Matching a Slater-type wavefunction with a minimal (STO-nG) Gaussian basis set I have on hand a high-precision wavefunction expressed in Slater orbitals.  I need to express it as accurately as possible in a minimal Gaussian basis set.

For background: I am currently working with STO-6G, but it is not good enough to describe the cusp condition with the accuracy I need. How can I get to a basis like STO-7G, or even beyond to something like STO-10G, preferably in a form that is compatible with QChem?

TL;DR

The procedure to represent a Slater-type orbital (STO) as a linear combination of Gaussian-type orbitals (GTO) is outlined in W. J. Hehre, R. F. Stewart and J. A. Pople, J. Chem. Phys. 1969, 51, 2657 and is a quite straight forward least square fitting.

Derivation

I'll be quoting/ rephrasing quite liberally and take the 1s orbital as an example. The STO is defined as
\begin{align}
\phi_\mathrm{1s}(\zeta_1, \mathbf{r}) &=
  (\zeta_1^3/\pi)^{1/2}\exp(-\zeta_1 \mathbf{r}).
  \tag1\label{STO-1s}
\end{align}

The general form of a Gaussian-type 1s orbital is defined as
\begin{align}
g_\mathrm{1s}(\alpha, \mathbf{r}) &=
  (2\alpha/\pi)^{3/4}\exp(-\alpha \mathbf{r}^2).
  \tag2\label{GTO-1s-general}
\end{align}

The STO is expressed as a sum of $K$ GTOs; the coefficients and exponents are obtained for a STO with $\zeta=1.0$ and the scaled uniformly. Hence we can write
\begin{align}
\phi_\mathrm{1s}'(\zeta, \mathbf{r}) &=
  \zeta^{3/2}\phi_\mathrm{1s}'(1.0,\zeta\mathbf{r}),
  \tag3\label{GTO-1s-zeta1}
\end{align}
where
\begin{align}
\phi_\mathrm{1s}'(1.0, \mathbf{r}) &=
  \sum_k^K d_{\mathrm{1s},k}\,g_\mathrm{1s}(\alpha_k,\mathbf{r}).
  \tag4\label{GTO-1s-lc}
\end{align}

In order to obtain the coefficients $d$ and exponents $\alpha$ the contracted GTO is fitted to the form of the STO via a least square method. Note that this is a quite unique approach, since most other basis sets are optimised to produce the minimal energy. It all comes down to minimising the integral
\begin{align}
I &=
  \int \left(
         \phi_\mathrm{1s}(1.0,\mathbf{r})-\phi_\mathrm{1s}'(1.0,\mathbf{r})
       \right)^2 \mathrm{d}\mathbf{r}.
  \tag5\label{min-int}
\end{align}

For $K=3$, i.e. STO-3G, the following values are obtained:
\begin{array}{llll}\hline
k & \alpha_{\mathrm{1s},k} & d_{\mathrm{1s},k}      & I\\\hline
  &                        &                        & 3.31\times10^{-4}\\
1 & 1.09818 \times10^{-1}  & 4.44635 \times10^{-1}  &\\
2 & 4.05771 \times10^{-1}  & 5.35328 \times10^{-1}  &\\
3 & 2.22766                & 1.54329 \times10^{-1}  &\\\hline
\end{array}

Implementation

Finally you need to scale the exponents to use the right exponent. From $\eqref{GTO-1s-zeta1}$ we see
$$\mathbf{r}\mapsto\zeta\mathbf{r}_{\zeta=1.0}$$
and from $\eqref{GTO-1s-general}$ we know $\mathbf{r}\mapsto\mathbf{r}^2$ and therefore the exponent
$$\alpha_\zeta\mapsto\alpha_{\zeta=1.0}\zeta^2.$$

Hydrogen for example has $\zeta=1.24$ (Tab. IX) and therefore the exponents that need to be implemented are:
\begin{array}{lll}\hline
k & \alpha_{\mathrm{1s},k} & d_{\mathrm{1s},k}      \\\hline
1 &       0.16885540       &      0.44463454        \\
2 &       0.62391373       &      0.53532814        \\
3 &       3.42525091       &      0.15432897        \\\hline
\end{array}

