Molten lithium carbonate conductivity Does anyone know or could make an educated guess as to the conductivity/resistivity of molten lithium carbonate? Ideally, around 750 degrees C.

The conductivity of molten lithium carbonate has been determined for vast range of temperatures (Ref.1). The abstract of which states that:

  The present communication reports the results of an investigation of the properties of surface tension, density, and electrical conductance for molten $\ce{Li2CO3}$, $\ce{Na2CO3}$, and $\ce{K2CO3}$ and some mixtures in the temperature range of $750^\circ –\pu{1000 ^\circ C}$. The surface tensions are approximately twice the values for the corresponding chlorides; the densities and electrical conductance are quite comparable to those of the chlorides. The ionic nature of the molten carbonates is examined from the physicochemical criteria based on these properties; and the mechanism of electrical transport is considered in the light of current theoretical concepts. Relative to $\ce{Na2CO3/K2CO3}$ mixtures, the surface tensions and partial molal volumes indicate but little deviation from the predictions for thermodynamically ideal mixtures.

The conductivity of molten lithium carbonate/metal carbonate mixture has also been determined for range of temperatures (Ref.2), the abstract of which states that:

  The electrical conductivities of the molten binary salt mixtures $\ce{Li2CO3–X2CO3}$ ($\ce{X: Na, K, Rb,}$ and $\ce{Cs}$) and $\ce{Na2CO3–Z2CO3}$ ($\ce{Z: K, Rb,}$ and $\ce{Cs}$) were measured using an ac two-probe technique and electrochemical impedance spectroscopy. The electrical conductivities of the eutectic compositions of $\ce{(Li_{0.52}Na_{0.48})2CO3}$ and $\ce{(Li_{0.62}K_{0.38})2CO3}$ at $\pu{923K}$ were $2.06$ and $\pu{1.31Scm−1}$, respectively. The conductivity of molten $\ce{Li2CO3–X2CO3}$ ($\ce{X: Na, K, Rb,}$ and $\ce{Cs}$) and $\ce{Na2CO3–Z2CO3}$ ($\ce{Z: K, Rb,}$ and $\ce{Cs}$) decreases with an increase in the ionic radius of $\ce{X}$ or $\ce{Z}$. We have succeeded in correlating the equivalent conductivity of a binary carbonate mixture $\Lambda$, with the composition ($x_\mathrm{A}$ and $x_\mathrm{B}$), the equivalent conductivities of the pure components $\ce{A}$ and $\ce{B}$ ($\Lambda_\mathrm{A}$ and $\Lambda_\mathrm{B}$), and two parameters ($\lambda_\mathrm{A}$ and $\lambda_\mathrm{B}$) as $\Lambda = x^2_\mathrm{A}\Lambda_\mathrm{A} + x^2_\mathrm{B}\Lambda_\mathrm{B} + x_\mathrm{A}x_\mathrm{B}(\lambda_\mathrm{A}x_\mathrm{A} + \lambda_\mathrm{B}x_\mathrm{B})$. The pair of interaction parameters ($\lambda_\mathrm{A}$ and $\lambda_\mathrm{B}$) can be estimated for a binary system of molten carbonates using the relationship between the cationic radii and the equivalent conductivities of the pure carbonates.

The table 107 in page 85 of government technical report (Ref.3) gives some insight to the conductivity (equivalent conductance ($\Lambda$)  and specific conductance ($\kappa$)) of molten lithium carbonate (m.p.: $\pu{618 ^\circ C}$ or $\pu{891 K}$; eq. wt.: $\pu{36.95 g/eq.}$) in different temperatures together with density ($\rho$) and viscosity ($\eta$) at the same temperature:

$$
\begin{array}{c|ccc}
\text{Temperature in }\pu{ K}  & \Lambda & \kappa & \rho & \eta \\
\hline
1010 & 82.88 & 4.097 & 1.8260 & - \\
1020 & 84.58 & 4.172 & 1.8222 & - \\
1030 & 86.31 & 4.249 & 1.8185 & - \\
1040 & 88.06 & 4.326 & 1.8148 & - \\
1050 & 89.83 & 4.404 & 1.8111 & 4.64 \\
1060 & 91.63 & 4.483 &  1.8073 & 4.34 \\
1070 & 93.46 & 4.563 &  1.8036 & 4.01 \\
1080 & 95.31 & 4.644 & 1.7999 & 3.67 \\
1090 & 97.19 & 4.726 &  1.7961 & 3.36 \\
1100 & 99.09 & 4.808 & 1.7924 & 3.10 \\
1110 & 101.02 & 4.892 & 1.7887 & 2.91 \\
1120 & 102.98 & 4.976 & 1.7850 & 2.83 \\
\hline
\end{array}
$$

The tabulated values of specific conductance ($\kappa$) for molten $\ce{Li2CO3}$ in this table were calculated from the data of Ref.1, Using the following quadractic equation for specific conductance (the precision is $3.00 \times 10^{-4}\; (0.0066\%))$:

$$\kappa = 0.9877 - 1.3529 \times 10^{-3} T + 4.3873 \times 10^{-6} T^2 $$

The uncertainty of the specific conductance values is estimated to be about $1.5\%$. The ref.3 listed this as the best equation for temperature range $\pu{1018-1118 K}$. However, the ref.3 did not provide the relevant units for these calculations and, unfortunately, I do not have access to the Ref.1.

The Ref.3 also provided three other equations for their calculations (I mention this only for interesting viewers):

$$\rho = 2.2026 - 0.3729 \times 10^{-3} T$$
The ref.3 listed this as the best equation for temperature range $\pu{1011-1115 K}$ (The uncertainty of the values is estimated to be about $0.2\%$).

$$\eta = - 5259.12 + 14.8091 T - 1.38581 \times  10^{-2 }T^2 + 4.31294 \times  10^{-6} T^3$$
The ref.3 also listed this as the best equation for the temperature range $\pu{1046-1122 K}$ exracted from Ref.4 (The uncertainty of the values is estimated to be about $3.0\%$).

$$\Lambda = 754.5 e^{-\frac{4438}{RT}}$$
The ref.3 listed this as the best equation for equivalent conductance (The uncertainty of the values is estimated to be about $0.15\%$). Here, the units are given as $\pu{ohm-1 cm2 equiv-1}$. The equation is derived from Arrhenius type equation, $\Lambda = A_\Lambda e^{-\frac{-E_\Lambda}{RT}}$. For $\ce{Li2CO3}$, $A_\Lambda = \pu{754.5 ohm-1 cm2 equiv-1}$ and $E_\Lambda = \pu{4438 cal mol-1}$.

References:

George J. Janz, Max R. Lorenz, “Molten Carbonate Electrolytes: Physical Properties, Structure, and Mechanism of Electrical Conductance,” J. Electrochem. Soc. 1961, 108(11), 1052-1058 (doi: 10.1149/1.2427946).
Toshikatsu Kojimaa,b,z, Yoshinori Miyazakia, Katsuhiro Nomuraa,* and Kazumi Tanimoto, “Electrical Conductivity of Molten $\ce{Li2CO3–X2CO3}$ ($\ce{X: Na, K, Rb,}$ and $\ce{Cs}$) and $\ce{Na2CO3–Z2CO3}$ ($\ce{Z: K, Rb,}$ and $\ce{Cs}$),” J. Electrochem. Soc. 2007, 154(12), F222-F230 (doi: 10.1149/1.2789389).
G. J. Janz, F. W. Dampier, G. R. Lakshminarayanan, P. K. Lorenz, R. P. T. Tomkins, In Molten Salts: Volume 1, Electical Conductance, Density, and Viscosity Data; National Standard Reference Data Series: National Bureau of Standards 15, US Department of Commerce, National Bureau of Standards, US Government Printing Office: Washington DC, 1968.
George J. Janz, Fumihiko Saegusa, “Molten Carbonates as Electrolytes: Viscosity and Transport Properties,” J. Electrochem. Soc. 1963, 110(2), 452-456 (doi: 10.1149/1.2425785).


What is the process or material that can enhance the conductivity of the graphene oxide? I am doing a research and experiment on using Graphene oxide for deicing. My purpose is to coat graphene oxide on the surface of the ship in the north and south pole where large amount of large size ice cubes and icebergs are present. So when voltage is applied to it, it can generate a large amount of heat to melt the ice but the amount of heat must not be significant enough to vaporize the water to steam.

So, the problem is: As you know graphene oxide has a very high resistance. So there must be some process or material that can enhance the conductivity of the graphene oxide. What is it? In other words, is there a process that can enhance the conductivity of graphene?

Addressing the problem of high resistance, we can identify that the resistivity derives from the oxidation of graphene, changing $sp^2$ carbons to $sp^3$ in the lattice, which increases the band gap, and creates an insulator. There are a number of answers to this question that could be correct, including simply mixing another, more conductive material in the synthesis, but since the question seems to imply that the material has already been synthesized and simply needs to be reduced, removing the oxygen, creating reduced graphene oxide (rGO). One simple approach is through thermal annealing, where GO is reduced by a heat treatment.

You can read more about the reduction of graphene oxide here.


Is propane-1,2,3-tricarbaldehyde the correct IUPAC name of CHO-CH2-CH(CHO)-CH2-CHO? We were asked this question in class:

  What is the IUPAC name of the compound $\ce{CHO-CH2-CH(CHO)-CH2-CHO}?$

I don't know for sure, but I think the answer should either be 3-formylpentan-1,5-dial or 2-(formylmethyl)-butane-1,4-dial. However, I am told that the correct answer is propane-1,2,3-tricarbaldehyde.

How can that be? I searched on the internet and found that we use 'carbaldehyde' as the suffix only when the $\ce{-CHO}$ group is attached to a ring. But here the given compound is an open chain compound, so I don't see why we're using 'carbaldehyde' as the suffix.

My course textbook mentions that this compound is named so "to give identical treatment to all aldehydic groups". But I haven't learnt that such a rule exists in IUPAC nomenclature.

What is the correct name of this compound and why?

From the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, let's cite one of the rules for usage of the -carbaldehyde suffix, the relevant one:

  P-66.6.1.1.2 The suffix ‘carbaldehyde’ is used when more than two $\ce{-CHO}$ groups are attached to an alkane.

  Example:

  $\ce{OHC-\overset{1}{C}H2-\overset{2}{C}H(CHO)-\overset{3}{C}H2-\overset{4}{C}H2-CHO}$
  butane-1,2,4-tricarbaldehyde (PIN)

Thus the correct name is propane-1,2,3-tricarbaldehyde.


A theoretical chemistry question about elements and their presence over time I'm not a chemist myself, but I can't think of a better place to get some insight about elements and chemistry than from people who live and breathe it here on stackexchange. This is not any kind of homework problem just my personal curiosity, and I'd appreciate any input you might have on it.
Imagine we are travelling back in time towards the big bang. Universes being pulled back together, the density of energy is larger, atoms still have their neutral state but energy is colliding at a much higher frequency.
Is it possible that in these times, the amount of energy would permit atoms larger than we have observed in nature or been able to create in the lab?
If we then travelled to a far future, hundreds of millions of years from now, when the energy of the universe is much more spread out, and atoms have had more time to decay and lose electrons, wouldn't there be, at some point, a time where we no longer observe certain heavier elements in nature?
At either of these points in time wouldn't you expect to find more ionic elements in nature?

Is it possible that in these times, the amount of energy would permit atoms larger than we have observed in nature or been able to create in the lab?

