = [HPO4^2-]}$. The $\ce{K+}$ ion is basically a spectator ion and doesn't influence the pH directly. So typically you'd start with a solution which has equal molarities of $\ce{KH2PO4}$ and $\ce{K2HPO4}$ and then adjust the pH with $\ce{H3PO4}$ and $\ce{KOH}$ using a pH meter to get $7.20$ "exactly."


How much heat is needed to break CO2 down to CO? I have found that because the oxygen is double bonded to carbon, it takes about $\mathrm{1500\ ^\circ C}$ to break it down from $\ce{CO2}$ to $\ce{C}$ and $\ce{O}$. But, I am curious, would it take less heat energy to break $\ce{CO2}$ down to $\ce{CO}$ and $\ce{O}$?

Heating CO2 at atmospheric pressure:

At 1000K it is still essentially all CO2.

At 2000K: 98% CO2, 1.4% CO, 0.7% O2

At 3000K: 44% CO2, 36% CO, 16% O2, 4% O

At 5000K: 50% CO, 50% O

only at even higher temperature does significant atomic C appear (9% at 6000K).

See Thermal decomposition of carbon dioxide in an argon plasma jet for more information.


Why does adding water to a saturated solution increase the number of ions present in the solution? In my book, it is stated that when some water is added to a test tube having a saturated $\ce{Ag2SO4}$ solution, with its solid at equilibrium, the number of $\ce{Ag}$ ions in the solution increases.

I don't understand it. If $\ce{Ag2SO4}$ solution is saturated, it means that no more solute can dissolve, and if we add water, the amount of dissolved solute should not change.

Also, if the amount of dissolved $\ce{Ag}$ increases, would the $K_\mathrm{sp}$ value also increase?

Consider a saturated solution of $\ce{Ag2SO4}$. The following equilibrium is attained.
$$ \ce{Ag2SO4 <=> 2Ag+  + SO4^2-}$$

Adding water accounts to increasing the volume of the solution. Hence more amount of solute can be dissolved since solubility of a salt depends on the amount of salt dissolved per unit volume. Therefore, there is an increase in the no. of ions produced in the solution. But as Avnish Kabaj mentioned, $[\ce{Ag+}]$ remains the same.

Also as MollyCooL said, $\pu{K_{sp}}$ doesn't change as it is a constant at a particular temperature.