For Q-Chem this is explained in the manual under 7.4.3 and needs to have the form
\begin{array}{llll}
\text{\$basis}\\
\ce{X} & 0\\
L & K & \text{scale}\\
α_1  & C_1^{L_\mathrm{min}} & C_1^{L_\mathrm{min}+1} &\dots& C_1^{L_\mathrm{max}}&\\
α_2  & C_2^{L_\mathrm{min}} & C_2^{L_\mathrm{min}+1} &\dots& C_2^{L_\mathrm{max}}&\\
\vdots    & \vdots & \vdots &\ddots & \vdots \\
α_K  & C_K^{L_\mathrm{min}} & C_K^{L_\mathrm{min}+1} &\dots& C_K^{L_\mathrm{max}}&\\
\text{****}\\
\text{\$end}\\
\end{array}
where $\ce{X}$ is the atomic symbol, $L$ is the angular momentum, $K$ is the degree of contraction, $\text{scale}$ is the scaling applied to all exponents, $\alpha_i$ are the exponents, and $C_i^L$ are the contraction coefficients.

As a practical example, the STO-3G basis set shall look like this:

$basis
   H   0
   S   3   1.00
      3.42525091             0.15432897
      0.62391373             0.53532814
      0.16885540             0.44463454
****
$end

For the actual fitting procedure, i.e. minimising $\eqref{min-int}$ you should consult a more technical oriented network page. It might be on topic at Mathematics, Computational Science, or even Mathematica depending on the question that troubles you. Please check their guidelines before asking.

General remarks

The most principle shortcoming of GTOs is that they cannot describe the cusp correctly. It's a trade-off you just make for faster integral evaluation. With STO-6G you are getting the orbital quite right already and there is a reason why we usually use STO-3G and not some higher contraction. It just offers the best cost : accuracy ratio.
This can be quite neatly seen in the graphical representation. The black curve  is the original STO, purple is STO-6G, cyan is STO-4G, and red is STO-2G plotted online at the wonderful FooPlot. Hack it here (I hope that works).

Going from STO-6G to STO-7G won't give you much more accuracy, maybe not even going to STO-10G. You probably have to go more into the direction of STO-20/30G. At that point it is probably more economic to switch to a program that can handle STO natively. From Wikipedia I see two, that handle those: ADF, and MolDS. The latter seems young, but is free.

Addendum:

If the FooPlot link does not work, here are the used formulae:

(1/pi)^(1/2)*exp(-abs(x))

0.679*(2*0.152/pi)^(3/4)*exp(-0.152*x^2)+
0.430*(2*0.852/pi)^(3/4)*exp(-0.852*x^2)

0.292*(2*0.088/pi)^(3/4)*exp(-0.088*x^2)+
0.533*(2*0.265/pi)^(3/4)*exp(-0.265*x^2)+
0.260*(2*0.955/pi)^(3/4)*exp(-0.955*x^2)+
0.0568*(2*5.22/pi)^(3/4)*exp(-5.22*x^2)

0.130*(2*0.0651/pi)^(3/4)*exp(-0.0651*x^2)+
0.416*(2*0.158/pi)^(3/4)*exp(-0.158*x^2)+
0.371*(2*0.407/pi)^(3/4)*exp(-0.407*x^2)+
0.169*(2*1.19/pi)^(3/4)*exp(-1.19*x^2)+
0.0494*(2*4.24/pi)^(3/4)*exp(-4.24*x^2)+
0.00916*(2*23.1/pi)^(3/4)*exp(-23.1*x^2)


What is the structure of trifluoromethanesulfonic acid in the gas phase? During an interesting discussion in our group’s seminar, the question arose what actually causes trifluoroacetic acid to be a much better nucleophile than trifluoromethanesulfonic acid. Part of the discussion was whether the oxygens of trifluoromethanesulfonic acid were equal or not.

Traditionally, $\ce{TfOH}$ is drawn as shown in the structure below on the left. The structure would assume the molecule having point group $C_\mathrm{s}$ due to the non-equality of the oxygen atoms only one of which bears a hydrogen. The triflate anion $\ce{TfO-}$ in solution, on the other hand, has three homotopic oxygens as can be seen in the structure on the right, leading to the more symmetric point group $C_\mathrm{3v}$.