Quite possibly, but we do not need to go back in time. In fact, the early moments after the Big Bang are the least likely time to find very heavy elements, as the primary elements produced initially are believed to have been almost exclusively hydrogen and helium.
Since those early moments, the universe has certainly expanded, but there are still places where energy and mass are highly concentrated, such as in supernovas and black holes. Whether or not very heavy elements are transiently created in these places is an open question.
The problem is that elements heavier than uranium tend to be unstable, so they break apart (nuclear fission) very quickly into lighter elements. In order to make very heavy elements, you need to combine two less heavy elements (nuclear fusion), but it's hard to get two heavy atoms together to fuse if they keep falling apart themselves. A good discussion of the possibility of very heavy element creation in stars is
Goriely, S. and Martinez Pinedo, G. (2015) The Production of transuranium elements by the r-process nucleosynthesis. Nuclear Physics A 944:158. DOI:10.1016/j.nuclphysa.2015.07.020

If we then travelled to a far future, hundreds of millions of years from now, when the energy of the universe is much more spread out, and atoms have had more time to decay and lose electrons, wouldn't there be, at some point, a time where we no longer observe certain heavier elements in nature?

The answer to this question relates to the first answer. Since heavy atoms are constantly being synthesized within stars, the elements that are decaying get replaced. But as the number of stars decreases over time, the rate of synthesis will necessarily decrease as well, and elements that are susceptible to fission will slowly decrease in abundance. The time scale for this, however, is much longer than even hundreds of millions of years. Our Sun, for example, is expected to last about another 4.5 billion years, and the final dying out of all of the stars in the universe is expected to take somewhere between 1 and 100 trilliion years. I encourage you to read up on ideas about the "heat death" of the universe if this is a topic you are interested in.

At either of these points in time wouldn't you expect to find more ionic elements in nature?

Since it takes energy to separate opposite charges from one another, ionization happens with high energy, so places like stars where there is a lot of energy will have a lot of ionic matter. Likewise, when high energy emissions from those places interact with other matter (like cosmic rays hitting the Earth's atmosphere), ionization will take place. So you're right that when the universe's energy is more concentrated, there were probably more ions.
In the distant future, the opposite is true. Since the universe decays to a low energy state, and ionization takes energy, we expect that oppositely charged matter will combine to make neutral atoms and molecules as the universe gets older. Some have theorized, however, that the expansion of the universe will mean that matter is so spread out that oppositely charged ions get too far away from each other to interact, so it is conceivable that the final state of the universe is separated ions, but that is just speculation.
And as noted in the comments, you may get a much better answer over at physics.SE.


What makes some chemicals form steep density gradients in solution? Caesium chloride and sucrose are both suitable for isopycnic centrifugation, and presumably not much else is. I'm unclear as to the mechanism by which they form useful density gradients. Wikipedia says,

  The $\ce{CsCl}$ molecules become densely packed toward the bottom, so even
  layers of different densities form.

(No doubt $\ce{CsCl}$ is molecular, despite being the chloride of a group I element, because $\ce{Cs}$ has such low electropositivity, but I digress.) But what is it about these molecules, not to mention those of sucrose, that causes their packing to be so strongly isopycnic in solution?

A determination was made that coordination number of cesium ions is 8 only in diluted solutions; in concentrated solutions it is near 4. The form of cation hydrate complexes depends largely on concentration and effect of counterions. https://inis.iaea.org/search/search.aspx?orig_q=RN:27004247

If the cesium hydrates are fairly stable, their volumes, and hence densities would vary with the amount of water bonded. It would seem that after centrifugation and separation, each portion with a uniform density would be a separate phase, with a different composition. Not a liquid crystal, or a crystalline liquid. Maybe a liquid glass? It should be possible to extract some of each layer and determine the total solids, or water content.

Sucrose presents a more complicated situation, but studies indicate degrees of hydration with relaxation times of a few picoseconds to 30 or so (https://www.sciencedirect.com/science/article/abs/pii/S0008621596910198), and so this could indicate hydrates of different stability that might be forced into a certain and constant composition - another liquid glass.

Another separation from a mixture that I have seen occasionally is when a mixture of different polymers in the same solvent evaporates. If the polymers are of sufficiently low molecular weight, or high mobility, or slow enough evaporation rate, they might separate into two layers. I suppose if the mobility is not high enough, the mixed polymer will evaporate with obvious inhomogeneity, or haze.


The weedkiller paraquat can be produced from methyl-pyridinium chloride using a cyanide catalyst: what is the mechanism? Paraquat is 4,4'dipyridinium chloride. There are two common ways to synthesise it starting with pyridine.

In one reaction that can be done on a large scale in water or liquid ammonia, methyl-pyridinium chloride is reacted with a base and catalytic amounts of sodium cyanide in the solvent to give a dimeric intermediate that oxidises easily into paraquat.

What is the chemical mechanism of the dimerisation reaction?

While the mechanism is not 100% certain (or wasn't when I worked on it some time ago), the current best guess is that it is similar to the well known Benzoin condensation:

This is based on the following intermediate:

Applied to the paraquat conditions this might go like this:

The neutral quinone-like intermediate crystallises out of solution and can be separated and then oxidised (it is very oxygen sensitive) to the desired paraquat product.

The details are partially guesswork, though, as it isn't the easiest reaction to investigate mechanistically.


Conversion of Z-matrix to Cartesian coordinates To begin with, I wrote a script that gets Cartesian coordinates of molecule as input in the below. These are $x,y,z$ coordinates of H2O2 molecule.

1 O          -1.7529   -0.5188    1.3324
2 O          -0.4737   -0.1091    0.7774
3 H          -2.2902    0.1678    0.8933
4 H           0.0636   -0.7957    1.2164

Then, the script constructs a Z-matrix with them, like this:

Z-mat :
O
O   1   1.45335189476
H   1   0.976176039452  2   96.5694760083
H   2   0.976131061897  1   96.5720363573   3   -179.995395182

Now I need to perform the reverse operation and use this Z-matrix as input and define $x,y,z$ coordinates for each atom. This is converting a Z-matrix to Cartesian coordinates.

My question is after setting first atom as 0,0,0

1 O          0   0   0

and the second one as 0,0,(distance from first) to put it on the z-axis

2 O          0   0   1.45335189476

How should I treat the 3rd and 4th atoms?

The 3rd atom must have coordinates that are something like this if read correctly:

3 H          0   distance*sin(angle)  z2+distance*cos(angle)

Taking z2 as the z-coordinate of atom 2. I am not sure if I should calculate this as z2 + distance.cos(angle) or z2 - distance.cos(angle) and what it depends on, if both are possible.

For the 4th atom, I use spherical coordinates

r, theta, phi = (0.976, 96.572, -179.995)

to calculate $x,y,z$ values from the formulas

x = r * sin(theta) * cos(phi)
y = r * sin(theta) * sin(phi)
z = r * cos(theta)

If there aren't any mistakes up to this point, how I will calculate Cartesian coordinates of the 4th atom using these $x,y,z$ values?

While searching I found TMPChem's work on GitHub and it does exactly what I want. However, in his work, there is a mathematical part that I don't understand:

# get local axis system from 3 coordinates
def get_local_axes(coords1, coords2, coords3):
    u21 = get_u12(coords1, coords2) #calculating vector between that points 1-2
    u23 = get_u12(coords2, coords3) #calculating vector between that points 2-3
    if (abs(get_udp(u21, u23)) >= 1.0):
        print('\nError: Co-linear atoms in an internal coordinate definition')
        sys.exit()
    u23c21 = get_ucp(u23, u21) # unit cross product
    u21c23c21 = get_ucp(u21, u23c21) # unit cross product
    z = u21
    y = u21c23c21
    x = get_ucp(y, z)
    local_axes = [x, y, z]
    return local_axes

What is "getting local axis system from 3 coordinates"?

Here is some context of how this function is used:

bond_vector = get_bond_vector(atom.rval, atom.aval, atom.tval)
disp_vector = np.array(np.dot(bond_vector, self.atoms[i].local_axes))
for p in range(3):
    atom.coords[p] = self.atoms[atom.rnum].coords[p] + disp_vector[p]

The bond vector definition is here:

def get_bond_vector(r, a, t):
    x = r * math.sin(a) * math.sin(t)
    y = r * math.sin(a) * math.cos(t)
    z = r * math.cos(a)
    bond_vector = [x, y, z]
    return bond_vector

Again, the only part I don't understand is # get local axis system from 3 coordinates. What is that function doing?

I will try to answer the question in bold from a mathematical perspective.

Basically, the first two lines in get_local_axes are to build the vectors $\vec r_{12}$ and $\vec r_{23}$. The next line builds the cross product $$\frac{\vec r_{23}\times\vec r_{12}}{|\vec r_{23}\times\vec r_{12}|},$$

The next vector is
$$\frac{\vec r_{12}\times(\vec r_{23}\times\vec r_{12})}{|\vec r_{12}\times(\vec r_{23}\times\vec r_{12})|},$$

This is a vector perpendicular to $\vec r_{12}$, and lies in the plane where $\vec r_{12}$ and $\vec r_{23}$ lie.

The next lines are defining the three axes of the system. $\vec e_z$ is just $\vec r_{12}$ normalized, $\vec e_y$ is chosen so that $yz$ is the plane where the three points (atoms) locate. and $\vec e_x$ is chosen so that $(\vec e_x, \vec e_y, \vec e_z)$ forms a orthonormal set and are the versors of a right-handed coordinate system. This system is the local axis system.

So this is the meaning of "getting local axis system from 3 coordinates".


How to derive these general formulae for number of stereoisomers of a compound with a possible plane of symmetry? These formulae are used if the molecule has a possible plane of symmetry. One such example would be:

Here the carbons marked with an asterisk are stereogenic centres (the asterisk is not used to mark isotopes). We can clearly see that if carbon number 2 (in the entire longest chain) and the carbon number 4 have opposite stereogenic configuration, then the molecule will be achiral because it will have a plane of symmetry. For example, (2R,3S,4S)-pentanetriol will be a meso compound with a plane of symmetry. This compound clearly satisfies the criteria we have set for the type of molecules being discussed.
Let us take another example.

Again, we see that if the molecule has the same geometrical configuration at the first and the third double bond, then the molecule has a plane of symmetry. For example, (2,7)-diphenylocta-(2Z,4E,6Z)-triene clearly has a plane of symmetry.
Our goal is to find the total number of stereoisomers such compounds can have in total. We can assume that a molecule does not have both, a double bond and a chiral centre.
In our class notes, we have written these formulae:

For geometrical isomers (i.e. in case of polyenes),

If 'n' is even (here n is the number of double bonds):$$\text{Number of stereoisomers} = 2^{n-1}+2^{n/2-1}$$
If 'n' is odd, then: $$\text{Number of stereoisomers} = 2^{n-1}+2^{(n-1)/2}$$

And for optical isomers (molecules with chiral centres):

If 'n' is even (here n is the number of chiral centres): $$\text{Number of enantiomers} = 2^{n-1}$$ $$\text{Number of meso compounds} = 2^{n/2-1}$$ $$\text{Total number of optical isomers} = 2^{n-1}+2^{n/2-1}$$
If 'n' is odd: $$\text{Number of enantiomers} = 2^{n-1}-2^{(n-1)/2}$$ $$\text{Number of meso compounds} = 2^{(n-1)/2}$$ $$\text{Total number of optical isomers} = 2^{n-1}$$

How to derive these formulae?

I managed to crack the formula for optical isomers with odd chiral centers, so I'll share my attempt here. Hopefully others may innovate on it and post solutions for other formulae.

Pseudo-chiral carbon atoms - an introduction
The Gold Book defines pseudo-chiral/pseudo-asymmetric carbon atom as:

a tetrahedrally coordinated carbon atom bonded to four different entities, two and only two of which have the same constitution but opposite chirality sense.

This implies that, in your case:

If chiral carbons 2 and 4 both have configuration R (or both S), then the central carbon 3 will be achiral/symmetric, because now "two and only two of its groups which have the same constitution" will have the same chirality sense instead. (Your approach by "plane of symmetry" is wrong. Find more details on this question)
Hence, there can be two stereoisomers (r and s) possible on the 3rd carbon due to its pesudochirality. But, there will be only one if both substituents on left and right have the same optical configurations.

Building up an intuition by manual counting
For optical isomers with odd number of chiral centers and similar ends, you can guess that, if there are $n$ chiral centers, then the middle ($\frac{n+1}2$-th) carbon atom will be pseudo-chiral. To build up an intuition, we'll manually count optical isomers for $n=3$ and $n=5$:
Case $n=3$
Take the example of pentane-2,3,4-triol itself. We find four (=$2^{n-1}$) isomers:
$$
\begin{array}{|c|c|c|}\hline
\text{C2}&\text{C3}&\text{C4}\\\hline
R&-&R\\\hline
S&-&S\\\hline
R&S&R\\\hline
R&S&S\\\hline
\end{array}
$$
As expected from the relevant formula, we find that the first two ($=2^\frac{n-1}2$) are meso compounds, and the remaining two ($=2^{n-1}-2^\frac{n-1}2$) are enantiomers.
Case $n=5$
Take the example of heptane-2,3,4,5,6-pentol:

We expect $16~(=2^{n-1})$ isomers, with the C4 carbon being pseudo-chiral. To avoid a really large table, we observe that the number of meso isomers is easily countable (<< number of enantiomers). Here is a table of those four (=$2^\frac{n-1}2$) meso isomers:
$$
\begin{array}{|c|c|c|c|c|c|}\hline
\text{C2}&\text{C3}&\text{C4}&\text{C5}&\text{C6}\\\hline
R&R&-&R&R\\\hline
R&S&-&S&R\\\hline
S&R&-&R&S\\\hline
S&S&-&S&S\\\hline
\end{array}
$$
Note that the total optical isomers are given by $2^{n-1}$ isomers (more on that below). Hence, the number of enantiomers is easily $12(=2^{n-1}-2^\frac{n-1}2)$.

A formula for the number of meso isomers
As you must have observed from the table, the sequence of optical configurations, when read from the fourth carbon atom, is exactly the same towards both left and right. In other words, if we fix an arbitrary permutation for the optical configurations of the carbon atoms on the left (say RSS), then we will get only one unique permutation of the optical configurations on the right (SSR).
We know that each carbon on the left has two choices (R or S), and there are $\frac{n-1}{2}$ carbon atoms on the left. Hence, the total number of permutations will be $2\times2\times2\cdots\frac{n-1}{2}\text{ times}=2^\frac{n-1}{2}$.
Since, our description ("the sequence of optical configurations, when read from the fourth carbon atom, is exactly the same on both left and right") describes meso isomers, we have hence counted the number of meso isomers, which is $2^\frac{n-1}{2}$.

A formula for the number of total isomers
We note that there are $n$ chiral carbons (including that pseudo chiral carbon). Again, each chiral carbon has $2$ choices. Hence, the maximum possible number of optical isomers is $2\times2\times2\cdots n\text{ times}=2^n$. This is the maximum possible, not the actual total number of isomers, which is much lower.
The reduction in number of isomers is because the string of optical configurations reads exactly the same from either terminal carbon. Example: RSsRS is the same as SRsSR. This happens because the compound has "similar ends"
Hence, each permutation has been over counted exactly twice. Thus, the actual total number of isomers is half of the maximum possible, and is  $=\frac{2^n}2=2^{n-1}$.

Conclusion
Hence, we have derived that, if 'n' (number of chiral centers) is odd for a compound with similar ends, then:

$\text{Number of meso isomers} = 2^{(n-1)/2}$
$\text{Total number of optical isomers} = 2^{n-1}$
$\text{Number of enantiomers} = 2^{n-1}-2^{(n-1)/2}$


Is there a paper glue that is NOT dissolved by acetone? Short version of the question: I need a spray-on glue that will stick paper to wood that isn't dissolved by acetone.

Long version: I am trying to glue paper on wood. It's a big sheet of paper and I need it stuck on throughout because then I cut the wood-with-paper-on-it into pieces with a laser cutter. Therefore, I'm using a spray glue that can cover the whole sheet of paper quickly and evenly. I have tried a couple different glues. After the pieces are cut, I want to spray an acrylic coating over the resulting pieces to seal them.

The process works except for one problem: if the acetone in the acrylic spray seeps through, it dissolves every glue I have tried. I try to go light with the acrylic, but more than 50% of my pieces end up with the paper coming loose from the wood. I can glue it back on later, but that means that the positioning of the paper is way less precise and the edge seal isn't there, making it susceptible to water damage.

Are there any glues anyone knows of that meet this challenge?
(I suppose, alternatively, is there an inexpensive sealant that doesn't use acetone? But I know the acrylic will hold up to the abuse these playing pieces are going to take, so I am hesitant to swap out that part of my process.)

Something you can try to use is tape to put under the glue. This may stop the acetone from seeping through. Not quite sure but maybe if you really need it done. Also, Elmer's glue does offer a spray on version regardless of what some say on your comments.


Are "gas phase" calculations the same as "vacuum" calculations? In both molecular mechanics/dynamics calculations and quantum mechanical calculations, the literature often refers to both gas-phase calculations and calculations done in vacuo. Is there actually a difference between these, or are they the same thing? If same, why would we call something a gas-phase calculation if there's no gas there?

I don't know the etymology of the phrase, but yes, "gas phase" calculations are the same as "vacuum" calculations. There is no difference.

Think of it this way: the only option other than a "gas phase" calculation is one where solvent is included, either as explicit atoms or implicitly via a mathematical model or even a solid-state/bulk-model calculation. This indicates that the solute is in the gas phase, and hypothetically not interacting with the environment in any way.


What is the maximum covalency of oxygen? My text says it to be 4. I believe it's 2. Is there any compound where covalency of 4 is observed?

We know the maximum covalent is at least 3 with such species as  $\ce{H3O+}$.  Also the pyrylium ion (https://en.m.wikipedia.org/wiki/Pyrylium_salt) has an oxygen covalency of 3 by forming a pi bond for the third bond.
Can we get 4-coordination?  As mentioned by @Rob, yes.   @Rob's comment states that basic beryllium acetate appears to have 4-coordinate, covalently bound oxygen.
And it check out.  Basic beryllium acetate, $\ce {Be4O(C2H3O2)6}$, has the four-coordinate oxygen in the center, then four beryllium atoms coordinated to this oxygen at the vertices of a tetrahedron, then an acetate group bridging each edge of the tetrahedron.  In addition to beryllium, zinc forms a similarly structured basic acetate, which can be obtained by heating normal zinc acetate (https://en.wikipedia.org/wiki/Zinc_acetate).
Might the central oxygen be mostly ionically bonded?  Magnesium and heavier alkaline earth metals, which form more strongly ionic bonds with oxygen, are not known to form such a basic acetate structure.  This corroborates the hypothesis that covalent bonding of the central, four-coordinate oxygen is involved.


What does the M in μM-Sux stand for? I am looking at some data regarding a study around a channel-blocking agent called suxamethonium.

The histograms of the open times of the channels at varying concentrations of suxamethonium are given. However, I am having a difficult time figuring out the dosage. Here are examples of dose listing formats:

200 μM-Sux
500 μM-Sux

etc.

I understand that μ in this context is to mean micro and I believe the suffix "Sux" is supposed to stand for suxamethonium. But what is the capital "M"? I looked it up in a scientific notation table and it stands for mega. Am I supposed to infer that this is a liquid (i.e. ML)?

You are right assuming that μ is a metric prefix, and that the number denotes the concentration of suxamethonium (suxamethonium chloride, I presume). However, this is supposed to be small caps "M" (as $\rm \small M$), not capital "M" (as $\rm M$). Also note that use of the term "molarity", as well as its notation $\rm \small M$, is discouraged [1, p.27]:

  The term molarity and the symbol $\rm \small M$ should no longer be used because they, too, are obsolete. One should use instead amount-of-substance concentration of $\ce{B}$ and such units as $\rm mol/dm^3$, $\rm kmol/m^3$,
  or $\rm mol/L$.

And, of course, $1\,\rm {\small M} = 1\, mol/L$.

Also, check out the answer to a similar question "What does the unit M stand for?"

Reference

Thompson, A.; Taylor, B. N. Guide for the use of the international system of units (SI). NIST Special Publication 811 2008.


In an isothermal process, how can the change in internal energy be 0? It was written in my textbook,

$$
\mathrm{d}U =
  \left(\frac{\partial U}{\partial T}\right)_V \mathrm{d}T
 +\left(\frac{\partial U}{\partial V}\right)_T \mathrm{d}V
$$
If the process is isothermal, $\mathrm{d}T = 0$. So, the equation reduces to:

$$
\mathrm{d}U = \left(\frac{\partial U}{\partial V}\right)_T \mathrm{d}V
$$

I was told, $\mathrm{d}U = 0$ in an isothermal process, does that mean $\mathrm{d}V = 0$, or  $\left(\frac{\partial U}{\partial V}\right) = 0$? But how is it possible that $\mathrm{d}V=0$ in an isothermal process, or $\left(\frac{\partial U}{\partial V}\right) = 0$?

If someone could explain what am I doing wrong, that would be very helpful.

The internal energy of an ideal gas is simply $$ U = \alpha nRT,$$
where $$\alpha = \frac{\text{degrees of freedom}}{2}$$
So, in an isothermal process,
$$
\Delta T = 0 \Longrightarrow \Delta U = \alpha nR\Delta T = 0,$$
and likewise any
$$\left(\frac{\partial U}{\partial P}\right)_T = \left(\frac{\partial U}{\partial V}\right)_T = 0.$$


How to inhibit Mn2+ autocatalysis in the oxalate-permanganate reaction? It is well known that permanganate in acidic solution can oxidize oxalate ions (or oxalic acid). The reaction produces $\ce{Mn^2+}$, which catalyzes this very reaction.

$$\ce{2MnO4^- + 16H^+ + 5C2O4^2- -> 2Mn^2+ + 8H2O + 10CO2}$$

As a part of a demonstration on chemical kinetics, I want to inhibit this autocatalysis somehow. (I intend to perform three reactions: one with normal autocatalysis, one with $\ce{Mn^2+}$ added, which will run faster, and one preventing the $\ce{Mn^2+}$ from catalyzing the reaction.)

At first I thought I might use EDTA to complex the manganese ions. Alas, it didn't help. Maybe the environment is too acidic. (The reaction was actually faster with EDTA, I don't know why.)

Perhaps I should try a different chelating agent? Perhaps I could precipitate the $\ce{Mn^2+}$ ions? What do you suggest?

At first the reaction is slow. But when we heat up, the reaction gets catalyzed due to formation of $\ce{Mn^2+}$ ions. It happens due to the fact that $\ce{Mn^2+}$ gets oxidised to $\ce{Mn^3+}$ ions. (Source).

$$\ce{4Mn^2+ + MnO4- + 8H+ -> 5Mn^3+ + 4H2O}$$

So, we have to inhibit the formation of $\ce{Mn^3+}$ ions so as to inhibit the autocatalysis reaction. For this,  we have to use  phosphonate and carboxylate based chelating agent that solubilize (hydr)oxide-bound $\ce{Mn(III)}$ . The following is some relevant information from the site:

  phosphonate-based chelators that solubilize (hydr)oxide-bound
  Mn(III) via ligand-promoted and/or reductive dissolution at
  circum-neutral pHs.

  pyrophosphoric acid (PP), methylenediphosphonic acid (MDP), and
  phosphonoacetic acid (PAA), ligand-promoted dissolution is
  predominant: from pH 6−8, initial dissolution rates and plateau
  concentrations for dissolved Mn(III) decrease in the order PP > MDP >
  PAA, and at pH 5, MDP reacts equally well (with birnessite) or more
  efficiently (with manganite) than PP, and PAA remains the least
  reactive chelator. For manganite reacting with an excess concentration
  of aminophosphonate/carboxylate-based chelators, the
  aminophosphonate-containing iminodimethylenephosphonic acid and
  glyphosate yield appreciable amounts of dissolved Mn(III), but the
  aminocarboxylate-based methyliminodiacetic acid yields solely
  dissolved Mn(II) via Mn(III) reduction.