One of the opinions voiced in the discussion was that even the gas-phase molecule $\ce{TfOH}$ adopted a $C_\mathrm{3v}$ point group, the hydrogen being in the same axis as carbon and sulfur and somewhat centred between the three oxygens. This makes perfect sense when thinking about it. Therefore, my questions are:

Has the structure of trifluoromethanesulfonic acid or a structurally strongly related acid been studied in the gas phase? (Please provide a reference)
Does the experiment suggest equivalent or non-equivalent oxygens?
How well can that idea be extended? e.g. should one consider the structure of formic acid to be $C_\mathrm{2v}$ symmetric rather than $C_\mathrm{s}$?
Do (solid-state) crystal structures or other experiments exist that confirm a different structure for solid $\ce{TfOH}$?
If solid and gas phase structures differ substantially, what about the structure of the molecule in liquid phase/solution?

Has the structure of trifluoromethanesulfonic acid or a structurally strongly related acid been studied in the gas phase? (Please provide a reference)

Yes. See Gas Phase Structure of Methyl Trifluoromethanesulfonate, CH3OSO2CF3, and Conformational Properties of Covalent Sulfonates Inorg. Chem., 1999, 38 (13), pp 3051–3055 and references cited therein, especially Electron Diffraction Investigation of the Molecular Structure of Trifluoromethanesulfonic acid (triflic acid) Z. Naturforsch. Teil A 1981, 36, 917

  Does the experiment suggest equivalent or non-equivalent oxygens?

Non-equivalent.

The 1981 paper finds

S=O 141.7 ( 1) pm
S-O 155.7 ( 2) pm

  How well can that idea be extended? e.g. should one consider the structure of formic acid to be C2v symmetric rather than Cs?

From the following gas phase study, I don't think that gas phase formic acid monomer is C2v.
http://actachemscand.dk/pdf/acta_vol_23_p2848-2864.pdf

  Do (solid-state) crystal structures or other experiments exist that confirm a different structure for solid TfOH?

Quoting from Trifluoromethyl Chlorosulfonate, CF3OSO2Cl: Gas Phase and Crystal
Structure, Conformation and Vibrational Analysis Studied by
Experimental and Theoretical Methods Inorg. Chem. 2003, 42, 7297-7303

"The structure of triflic acid, HOSO2CF3 (trifluoromethansulfuric acid), was also studied by GED, and an anti orientation
was assumed in the analysis of the diffraction pattern. [reference 6: the 1981 reference above] In
contrast, an X-ray investigation of 2HOSO2CF3 [dot] H2O [ie. the hemihydrate] resulted
in gauche orientation of the O-H bond,[reference 7] in accordance with
theoretical calculations. [references 8 and 9]"


What is the inert pair effect? I was reading about the p-block elements and found that the inert pair effect is mentioned everywhere in this topic. However, the book does not explain it very well. So, what is the inert pair effect? Please give a full explanation (and an example would be great!).

The inert pair effect describes the preference of late p-block elements (elements of the 3rd to 6th main group, starting from the 4th period but getting really important for elements from the 6th period onward) to form ions whose oxidation state is 2 less than the group valency.

So much for the phenomenological part. But what's the reason for this preference? The 1s electrons of heavier elements have such high momenta that they move at speeds close to the speed of light which means relativistic corrections become important. This leads to an increase of the electron mass. Since it's known from the quantum mechanical calculations of the hydrogen atom that the electron mass is inversely proportional to the orbital radius, this results in a contraction of the 1s orbital. Now, this contraction of the 1s orbital leads to a decreased degree of shielding for the outer s electrons (the reason for this is a decreased "core repulsion" whose origin is explained in this answer of mine, see the part titled "Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?") which in turn leads to a cascade of contractions of those outer s orbitals. The result of this relativistic contraction of the s orbitals is that the valence s electrons behave less like valence electrons and more like core electrons, i.e. they are less likely to take part in chemical reactions and they are harder to remove via ionization, because the s orbitals' decreased size lessens the orbital overlap with potential reaction partners' orbitals and leads to a lower energy. So, while lighter p-block elements (like $\ce{Al}$) usually "give away" their s  and p electrons when they form chemical compounds, heavier p-block elements (like $\ce{Tl}$) tend to "give away" their p electrons but keep their s electrons. That's the reason why for example $\ce{Al(III)}$ is preferred over $\ce{Al(I)}$ but $\ce{Tl(I)}$ is preferred over $\ce{Tl(III)}$.