If we put an metal electrode in water with no metal ions present potential should tend to minus infinity? If we consider an equilibrium between metal electrode and its ions in solution we know that equilibrium potential is proportional to activity of ions in solution from Nernst equation. Since this dependence is logarithmic, what we conclude is that when activity of ions tends towards zero, potential tends to minus infinity. If we do place a metal electrode in pure water, would its potential decrease continously if we disregard any kind of other processes on it like corrosion which create their own potential?

Yes. The potential is in principle infinite if the water is pure. But in reality, when a metallic plate is dipped into pure water, you cannot prevent one or two atoms from being oxidized by air and pass into solution. On the other hand, the same thing can be said about $\ce{O2}$ producing one or two $\ce{OH-}$ ions. As a consequence the potential goes quickly from infinite to a finite value.


Anomalous Electronic Configuration of Thorium The electronic configuration of thorium ($Z=90$) is $5\mathrm f^0 6\mathrm d^2 7\mathrm s^2$. But, according to the aufbau principle, the electrons should first enter the $\mathrm f$ subshell and not the $\mathrm d$ subshell. This is not a special case of a half-filled subshell or a fully-filled subshell like we see in the respective cases of chromium and copper. Then why do we write the electronic configuration like this, first filling the $\mathrm d$ subshell and not the $\mathrm f$ subshell?

P.s I came across the anomalous configuration of thorium here: https://en.wikipedia.org/wiki/Thorium#cite_ref-Wickleder5960_19-0

But it doesn't state the reason for the anomaly.

For first four actinide elements $\ce{Th}$, $\ce{Pa}$, $\ce{U}$ and $\ce{Np}$, the difference in energy between $\mathrm{5f}$ and $\mathrm{6d}$ orbitals is small. Thus in elements electrons may occupy the $\mathrm{5f}$ or $\mathrm{6d}$ levels or sometimes both.

Later in actinide series the $\mathrm{5f}$ do become appreciable lower in energy, thus from $\ce{Pa}$ onward the $\mathrm{5f}$ shells fill in regular way.


Which chemicals could be used to dissolve Polybutylene Terephthalate (PBT), but not dissolve brass? This is for a recycling project, I am looking to recover brass pins from electronic connectors made of glass reinforced Polybutylene Terephthalate (PBT).

The objective is to find a chemical which would be able to dissolve Polybutylene Terephthalate (or PBT, a common type of plastic) so that i can separate the brass pieces inside, without dissolving them too. I think that if it dissolves PBT the glass reinforcement will become dust so it won't be an issue.

So far my search has been unsuccessful, either it was acids which were too strong and would also dissolved] the brass, or common solvent which had no effect on the PBT (tested Acetone). Even 30% Caustic soda doesn't attack the PBT.

I found a method with high temperature methanol under pressure into a closed stainless steel shape, but it's way too complex for me to make. Similarly chemicals which are too cancerogenous are not worth it (at least given that i am not a specialist capable of handling the most dangerous ones).

If anyone has an answer or some ideas i'd love to hear it!

Thanks a lot!

PBT is a thermoplastic and its melting point is only 223 °C. Depending on its composition, brass doesn't melt until 900 °C or so. High enough heating may change the structural properties of the brass, but perhaps there is a temperature to which you can soften the plastic enough to detach the pins without damaging them.

Unfortunately, brass is easily attacked by many chemicals and PBT is resistant to many solvents, but something like hot DMSO or N-methyl-2-pyrrolidone may swell the polymer enough to separate the parts.


Why are overcoats used on aluminum if aluminum oxide forms on surface anyway? If a thin layer of aluminum oxide is formed on the surface of aluminum to prevent further oxidization, why are optical components using extra "overcoat" materials to prevent dulling of the metal over time?

Sapphire is an aluminum oxide which is transparent when in crystalline form. However when aluminum metal corrodes the aluminum oxide takes up more volume that the aluminum metal from which it was formed. So the aluminum oxide can't form the same crystalline structure as aluminum metal. Thus instead of one nice nice uniform crystalline sheet of aluminum oxide the oxide forms small crystals. Such an oxide coating reduces the reflectivity of the mirror. (Thinking of it like dropping table salt on a mirror.)

The overcoating materials keep oxygen away from the aluminum metal so that the oxide coating doesn't form.


Non-bonded orbitals in water General Chemistry perspective: Looking at the molecular orbitals of water, we can see that the oxygen is $sp^3$ hybridized. Oxygen forms two sigma bonds with hydrogens, and there are two lone pair orbitals in the molecule. There are five occupied molecular orbitals: The core $1s$ of oxygen, and the four $sp^3$ orbitals, out of which two are lone pair orbitals.

Group Theory and Quantum Chemistry perspective: Water belongs to $C_{2v}$ point group. There are four irreducible representations, and the atomic orbitals belong to these irreducible representations (or linear combinations of these irreducible representations). Hence, we get five occupied molecular orbitals of symmetries $a_1, b_1,$ and $b_2$.

There is only one non-bonding orbital ($1b_1$) in which the AOs of hydrogens do not participate. Where is the second lone pair orbital of water?

General Chemistry perspective: Looking at the molecular orbitals of water, we can see that the oxygen is $\mathrm{sp}^3$ hybridized.

No. One could say that, but $\mathrm{sp}^2$ is equally possible. In fact, it may seem more likely, we’ll get to that in a second.

Group Theory and Quantum Chemistry perspective: Water belongs to $C_{2\mathrm{v}}$ point group. There are four irreducible representations, and the atomic orbitals belong to these irreducible representations (or linear combinations of these irreducible representations). Hence, we get five occupied molecular orbitals of symmetries $\mathrm{a}_1, \mathrm{b}_1$, and $\mathrm{b}_2$.

Yes, and you did a nice job of drawing them.

There is only one non-bonding orbital ($1\mathrm{b}_1$) in which the AOs of hydrogens do not participate. Where is the second lone pair orbital of water?

And this is where it gets interesting.
I’ll once again point to Professor Klüfers’ web scriptum for the basic and inorganic chemistry course in Munich, section about localising molecular orbitals. If you don’t understand German, all you need to do is look at the pictures and understand that wenig means a little in this context.
You see, to do a discussion such as ‘this is the bonding electron pair’ we need to localise molecular orbitals. Specifically by linear combining them. It’s almost like in the linked pictures: $\Psi_2 (1\mathrm{b}_2) + \Psi_3 (3\mathrm{a}_1)- \mathrm{some~} \Psi_1 (2\mathrm{a}_1) = \unicode[Times]{x3C3}_1$. Invert the signs: $\Psi_2 (1\mathrm{b}_2) - \Psi_3 (3\mathrm{a}_1)  + \mathrm{some~} \Psi_1 (2\mathrm{a}_1) = \unicode[Times]{x3C3}_2$. These are your two bonding orbitals to hydrogen. We have some $\Psi_1$ remaining (and, theoretically, also some $\Psi_2$ and $\Psi_3$, since we didn’t use all of them). That (remaining linear combination) is our second non-bonding orbital.

Now why is $\mathrm{sp}^3$ a bad description? You see, we arrived at one lone pair that is definitely $\unicode[Times]{x3C0}$-shaped or p-shaped. The second is somewhat p-ish but also somewhat s-ish, so maybe some sp-hybrid. But there is no mistaking that $1\mathrm{b}_1$ is in no way influenced by all the others. It is antisymmetric to the $\ce{H-O-H}$-plain of symmetry while all the others are symmetric. $\mathrm{sp}^2$ describes this a lot better.


Why aren't chaperones considered catalysts? I'm reading about protein folding on Wikipedia and I stumbled on a bit about a class of proteins called chaperones that aid in the folding of proteins by:

  ...reducing possible unwanted aggregations of the polypeptide chain that might otherwise slow down the search for the proper intermediate...

However, they aren't considered to be catalysts because they do not actually speed up the folding process. My thought is that without these chaperones, the polypeptides chain would go through a long process of finding a proper intermediate and therefore should be considered catalysts.

Where am I going wrong in my thinking?

I don’t think you are wrong. Actually molecular chaperones have been named enzymes in some texts. The molecular chaperones comprise several unrelated classes of proteins that have rather different functions.

  Most molecular chaperones are ATPases (enzymes that catalyze ATP
  hydrolysis), which bind to unfolded polypeptides and apply the free
  energy of ATP hydrolysis to effect their release in a favourable
  manner.

It is also made evident in the way some of them function (they possess some ATPase activity), it suggests they’re kind of  catalysts  enzymes:

  They function in an ATP-driven process to reverse the denaturation and
  aggregation of proteins (processes that are accelerated at elevated
  temperatures), to facilitate the proper folding of newly synthesized
  polypeptides as they emerge from the ribosome, to unfold proteins in
  preparation for their transport through membranes.

Two classes of molecular chaperones have been well studied. Both are found in organisms ranging from bacteria to humans.

The heat shock proteins 70 (Hsp70)

The first class is a family of proteins called Hsp70. They function in an ATP-driven process to reverse the denaturation and aggregation of proteins (processes that are accelerated at elevated temperatures), to facilitate the proper folding of newly synthesized polypeptides as they emerge from the ribosome, to unfold proteins in preparation for their transport through membranes and to subsequently help them refold.

Chaperonins

Are also heat shock proteins that form large, multisubunit, cagelike assemblies that are universal components of prokaryotes and eukaryotes. They bind improperly folded globular proteins via their exposed hydrophobic surfaces and then, in an ATP-driven process, induce the protein to fold while enveloping it in an internal cavity, thereby protecting the folding protein from nonspecific aggregation with other unfolded proteins.

Other chaperones e.g. Hsp90 proteins (have a regulatory role in that they induce conformational changes in nativelike substrate proteins that result in their activation or stabilization) and Trigger factor (prevents the intra- and intermolecular aggregation of newly synthesized polypeptides as they emerge from the ribosome by shielding their hydrophobic segments) However, unlike most other chaperones, trigger factor does not bind ATP.

Enzymes are more or less catalysts in biological systems, so yes its right to synonimise them with catalysts.

References

Biochemistry (Voet and Voet) Section 9.2 Folding Accessory Proteins, Molecular Chaperones: The GroEL/ES System.
Lehninger Principle of Biochemistry Chapter 4, Protein Denaturation and Folding.
Biochemistry R. Garret, C. Grisham. Chapter 31 Completing the Protein Life Cycle: Folding, Processing, and Degradation


Inherent error in mass and volume? In one of my chem classes I had to do a lab that involved finding the mass and volume of a substance.  I did three trials and for my mass I got 9.3g, 9.3g, and 9.4 g.  For my volume, I got 9.7ml, 9.9ml, and 9.9 ml.

I am being asked to find the inherent instrument error calculation.

For volume, I used a graduate cylinder which has 1 % vol/reading as an inherent error.

For mass I used a one place electronic balance which has error of 0.1g/reading

How would I find the error in mass and the percent error in mass?

Is the percent error formula you are looking for something like the formula this website service offers? Your lab instructor should have given you the formula, perhaps you missed that. Do you have a reliable lab partner you can ask? The error in mass would probably be something to do with how far each measurement deviates from the average.

If for example I used a 250 ml Erlenmeyer flask which had marks at only 25ml increments. I would then say that I could make a 'guess' as to what the volume is at if I were trying to measure something less than 25ml or between two marks. But the error for that instrument would then be somewhere around +/- 12.5ml error. My general-chem teacher was fine with that, maybe you should ask yours?

I was told that the convention is to expect at least 2% error in the last significant digit if not explicitly defined. When taking a measurement with a device like a graduated cylinder, you always measure down to the most precise "known" value, and then the last significant digit is your own judgement.


How do I store HCl safely? I'm in a textile unit, where we use different chemicals. My question is what type of material should I make my tank out of to store my hydrochloric acid?

This question is related: Handling and storing sodium hydroxide, hydrochloric acid, hydrogen peroxide and isopropyl alcohol

In the industries the handling of chemicals maybe periodic and, with routine, one tends to forget the hazardous and risks related to this. There's a lot of information on the web (http://water.usgs.gov/admin/memo/policy/wrdpolicy94.006.html):

STORAGE: Hydrochloric acid should be kept in a tightly closed container and
stored in chemical area that is compatible with other chemicals.
Store in a secure, well-ventilated area, that is well marked, and
away from the general work population.  Do not store near oxidizing
materials.

This manual should be helpfull too: http://www.jsia.gr.jp/data/handling_02e.pdf

1. Types of Containers:
The shipping containers used for hydrochloric acid are those rail tanks be tightly sealed, such as acid-resistant bottles (including polyethylene containers, etc.), and steel drums tank trucks, tank cars, and tankers with corrosion resistant linings

If you keep reading the manual there is valuable information about handling, transportation, and rules to keep the human safety.


Rate of EAS in chlorobenzene and fluorobenzene In both chlorobenzene and fluorobenzene, both the halogens show a −I effect where fluorine has more electron withdrawing effect than chlorine.

Now, if we consider the mesomeric/resonance effect, both show +M/+R effect. On comparison of resonance effect and inductive effect, resonance effect dominates over inductive effect. Then:

Why does −I effect dominates over +M/+R effect of chloro and fluoro group in the case of chlorobenzene and fluorobenzene? Give reason.

I know that the −I effect dominates over the +M/+R effect of the chlorine and fluorine substituents in case of chlorobenzene and fluorobenzene. Now, on the basis of this chlorobenzene should be more reactive for electrophilic aromatic substitution reaction than fluorobenzene (as −I effect: $\ce{-F~ >~ -Cl}$). But, it is not true though vice-versa is true.

Why is fluorobenzene more reactive for electrophilic aromatic substitution reaction than chlorobenzene? Give reason.

We need some data to answer your question. As presented in this earlier answer (which is also relevant to this question), here are the relative rates for electrophilic aromatic nitration of the halobenzenes compared to benzene itself.

\begin{array}{|c|c|c|c|} \hline
\ce{Ar-X} & \text{Relative Rate}  \\
\hline
\ce{Ar-H} & 1.0 \\ \hline
\ce{Ar-F} & 0.11 \\ \hline
\ce{Ar-Cl} & 0.02 \\ \hline
\ce{Ar-Br} & 0.06  \\ \hline
\ce{Ar-I} & 0.13 \\ \hline
\end{array}

You are correct that inductive and resonance effects operate in opposite directions here. All of the halogens are more electronegative than hydrogen so they inductively remove electron density from the aromatic ring and tend to slow the reaction down. Whereas resonance effects would donate electron density into the aromatic ring and speed the reaction up.

However it is not correct to say

  On comparison of resonance effect and inductive effect, resonance
  effect dominates over inductive effect

Here is a series where inductive effects play a larger role than resonance effects.  This is why all of the halobenzenes react more slowly than benzene itself.

In fluorobenzene the overlap between carbon 2p and fluorine 2p orbitals is very effective, but the carbon 2p orbital cannot overlap as effectively with higher p orbitals (e.g. chlorine 3p, bromine 4p, etc.). It is the stronger resonance contribution from this effective electron donating resonance interaction with fluorine that makes fluorobenzene more reactive towards electrophilic aromatic substitution than chlorobenzene.

The electrophilic reactivity of the other halobenzenes is also controlled by a blend of resonance and inductive effects.  As we move down the series from chlorine to iodine the electronegativity decreases and the strength of the electron withdrawing inductive effect also decreases - this should speed up the rate of reaction. However the strength of the resonance effect also decreases due to less effective orbital overlap between halogen and carbon - this should slow the reaction down. Looking at the data we can see that the reaction rate slowly increases as we move from chlorine to iodine, suggesting that the inductive effect is still the primary effect with resonance, as expected due to poorer overlap, still playing a secondary role.


How to study (by heart) all the equations in inorganic chemistry? Physical chemistry is easy to study as I have to just understand the concepts (like in physics) and apply some equations for calculations. In organic chemistry, reaction mechanisms helpsin remembering a lot of reactions and also doing conversion problems help a lot.

But, in the case of inorganic chemistry it is very difficult for me to remember those equations. My high school syllabus includes a lot of preparation methods of different compounds and reactions showing their chemical properties. But, unlike organic chemistry there is no problems based on it for deep understanding. So, what is actually needed is a good memory power. But, I'm not able to remember all those equations. So, how should I study them?

In organic chemistry, everything is based around C, H, O and N. Therefore to understand reactivity (somehow), there is just limited number of options.

In contrast, inorganic chemistry deals with the rest of periodic table, and each of the elements has its special properties. But there are general trends as well, but are more difficult to decipher and are often expressed by set of typical reactions, you should learn.

Usually they concern the stability of (oxidation) states, how to go from less stable to more stable one to obtain the desired compound and few processes how tho achieve the least stable state from the one found in nature.

It is difficult to state some simple rules, but there are several most typical anions and cations, and the chemistry of given element is just interactions with them. During the learning it helped me to have "all" the reactions (several hundreds) on a paper and for each one try to understand what happens and if there are others in the set, which are similar, and why. This helps you to spot the similarities. And - it is better to have more reactions then less, because then you have higher probability to spot the trends.


GCMS Peak Identification: Online or Open Source / Free Programs? Suppose I have a GCMS peak graph for a component, what are possible options to figure out what the molecule may be?

Any tips about good software that will do this? I do know that for many compounds there is a hit in the NIST library, but how does one go about automatically checking the library?

Google came up with OpenChrom. Has anyone used this? The peak library in it is unfortunately not accessible in the free version.

Have you tried the Golm Metabolome Database?
It is a collection of GCMS mass spectra. The focus is on metabolites.

However, on their webpage is the entry "Decision Trees". If you follow the link, there is the option to enter your own mass spectra information and receive a prediction of functional groups in the measured substance.
This might be of help to identify your molecule.
There are more details in the Wikipedia entry.


What are bond lengths and angles for picric acid? I am working on a project in which I am required to create a model of picric acid. I do not understand how to find the bond lengths, bond angles, shape, and atom size of this molecule. Any help in finding these measurements or in showing me how to find them would be amazing!

I would suggest to take a crystal structure available for free from Crystallography Open Database, COD#5000146. The entry webpage already has a JSmol script loaded, so you should see the molecular structure loaded on a black area. Selecting RMB (right mouse button) -> Measurements -> Show Measurements allows you to measure linear and angular parameters with mouse clicks.

Alternatively, you can download the CIF-file (5000146.cif) and load it with any crystallographic viewer if you prefer to work offline. For beginners I'd recommend Mercury, there is a free version available for Windows, macOS and Linux:

N. B. The asymmetric unit in this structure contains two molecules of picric acids – with and without H-atoms. Obviously, you need the one with the protons assigned.


Meaning of 'constant T and p' In chemical thermodynamics, the condition "constant temperature and pressure" appears almost everywhere. However, I felt a small ambiguity in the phrase so I wanted to clear my doubts. When we say "constant $T$ and $p$" in chemical thermodynamics, which of these two statements do we actually mean?

$T$ and $p$ are the same at the beginning and end of the reaction
$T$ and $p$ are the same throughout the entire reaction

If we care about path-dependent variables, then there is a major difference between the two meanings.

As a chemical engineer, my perspective on this differs from those of the other responders.  I regard a process occurring at constant temperature and pressure as one in which

The system is held in contact with a constant temperature over the portion of its interface with the surrounding through which heat transfer is occurring, not necessarily the initial temperature of the system, throughout the process

The external pressure (or more precisely, the external compressive stress) at the portion of the system interface with its surroundings at which displacement work is being done is held constant (not necessarily at the initial pressure of the system) throughout the process.


Phase description for a substance dissolved in a solvent other than water? I'm trying to write an equation for which I have $\ce{CuI}$ dissolved in acetonitrile. Usually if you have a salt dissolved in water you can denote that using $\ce{(aq)}$, but is there a notation for substances dissolved in a solvent other than water?

IUPAC “Green Book” recommends to use $\mathrm{sln}$ for denoting a solution in general [1, p. 54], referring to earlier Recommendations 1981. Appendix No. IV to Manual of Symbols and Terminology for Physicochemical Quantities and Units [2, pp. 1240–1242].
This has been extensively covered in the following posts:

What is the standard way to denote physical states in a chemical reaction? (Meta)
Phase abbreviations for non-aqueous solutions

On the contrary, abbreviation $\mathrm{sol}$ is used to denote a process (typically a sub- or superscript to a symbol for a thermodynamic quantity), not state of matter [1, pp. 59–60]:

2.11.1 Other symbols and conventions in chemical thermodynamics
(i) Symbols used as subscripts to denote a physical chemical process or reaction
These symbols should be printed in Roman (upright) type, without a full stop (period).
$$
\begin{array}{ll}
\ldots \\
\text{solution (of solute in solvent)} & \mathrm{sol} \\
\ldots
\end{array}
$$

Another reason not to use $\mathrm{sol}$ for denoting a state of aggregation is its ambiguity: $\mathrm{sol}$ is often used as an abbreviation for “solid”, as listed in Appendix 10-2 Abbreviations, Acronyms, and Symbols in ACS Style Guide [3, p. 197]

$$
\begin{array}{ll}
\ldots \\
\text{sol} & \text{solid} \\
\text{soln} & \text{solution} \\
\ldots
\end{array}
$$

Also, specialized literature on electrochemistry of non-aqueous solutions uses $\mathrm{sv}$ to denote solvation process separately, see e.g. [4, p. 27]:

$ΔH^\circ_\mathrm{sv},$ $ΔS^\circ_\mathrm{sv},$ $ΔG^\circ_\mathrm{sv}$: Enthalpy, entropy, and Gibbs energy of solvation of the electrolyte

With this obligatory standardization part, the simplest reaction can be written as
$$\ce{CuI(s) ->[CH3CN] Cu^+(sln) + I^-(sln)}$$
However, copper(I) is extremely stable in acetonitrile due to solvation [4, p. 33]:

Interactions by Back-Donation from $\mathrm{d^{10}}$-Cation to Solvent Molecules
Acetonitrile (AN) has relatively small DN and usually solvates rather weakly to
metal ions. However, it solvates very strongly to $\ce{Cu+},$ $\ce{Ag+}$ and $\ce{Au+},$ which are univalent $\mathrm{d^{10}}$-metal ions.
This is because these metal ions have an ability to back-donate their electrons into a π*-antibonding orbital of the $\ce{CN}$ group of AN, as shown by $\ce{CH3C#N\overleftarrow{\ce{:\bond{->}A}}g+}.$
As a result, $\ce{Cu+}$ and $\ce{Ag+}$ in AN are stable and not easily reduced to metal, while the weakly solvated $\ce{Cu^2+}$ is very easily reduced to $\ce{Cu+}$ … , making $\ce{Cu^2+}$ in AN a strong oxidizing agent.

and may theoretically be coordinated by up to five acetonitrile ligands [5] with C.N. 4 being the most common unless there are other ligands.
Copper(I) and iodide don't seem to form a complex in acetonitrile [6, p. 194]:

Also in tetraethyl ammonium iodide and thiocyanate solutions, the dissociation is seemingly complete, … while a slight association is indicated for bromide … and, especially, chloride.