Average or individual molar heat capacity? In finding enthalpy change $\Delta H$ during adiabatic, reversible expansion of  mixture of two ideal gases $n_1$ mol of $\ce{A}$ ($C_{\mathrm{m},v}=\frac{3}{2}R$) and $n_2$ mol of $\ce{B}$ ($C_{\mathrm{m},v}=3R$) taken in a container and during this temperature change is $T_\mathrm{i}$ to $T_\mathrm{f}$. Then what should be the value of $\Delta H$?

It can be $$\Delta H=\left(n_1\right)\frac{5}{2}R\left(T_\mathrm{f}-T_\mathrm{i}\right)+\left(n_2\right)4R\left(T_\mathrm{f}-T_\mathrm{i}\right)$$ taking individully for each gas. Or let say average molar heat capacity of gaseous mixture at constant pressure be $C_{\mathrm{m},p,\mathrm{avg}}$ then it can be $$\Delta H=\left(n_1+n_2\right)C_{\mathrm{m},p,\mathrm{avg}}\left(T_\mathrm{f}-T_\mathrm{i}\right)$$ so please help which is correct one. As per suggestion both are same.
$$C_{\mathrm{m},p,\mathrm{avg}}=\frac{\left(n_1\right)\frac{5}{2}R+\left(n_2\right)4R}{n_1+n_2}$$

They both are same.

If you plug $C_{\mathrm{m},p,\mathrm{avg}}$ in your second equation, you will get your first equation. The average $C$ is calculated as a mole fraction weighted average.


Fe with different charges Taken these reactions:

  \begin{align}
\ce{2\overset{0}{Fe} + 3Cl2 &-> 2\overset{+3}{Fe}Cl3} \tag{1}\\
\ce{\overset{0}{Fe} + 2HCl &-> \overset{+2}{Fe}Cl2 + H2} \tag{2}
\end{align}

Why does $\ce{Fe}$ have +3 charge in the first example and +2 charge in the second?

Iron exists in the +2 oxidation state (iron(II)), also called ferrous iron, as well as the +3 oxidation state (iron(III)), also called ferric iron.  There are sever other oxidation states of iron that are not pertinent to your reactions.

  In your first reaction, the strongly oxidizing $\ce{Cl}$2
  oxidizes the neutral, metallic iron to it's highest stable oxidation
  state, iron(III).

  In your second reaction, $\ce{HCl}$ oxidizes iron to the lower stable
  oxidation state, iron(II).


How does pH of soda drink vary with time Basically I have done an experiment where I measure the pH of my soda drink at a specific temperature over $\pu{45 min}$. At every temperature, the $\mathrm{pH}$ of my soda drink increases, but much faster for higher temperatures.

For instance, I measured at $\pu{25 ^\circ C}$ how fast the pH increases of my soda drink over $\pu{45 min}$. Next I measured at $\pu{35 ^\circ C}$ how fast the $\mathrm{pH}$ increases of my soda drink over $\pu{45 min}$. Continued for the next 4 temperatures.

I plotted the graph of how $\mathrm{pH}$ increases with time for 6 temperatures (getting me 6 graphs), but I have difficulty finding out the relationship I have is linear or exponential. The current trend I have only seems to be linear from the $\pu{45 min}$ time period I performed my experiment for, but I was wondering, if it just looks linear because my experiment was faulty, or it is actually supposed to be linear in relationship. I basically do not know if there is a theory which states that the relationship I am looking for is or is not linear.

I will appreciate the assistance.

I will make few simplifications:

I will neglect presence of other acids like citric or phosphoric acid, considering just water and carbon dioxide.
I will use the simplified exquation of $\mathrm{pH}$ of a weak acid (  :

$$\mathrm{pH}=\frac 12 (\mathrm{p}K_\mathrm{a} - \log {[\ce{CO2}]}) \tag{1}$$

I will  consider $\ce{CO2}$ escaping as the exponential process of the 1st order kinetics.

$$[\ce{CO2}]=A + B \cdot \exp{(-Ct)} \tag{1a}$$

I will  neglect the final equilibrium $\ce{CO2}$ concentration, compared to initial one:

$$[\ce{CO2}]=B \cdot \exp{(-Ct)} \tag{1b}$$

We can realize that the exponential function for $\ce{CO2}$ escaping kind of nullifies the logarithm function of $\mathrm{pH}$ definition:

$$\mathrm{pH}=\frac 12 \left(\mathrm{p}K_\mathrm{a} - \log {\left(B \cdot \exp{(-Ct)}\right)}\right) \tag{2}$$

$$\mathrm{pH}
\simeq \frac 12 \left(\mathrm{p}K_\mathrm{a} - \log {B} + \frac1{2.303}Ct\right) \tag{3}$$

Therefore, $\mathrm{pH}=f(t)$ is approximately linear.