To sum it up, upon dissolving of copper(I) iodide in acetnitrile the formation of tetrakis(acetonitrile-N)copper(I) is the most likely scenario so that the final reaction can be written as
$$\ce{CuI(s) + 4 CH3CN(l) ->[CH3CN] [Cu(CH3CN)4]^+(sln) + I^-(sln)}$$
It should be noted that in a presence of better coordinating ligands $\ce{L}$ composition of the complex may, of course, vary $(\ce{[Cu(CH3CN)_xL_y]^z+}).$
Also, copper(I) iodide may form various coordination polymers based on $\ce{(CuI)_n}$ cluster core [7, 8, 9].
References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
Cox, J. D. Notation for States and Processes, Significance of the Word Standard in Chemical Thermodynamics, and Remarks on Commonly Tabulated Forms of Thermodynamic Functions. Pure and Applied Chemistry 1982, 54 (6), 1239–1250. https://doi.org/10/d783th. (Free Access)
The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. ISBN 978-0-8412-3999-9.
Izutsu, K. Electrochemistry in Nonaqueous Solutions; John Wiley & Sons, 2005. ISBN 978-3-527-60065-6.
Vitale, G.; Valina, A. B.; Huang, H.; Amunugama, R.; Rodgers, M. T. Solvation of Copper Ions by Acetonitrile. Structures and Sequential Binding Energies of $\ce{Cu + (CH3CN)_x},$ $x = 1−5,$ from Collision-Induced Dissociation and Theoretical Studies. J. Phys. Chem. A 2001, 105 (50), 11351–11364. https://doi.org/10/ckvsxw.
Ahrland, S.; Nilsson, K.; Tagesson, B.; Haaland, A.; Schilling, B. E. R.; Seip, R.; Taugbøl, K. Thermodynamics of the Copper(I) Halide and Thiocyanate Complex Formation in Acetonitrile. Acta Chem. Scand. 1983, 37a, 193–201. https://doi.org/10/bqvn5f.
Knorr, M.; Guyon, F.; Khatyr, A.; Strohmann, C.; Allain, M.; Aly, S. M.; Lapprand, A.; Fortin, D.; Harvey, P. D. Construction of $\ce{(CuX)_{2n}}$  Cluster-Containing ($\ce{X} = \ce{Br}, \ce{I}$; $n = 1, 2$) Coordination Polymers Assembled by Dithioethers $\ce{ArS(CH2)_mSAr}$ ($\ce{Ar} = \ce{Ph},$ $\ce{\textit{p-}Tol}$; $m = 3, 5$): Effect of the Spacer Length, Aryl Group, and Metal-to-Ligand Ratio on the Dimensionality, Cluster Nuclearity, and the Luminescence Properties of the Metal–Organic Frameworks. Inorg. Chem. 2012, 51 (18), 9917–9934. https://doi.org/10/f37kdx.
Knorr, M.; Khatyr, A.; Dini Aleo, A.; El Yaagoubi, A.; Strohmann, C.; Kubicki, M. M.; Rousselin, Y.; Aly, S. M.; Fortin, D.; Lapprand, A.; et al. Copper(I) Halides $(\ce{X} = \ce{Br}, \ce{I})$ Coordinated to Bis(Arylthio)Methane Ligands: Aryl Substitution and Halide Effects on the Dimensionality, Cluster Size, and Luminescence Properties of the Coordination Polymers. Crystal Growth & Design 2014, 14 (11), 5373–5387. https://doi.org/10/f6pnff.
Knorr, M.; Bonnot, A.; Lapprand, A.; Khatyr, A.; Strohmann, C.; Kubicki, M. M.; Rousselin, Y.; Harvey, P. D. Reactivity of $\ce{CuI}$ and $\ce{CuBr}$ toward Dialkyl Sulfides RSR: From Discrete Molecular $\ce{Cu4I4S4}$ and $\ce{Cu8I8S6}$ Clusters to Luminescent Copper(I) Coordination Polymers. Inorg. Chem. 2015, 54 (8), 4076–4093. https://doi.org/10/f7cjk2.


Why do soft drinks fizz? I know it has something to do with carbon dioxide, but what exactly happens and causes the characteristic sound we hear (especially after we shake the bottle)? How do we manage to dissolve a non-polar substance ($\ce{CO2}$) in a polar solvent ($\ce{H2O}$)?

Because $\ce{CO_2}$ isn't exactly what's being dissolved. When it encounters water, it immediately encounters the following potential reaction:

$$\ce{CO2~(g) + H2O~(l) <=> H2CO3~(aq)}$$

Carbon dioxide is non polar, yes, but carbonic acid readily dissolves (and dissociates, but I didn't care to show that) in water.

The sound we hear is because when we open the bottle, the equilibrium above shifts to the left, creating more carbon dioxide that is being released. Why does the equilibrium shift? It has to do with the pressure in the bottle.

When the bottle is closed, it's under a certain set pressure. When it's closed at the same temperature and pressure for a long time, the reaction above reaches an equilibrium state, where the forward and backward versions of the reaction are happening at the same rate. However, when the bottle is opened, the pressure inside the bottle decreases, as the gas in the bottle suddenly has a lot more space to expand into. A decrease in pressure favors the side of an equilibrium that has more gaseous molecules, which in this case means the carbon dioxide. Ergo, when you open the bottle, the gas is formed, escapes from solution, and rapidly expands outward, making the infamous popping noise.


Reaction of Grignard reagents with esters Suppose I have a phenyl Grignard and methyl benzoate.

If I react the two, can I expect benzophenone and a triphenylmethoxide ion as my products?

I ask because I know that Grignards react with ketones and aldehydes. So as the benzophenone is generated, couldn't more Grignard attack the benzophenone as well as the original reactant?

Expected products and mechanism shown below:

Reaction of a Grignard reagent with an ester is a standard method for producing tertiary alcohols where at least two of the substituents (the "$\small\ce{R_2}$" group attached to the Grignard) are the same.  Judicious choice of the starting ester allows for the preparation of a tertiary alcohol where all 3 substituents are the same. Alternately, you could react 3 equivalents of the Grignard with a cheap dialkyl carbonate (a "double" ester, if you will) to obtain a tertiary alcohol where all 3 groups are the same.

Typically a slight excess of Grignard reagent is used and the tertiary alcohol is produced after acid work-up.  If an excess of Grignard is not used then a mixture (ugh!) of ester, intermediate ketone and alcohol may result.


Do Bases and Acids actually react with each other? Let's say we have $5$ $\ce{HCl}$ molecules and we dump them into a container full of water. It's a strong acid so all $5$ $\ce{H+}$ ions would react with water and create $\ce{H3O+}$. It reacts with the water, right?

Now, we add $5$ $\ce{NaOH}$ molecules down there. They all react with some water molecules of their own as well and create $5$ $\ce{OH-}$. So now we have $5$ $\ce{H3O+}$ and $5$ $\ce{OH-}$. Those two don't react with each other, correct? They just react with the water molecules and balance each other out, giving us a neutral solution.

I am unsure of this. Is this true? Thanks in advance.

$$\ce{NaOH + HCl -> NaCl + H2O}$$

$$\ce{Na+ + Cl- + OH- + H+ -> Na+ + Cl- + H2O_{(l)}}$$
Both sides $\ce{Na+}$ ans $\ce{Cl-}$ cancel out remaining reaction

$$\ce{H+ + OH- <=> H2O_{(l)}}$$

Any strong acid and strong base reaction the remaining reaction is hydrolization of water.


Why are d-d electronic transitions forbidden and weakly absorbing? Why do they occur at all? I am unable to understand why d-d electronic transitions are forbidden? Why are they weakly absorbing and apart from that, why do they occur at all?

In order for an electronic transition to be allowed (occur with strong intensity), certain "selection rules" must be obeyed.  You may already be familiar with the rule that the electron spin quantum number cannot change during a transition. It is this rule that forbids singlet-to-triplet absorption or emission (emission follows the same rules as absorption) and consequently makes phosphorescent emission so much weaker than fluorescent emission.

$\ce{\Delta S = 0}$ the Spin rule

Another rule, often referred to as the "LaPorte" rule or the "Orbital" rule, states that in molecules with a center of symmetry (centrosymmetric molecules), transitions within a subshell are forbidden when $\ce{\Delta l =0}$.

$\ce{\Delta l = \pm 1}$ the LaPorte rule

The orbital angular momentum quantum number "l" describes the type of orbital (s, p, d, f) within the subshell.  In centrosymmetric molecules, the various s, p, d, f orbitals cannot mix within a subshell.  For example, the 4d orbital cannot mix with the 4p orbital in a centrosymmetric molecule, such mixing being symmetry forbidden.  Octahedral complexes can be centrosymmetric.  In such a case 3p->3p, 3d->3d transitions would be forbidden by LaPorte's rule.  Tetrahedral molecules do not have a center of symmetry and p-d orbital mixing is allowed, so in the case of tetrahedral molecules 3p->3p and 4d->4d transitions may appear stronger because a small amount of another orbital may be mixed into the p or d orbital thereby removing the violation of LaPorte's rule.

If we can do something to remove the centrosymmetric nature of an octahedral complex, then we "relax" LaPortes rule and weak transitions can be observed.  For example, if a molecular vibration removes the molecular center of symmetry, then a d->d transition can occur if light is absorbed at that instant.  Things such as spin-orbit coupling, vibronic coupling, or ligands that can mix with the metal's d orbitals would all allow relaxation of LaPorte's rule and thereby permit weak transitions to be observed.


Is oxidation always thermodynamically favorable? In biological systems, energy is often extracted from molecules with reduced carbon backbones — the oxidation of those molecules has $\Delta G < 0$ and that energy can be captured, often in the reduction of other molecules. I'm wondering if it is always the case that oxidation is thermodynamically favorable and reduction thermodynamically unfavorable.

I suppose I'm not asking if the oxidation of any molecule is favorable (I'm pretty sure that's not the case), but for redox reactions that do occur, is it ever the case the oxidation half reaction is thermodynamically unfavorable and the reduction half reaction thermodynamically favorable?

As the term redox implies you have an oxidation and a reduction in every redox reaction.
It makes no sense to just talk about a half reaction, but you have to look at the cumulated energy difference of educts and products.
Now you have to remember two things:

You can easily convert between the energy and potential difference of a redox reaction by multiplying with the exchanged charges:
$$\Delta G= Uzq$$
$U$ is the potential (usually measured in Volt), $z$ the number of exchanged elementary charges and $q$ the elementary charge (usually measured in Coulomb).
Energy differences are transitive. That means:
$$\Delta G (A\rightarrow C) = \Delta G (A \rightarrow B) + \Delta G (B \rightarrow C)$$
There is no absolute zero point for the energy in chemical reactions. You can only talk about differences.

What chemists did, was to define a Standard electrode potential.
With this it is possible to measure half reactions compared to this standard. Because of sloppy naming, people call this the potential of the half reaction. But strictly speaking it is just the difference to the standard electrode potential. Since differences are transitive that is not a problem when working with potentials.

This means: The oxidation can be unfavoured compared to the standard electrode potential if the reduction is even more favoured.


Copolymer purity, what does AN / MA / MS mean? I would like to buy some polyacrylonitrile (PAN) powder from GoodFellow and I have the choice between these two products:

AN316010, quality : Copolymer (99.5% AN / 0.5% MA)
AN316020, quality : Copolymer (93.9% AN / 5.8% MA/ 0.3 % MS)

What I do not understand is the meaning of these three acronyms. What is the abbreviation hidden behind AN, MA and MS?
This may be an obvious question but I was unable to find any answer so far. I guess this may be somewhat linked to a certain degree of purity of the product, but since polymers can also be sold with different Monomer / Polymer ratios, I wanted to be sure...

My best guess for now is that AN may be AcryloNitrile, but I have no idea as for MA and MS... MA may be Methacrylic Acid, but could also be Methylmethacrylate Acrylonitrile or Maleic Anhydride (see here). Any help to answer this question is more than welcome.
Although in this particular case I may simply contact the supplier to have my answer I also wonder:
Is there a general way to solve these chemical identification issues or is it up to each supplier to name these types of compounds as they see fit?

I contacted GoodFellow directly and they gave me the following answer:

93.9% AN: 93.9% AcryloNitrile
5.8% MA: 5.8% Methyl Acrylate
0.3% MS: 0.3% Methyl Sulfonate

I guess different manufacturers may name their products as they see fit. In particular, for polymers that can share the same abbreviations, only the context and a good background knowledge of their synthesis pathway may allow one to decide...
There are, however, IUPAC guidelines for polymer abbreviations available here[1]. In addition several other websites offer their own abbreviation list e.g. here and there.

[1]. J. He et al., Abbreviations of polymer names and guidelines for abbreviating polymer names (IUPAC Recommendations 2014), 2014. DOI: 10.1515/pac-2012-1203


What does the Pi stand for in the reaction for biological nitrogen fixation? In the reaction for biological nitrogen fixation, what does the Pi stand for in the products?
$$\ce{N2 + 8H + 8e- + 16 ATP -> 2NH3 + H2 + 16ADP + 16 P_{i}}$$

It stands for inorganic phosphate (Pi). When ATP is broken down into ADP, energy is released along with a phosphate. You can visualise this if you look at the structure of ATP. ADP is then also able to be broken down into AMP and more Pi


Decrease in Surface area leads to Increase in load capacity for mesoporous material? So at the bottom of page 86 it says that when SBA-15 is functionalized with Glutaraldehyde it results in a substantial decrease in the surface area and pore diameter of the mesoporous material.
This paper rationalizes this by saying that the bridge chain insertion causes an increase in hydrophobicity that improves its ability in loading ibuprofen.  Can someone please why this is the case!!?? thank so much it just seems very conterintuitive.

http://ac.els-cdn.com/S0927776514001751/1-s2.0-S0927776514001751-main.pdf?_tid=cea489ec-ec87-11e6-a9e2-00000aab0f01&acdnat=1486398051_2d9403232650bb1a93cf30c41e9a1d72

Although I only have access to the abstract of the link you provided (unless you want to spot me $40 ;) they have characterized their glutaraldehyde-functionalized SBA-15 surface using at least 7 different analytical techniques, so I'm pretty convinced that the surface area, pore diameter and hydrophobicity are as they say.

The ~50% increase in surface adsorption in spite of the 13-fold decrease in surface area just illustrates the fact that surface adsorption is a function of two parameters: the affinity of the compound for the surface in question and the area of the surface available for adsorption.  It seems that for the system you've described, these two parameters simply switched from high-surface area / low adsorption affinity to low-surface area / high adsorption affinity, with the adsorption affinity "winning out" over the decreased surface area.


Is trans-2-butene IR active with regard to its C=C bond? To be IR active, the vibration of a bond must result in a substantial change in dipole moment. Since trans-2-butene is symmetrical, will the C=C stretch show up on IR?

Something tells me no because if the C=C bond vibrates, there will be no net change in the bond dipole moment, unlike in cis-2-butene. Any polarization one way in trans-2-butene is cancelled out by a polarization in the opposite direction in trans-2-butene. Cis-2-butene, however, has a permanent dipole and any vibration only exacerbates this dipole.

The $\ce{C=C}$ stretch will not show in IR. The frequency is $\nu_{\ce{C=C}}=1700~\mathrm{cm^{-1}}$ and has A gerade symmetry (DF-BP86/def2-SVP). There is no change in the dipole moment.

However, some bending frequencies involving the $\ce{C=C}$ double bond will show up. For example $\nu_{\ce{HC=CH}}=963~\mathrm{cm^{-1}}$ with a medium intensity.


What is the expected product in the reaction of hydrogen chloride with allylbenzene? Shouldn't it be 4th option since since the carbon attached to the ring is secondary as well as resonance stabilized?

I've searched SciFinder for this and related reactions (allyl benzene + H-X). In the results, there are no examples of the halide at the benzylic position in the product, and there are many examples of the halide at the secondary postion arising from the more substituted end of the alkene. One such example is below.

It is certainly warranted to consider both the direct addition and the product from rearrangement (since a resonance stabilized, benzylic carbocation would be formed as an intermediate), and if one were performing the experiment in the lab, it would be a good idea to look for both products. It seems that in practice, the addition of halide to the secondary carbocation is faster than the hydride shift. Predictions about kinetics like this are very difficult, and exam questions that require them could be considered unfair unless there is an overwhelming and predictable rationale for one product over another.


Analytical solution of the Schrödinger Equation for AB^+ systems According to Wikipedia, there are not many problems for which the Schrödinger equation can be solved analytically. [1] The $\ce{H2+}$ ion is probably the most complex molecule that can be treated this way.

As far as I know, every diatomic homonuclear system like this can be solved, e.g., $\ce{U2+}$.

Up to this point the molecules always consisted of two equal atoms which makes me curious about the following:

Is the Schrödinger equation not only solvable analytically for diatomic homonuclear systems $\ce{A2+}$ but also for diatomic heteronuclear systems like $\ce{AB+}$?

First, one has to be really careful with the words: when it is said that the Schrödinger equation can be solved the dihydrogen cation completely analytically, several remarks has to be made.

There are few different Schrödinger equations out there, so it is better to explicitly mention that one which can be solved analytically for $\ce{H2+}$ is the electronic Schrödinger equation. So, first of all, we are talking about the time-independent Schrödinger equation here, and secondly, we employ the Born-Oppenheimer approximation and solve only the electronic problem. No claim is maid with respect to analytical solutions of the whole time-independent Schrödinger equation without any approximation introduced, as well as regarding analytical solutions of its time-dependent counterpart.
The second remark is about the meaning of the word "analytical", or "analytic". The definitions taken from the Wikipedia are as follows:

  An equation is said to have an analytic solution if and only if at
  least one solution can be expressed as an analytic expression.

  An analytic expression is a mathematical expression constructed using well-known operations that lend themselves readily to calculation.

These definition seems to be rather vague for me, especially taking into account that,

  [..] the set of well-known functions allowed can vary according to context [...]

so we have to be careful relying on some intuitive notions of "analytic expression" and "analytic solution".

Having said so, and in hope that my intuitive notions of "analytic expression" and "analytic solution" are not miles away from the truth,  it is important to note that $\ce{H2+}$, for which analytic solutions are known, is (a special case of) a three-body system. And since there is no claim about analytic solutions for a four-body system, a five-body one, etc., from the beginning we have to restrict ourselves for three-body systems.

From that viewpoint the notation $\ce{AB+}$ used by OP is misleading, because the only three-body chemical system of $\ce{AB+}$ kind is $\ce{H2+}$. Already for the system composed of $\ce{H}$ and $\ce{He}$ the charge has to be +2 since we have to remove 2 electrons so that we end up with just one of them which together with 2 nuclei constitutes a three-body system, and not a four-body one.

Now, what about analytical solutions for $\ce{AB^{n+}}$ systems where $n$ is chosen appropriately so that the system is a three-body one? Well, I think, as for $\ce{H2+}$, there are analytic solutions for electronic Schrödinger equation in the Born-Oppenheimer approximation. The reason is pretty simple in fact: the clamped nuclei model reduces the three-body  problem to a single-body problem of the motion of a single electron within the fixed nuclear potential.

Why there are no related references (at least I could not find one)? It is very likely because chemist has little or no interest in three-body chemical systems of $\ce{AB^{n+}}$ type in general, and specifically, little or no interest in analytic solutions for these systems. Except for the simplest $\ce{H2+}$ system which serves as a primary model for the concept of the one-electron bond.


Why is porphyrin aromatic? I was asked which of the following were aromatic:

My understanding of aromatic nature is, that the molecule must be cyclic, planar, every atom in the ring must be sp2-hybridized and the molecule must have [4n+2] π electrons, where n∈N (Hückel’s rule). The molecular orbitals also have to have the correct orientation and phase for aromaticity.

A and B are aromatic since they follow conditions as above(including empty orbital of boron in conjugation.)
azulene's aromaticity is due to the “fusion” of
two aromatic ions, both with 6 π electrons, cycloheptatrienyl (tropylium) cation and cyclopentadienyl anion.

Based on the above My answer was a,b, and c .

Given answer : a,b,c,d.

I checked for d. It was aromatic as described in this link. 1. The diagram indicates only π electrons along the blue need to be considered amounting to 18 π electrons in conjugation.

My question is why only consider π electrons in blue and why not others? How do we explain the aromaticity of porphyrin?

References

https://en.wikipedia.org/wiki/Porphyrin

The aromatic properties are evident from the heats of combustion, the fact that all bonds have the same length, the ability of porphyrins and metalloporphyrins to undergo electrophilic substitution reactions, the stability of the porphyrin nucleus in electron impact mass spectrometry and the characteristic chemical shifts observed in proton NMR spectra1.

The 18π electron substructure (shown in blue in your figure) was thought to be responsible for the aromatic characteristics. In support of this view it was noted that the outermost bonds in the pyrrolenine rings(the not blue double bonds in your figure) take part in cycloaddition reactions(Fig.1, taken from ref.2) and therefore have a significant double bond character suggesting that their electrons are more localized.

This 18π model is also called "the diaza[18]annulene model" because it was speculated that the aromatic pathway found in porphyrins resembles [18]annulene(fig.2, ref.1). You can also note the similarities in the spectra of [18]annulene(fig.3, ref.3) and porphyrins: both have a major peak in the UV region and a less intense absorption at longer wavelengths.

Other supportive data to the the diaza[18]annulene model are proton NMR studies all mentioned in ref.1

https://www.worldscientific.com/doi/abs/10.1142/S1088424611004063
http://www.scielo.br/pdf/jbchs/v15n6/22664.pdf
Chem. Commun., 2016, 52, 4710


Calculate voltage, cost and time of nickel coating To nickel with a bath of $\ce{NiSO4}$ a cathodic current of $\pu{0.3 A/cm2}$ is applied with a square plate of side $\pu{5 cm}$ is used to nickel a certain substrate. Due to the formation of $\ce{H2}$ the current efficiency is $75\%.$ If the resistance of the solution is $\pu{0.4 \Omega}$ and the price of the current $\pu{1 c€/kWh},$ determine the voltage to be applied, the energy cost and the
  required time to nickel a total surface of $\pu{2 m2}$ with a coating thickness of $\pu{0.03 cm}.$
  Data: $M(\ce{Ni}) = \pu{58.71 g/mol}$; $\delta(\ce{Ni}) = \pu{8.9 g/cm3}.$

First I’m going to show you what I got

I’m sure this is wrong because I didn’t use given data. But I don’t know where should I use mass of nickel.

I’ve never seen this type of exercise before and I don’t where should I start from.

The volume of nickel plated is surface area times the thickness.
That's $2m^2 \times \pu{0.03 cm} $.

We'll use cubic cm as the units so it is: $2m \times 1m \times \pu{0.03 cm} = \pu{200 cm} \times \pu{100 cm} \times \pu{0.03 cm} = \pu{600 cm3} $

This helps to find the mass as the density is given as $\pu{8.9 g/cm3}$.

The mass = density $\times$ volume = $ \pu{8.9 g/cm3} \times \pu{600 cm3} = \pu{5.34 kg}$

Convert this to moles of Nickel: $\displaystyle \frac{5.34 \times \pu{10^3 g}}{\pu{58.71 g/mol}} = \pu{90.96 mol} $.
We'll come back to this.

Now for the current, $\pu{0.3 A/cm2}$ applied to a $5 \times \pu{5 cm}$ square plate, meaning that the total current on the plate at a given time is $\pu{0.3 A} \times \pu{25 cm2} = \pu{7.5 A}$ or $\pu{7.5 C/s}$. However, only a fraction of that actually reduces the nickel ($75\%$) so $\pu{5.625 A}$.

The voltage is then $ V = IR = 5.625 \times 0.4 $

The time calculation involves the actual nickel.
$I = \displaystyle \frac{q}{t}$
Where $q =$ total charge

We have $\pu{90.96 mol}$ of Nickel reduced.

$\ce{Ni^{2+} + 2e^- -> Ni}$

That's $\pu{2 mol}$ of electrons per mole of Nickel. So we need $\pu{181.92 mol}$ of electrons.

The charge per mol of electrons is Faraday's constant of $\pu{96485.3 C/mol}$
The total charge is just $\pu{181.92 mol} \times \pu{96485.3 C/mol}$.

Then we rearrange the formula of current for time $\displaystyle t = \frac{q}{I} = \frac{\pu{181.92 mol} \times \pu{96485.3 C/mol}}{\pu{5.625 C/sec}}$
$\text{time} = \pu{3120463 seconds}$ or $\pu{866 hours}$.

That's sort of the basic idea I'd imagine. I may have made a mistake somewhere in between. The time though while seems long is in line with this:
https://sciencing.com/calculate-electroplating-7597391.html

There, they plated just one mole of Cu and with a higher current. That took 2 hours. We have a lot more of $\ce{Ni}$ and smaller current.

Then $\text{Energy} = \text{power} \times \text{time} = VIt$ and you can calculate the cost from there.


Calculating enthalpy per gram mole of oxygen bond in haemoglobin A solution of $\pu{5 g}$ of haemoglobin in $\pu{100 cm^3}$ of solution shows a temperature raise of $\pu{0.031 ^\circ C}$ for complete oxygenation. Each mole of haemoglobin binds $\pu{4 mol}$ of oxygen. If the heat capacity of the solution is $\pu{4.18 J K^{-1} cm^{-3}}$, calculate $\Delta H$ per g of oxygen bond.

Attempt:

for $\pu{100 cc}$, heat capacity is $\pu{418 J K^{-1}}$
for $\pu{\frac{5}{64000}mol}$ of haemoglobin, $C\Delta T=\pu{12.952 J}$ is released
hence for $\pu{1 mol}$, $\pu{165862.4 J}$ energy is released
This energy is released when four bonds are formed with oxygen. Hence for 1 bond, $\pu{41.47 kJ}$ energy is released.

The answer given in the book is $\pu{-41.47 kJ}$.

The heat flows from each Hb molecule during the reaction at constant p and T, so $q_{p, rxn} = \Delta H_{rxn}\cdot\Delta X$.

The energy of the reaction is $\Delta H_{rxn}$ as per mole of reaction.

The reaction itself is Hb + 4O$_{2}\rightarrow$ Hb$\cdot$(O$_{2}$)$_{4}$.

The reaction is exothermic.

The number of moles of Hb present is $$n_{\rm Hb} = {5\over 64,000} = 0.078\;{\rm mmol},$$

all of which reacts: $\Delta X = -\Delta\,n_{\rm Hb} = n_{\rm Hb}$.

That much heat raises the temperature of the water (assuming all the heat is contained in the water):

$$0 = \Delta H_{\rm water} + q_{p, rxn} = C_{p}\Delta T + q_{p, rxn}$$
from which it follows that
$$q_{p, rxn} = -C_{p}\Delta T = -100\times 4.18\times 0.031 = -13\;{\rm J}$$
and
$$\Delta H_{rxn} = {q_{p}\over\Delta X} = {-13\;{\rm J}\over 0.078\;{\rm mmol}}= -167\;{{\rm kJ}\over{\rm mol-rxn}}$$
There are 4 moles of O$_{2}$ used so
$$\nu_{\rm O_{2}} = 4\;{{\rm O_{2}}\over{\rm rxn}}$$
and on an oxygen basis
$$\Delta H_{rxn} = -42\;{\rm kJ}$$


How can this ketone show Cannizzaro reaction? According to answer key, Cannizzaro reaction of 3-hydroxy-2,2-dimethyl-1-phenylpropan-1-one (a ketone) takes place but how is that possible when we know only aldehydes show Cannizzaro reaction?

There are a number of discussions on this site regarding the Cannizzaro reaction. In strong base there are competing reactions (aldol self-condensation) that can occur with ketones that have at least two hydrogens vicinal (next to) the ketone. Even aldehydes suffer the same fate. That is why the aldehydes are usually aromatic. But I believe in this case reduction may well be facilitated by an intramolecular hydride transfer. Intramolecular reactions are, as a rule, faster than intermolecular reactions owing to the high concentration of the reacting species. The keto alcohol 1 forms alkoxide 2 which reversibly adds to formaldehyde. Intramolecular hydride transfer occurs in intermediate 4. The formate group is saponified (5 ---> 7) forming the diol 9 and sodium formate 8. The deprotonation 7 ---> 8 should not be taken literally. Base and water intercede. Just saving space. See transition state for reduction below reaction mechanism.


Synthases vs Lyases - nomenclature Lyases are a class of enzymes that break down (lyase: ly- = lysis, -ase = enzyme) bonds like $\ce{C-C, C-S}$ and so on (except hydrolysis), and can eliminate molecules like $\ce{H2O}$ or $\ce{CO2}$. They are referred to as synthases, respectively, some authors categorize synthases among lyases.

My question is, (if word "synthesis" still means joining two substances into one more complex by creation of bonds between them), why are synthases placed among lyases if lyases actually break down bonds and synthases are (by their name) supposed to join, create bonds? Respectively, why was their name derived from 'synthesis' if their role actually is to break down?

Lastly I have even seen that JCBN (Joint Commission on Biochemical Nomenclature) dictated that "synthase" may be used for any enzyme catalysing synthesis, and that even without paying attention to whether the enzyme uses NTP like ATP (ligases = synthetases) or not (synthases).

I'd like to let everybody know that I understand the difference between synthases and synthetases, and that this difference isn't my point of question. I also do not controvert the fact that in the literature synthases are referred to as lyases, but somehow I kind of miss logic of naming enzymes responsible for breakdown after 'synthesis' which is actually opposite reaction than analysis. Until today I thought 'synthesis' stands for joining (syn-) and 'analysis' for breakdown into simpler products.

Here I attach in the images the answers from Wikipedia and forums for you to read.

Enzymes always catalyze reactions in both directions (all catalysts do). When an enzyme is first discovered and named, it is not always clear which direction is physiologically relevant (and sometimes both are).

There is a nice answer on StackExchange Biology talking about the enzyme pyruvate kinase, which acts as a phosphatase (i.e. in the opposite direction than the name suggests) in glycolysis. In the lab, you can study the reaction in both directions, depending on the concentrations of species present.

For the class of enzymes you are talking about, the same idea applies. A lyase can work in the other direction, as a synthase, and vice versa. If you want to classify enzymes in terms of mechanism and structure, there is no reason to separate the lyases from the synthases. They are expected to be highly similar or identical (if we are just using two names for the same enzyme).


What type of reactions are used in this synthesis? The compound $\ce{(CH3)2CHCH2NH2}$ can be synthesised by the following route.
$$\ce{CH3CH=CH2 ->[1] CH3CHBrCH3 ->[2] (CH3)2CHCN ->[3] (CH3)2CHCH2NH2}$$
What types of reaction are used in stages 1, 2 and 3?

A 1=substitution; 2=addition; 3=reduction

B 1=substitution; 2=addition; 3=hydrolysis

C 1=addition; 2=substitution; 3=reduction

D 1=addition; 2=addition; 3=reduction

E 1=addition; 2=substitution; 3=hydrolysis

Could someone please explain to me how they worked out what reactions were used in the stages? I have been able to figure out stage 1 and 2, but I am really stuck on stage 3, I think it should be addition but this isn't one of the answers.

Most reactions can be classified in a number of different ways, just like many compounds can be classified in a number of different groups. For each transformation, your question only allows for two choices, so you either work out which one works or you just need to identify the wrong choice.

The first reaction is to be classified as either a substitution or an addition. If we look at the structures of reactant and product, we realise that all the reactant’s atoms are present in the product — but an additional $\ce{HBr}$ have been added.

That should give you more than enough clues that it has to be an addition.
The second reaction is to be classified as either an addition or a substitution. Again, just by checking the atoms present in reactant and product, we realise that we lost the bromine but gained a nitrile group.

‘Loss and gain’ is definitely not an addition but a substitution.
The third reaction is to be classified as either a hydrolysis or a reduction. It does get a bit more tricky here. To identify a reduction, you would need to do one of the following:

allocate oxidation states and determine whether they have changed
check if formally an oxygen has been removed
check if formally $\ce{H2}$ has been added.

Identifying a hydrolysis is easier. It is a reaction with water that can occur in one of two ways:

the original molecule is split into two fragments, one of which contains a hydrogen, the other a hydroxy group
the molecule has formally added $\ce{H2O}$

Just by counting carbon atoms, we realise that the molecule has not fragmented, and by identifying atom symbols we can see that water has not formally been added. Thus, it cannot be a hydrolysis, it must be a reduction. In fact, two of the three reduction identification conditions are met, but I’ll leave the proof up to you.

Note that reduction is not the only way in which you can classify reaction 3. You could also call it:

hydrogenation
hydrogen addition
reductive nitrile cleavage
saturation
and probably others.


What is the lower bound to the temperature at which a fire can burn? To have a fire, you need 1) oxygen, 2) fuel and 3) heat. You can have different fuels that burn at different temperatures, but you always need some amount of heat to have a fire.

If you can use any fuel, what is the lowest temperature at which you can have a fire with an open flame? Is it possible to have a fire that's so cold that you can hold your hand in it without burning yourself?

Fires which burn below about 400°C are known as cool flames, and this is a well known and widely researched phenomenon. Most hydrocarbons and many alcohols can produce cool flames, and the conditions under which they burn depends on the oxygen content available. The most common encounter that most people have with the cool flame effect is through incomplete combustion of a petrol engine, leading to engine knocking.

The lowest recorded cool flame temperatures are between 200 and 300°C; the Wikipedia page references n-butyl acetate as 225°C. You can read a lot more about cool flames on that page. One thing to note is that cool flames are very hard to visibly detect at lower temperatures - both heat and light being two of the by products of the combustion process.

NASA have conducted a number of experiments in space on low temperature combustion of nano droplets of heptane, and this is an interesting area of science still being investigated.

An interesting point to make relates to your question of whether you could hold your hand in the flame and not burn yourself. Clearly the answer is no, but for some cool flames it may take noticeably longer to feel the pain of heat. Paper starts to discolour at about 150°C, but may not burn as ignition temperature is between 220-250°C. For reference, candle flames are about 600-1400°C.


Elimination of tertiary alkyl halide with lithium aluminium hydride to give alkenes I'm confused about the mechanism and thus cannot decide which alkene would be the product. Reaction with primary and secondary alkyl halides gives alkanes. What about this case?

The reaction is E1 due to the formation of most stable tertiary carbocation, as such the double bond will be on the most substited side to give most substituted alkene as a major product.