It is not linear if we consider all factors, especially if other acids are present.

Additional requested clarification:

The equation (1) is quite notoriously known simplified equation for $\mathrm{pH}$ of weak acid. It can be directly derived from definition of a dissociation constant of an acid:

$$K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}} \tag{4}$$

involving 2 simplifications:

solution is acidic enough to ignore water auto-dissociation:

$$K_\mathrm{a} \simeq \frac{\ce{[H+]}^2}{\ce{[HA]}} \tag{5}$$

solution contains enough of a weak acid, so it's majority is not dissociated:

$$K_\mathrm{a} \simeq \frac{\ce{[H+]}^2}{c} \tag{6}$$

where

$$c=\ce{[HA]}+\ce{[A-]} \tag{7}$$

Then

$$\ce{[H+]}=\sqrt{ K_\mathrm{a} \cdot c} \tag{8}$$

After logaritmization of (8), we get (1).


Why does aqeous fluorescein shine when exposed to UV light? I know that (aq) fluorescein "absorbs" UV light and releases light in the visible spectrum , but how does it do that. It is a really interesting and useful organic chemical and I just wanted to know what goes on in it so that it lights up.

The process is called fluorescence. Some substances, such as calcium sulfide, $\ce{CaS}$, absorb light and release it slowly in a process called phosphorescence, glowing for some time after the source of light is removed.


Do Faraday's gold colloidal solutions still exist? The Royal Institution seems to claim that they still possess some of Faraday's original gold colloids: "These liquids are some of the first examples of metallic gold colloids, made by Michael Faraday over 150 years ago."

However, in a publication in Gold Bulletin (D. Thompson, Gold Bull. 2007, 40(4), 267-269.), the following is stated: "Faraday's original gold colloids were very stable and at least one example survived until the second World War, when it was accidentally destroyed by enemy action [...]."

Do Faraday's gold colloidal solutions still exist?

Please note that I haven't seen the sample myself!

However, the Royal Institution claims that a bottle with the collidal gold solution is on display at Michael Faraday's Magnetic Laboratory (see bottle at the bottom of the image) in the museum of the Royal Institution.

Update

This might be nitpicking and English isn't my native language, but David Thompson does not explicitly state in his article in Gold Bulletin that all samples are lost.

On the other hand, Ryan D. Tweney in his article Discovering Discovery: How Faraday Found the First Metallic Colloid, published in Perspectives on Science, 2006, 14, 97-121 (DOI) not only describes the finding of 700 surviving microscopic slides in Faraday's reconstructed magnetic with gold film samples and numbered in Faraday's hand and that the numbers correspond to the numbering in Faraday's 1856 Diary (lab journal) but continues:

  It had long been known that some of Faraday's gold colloids survive from the 1856 research: Four bottles of these are on display at the Royal Institution in London […]. These few colloids are all that survive of many dozens referred to in the Diary.


Can I use the Henderson–Hasselbalch for calculating the pH of non-buffer systems? Can I use the Henderson–Hasselbalch equation on reactions that are not buffers? When I read up on the derivation of the equation, I noticed that there is absolutely no assumption on the relative concentrations of the acid and conjugate base at least for a monoprotic acid.

Here's a quick derivation of the Henderson Hasselbalch equation:
The reaction:

$$\ce{HA(aq) + H2O(l) <=> A-(aq) + H3O+(aq)}$$

$$K_\mathrm{a} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$

Then I take the logarithm of each side, resulting in

$$\log K_\mathrm{a} = \log\frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]} = \log\frac{[\ce{A-}]}{[\ce{HA}]} + \log [\ce{H3O+}]$$

Consequently,

$$-\mathrm{p}K_\mathrm{a} = -\mathrm{pH} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

Can I use the Henderson-Hasselbalch equation on reactions that are not buffers?

If you know the equilibrium concentrations of weak acid and conjugate base, you can use the equation. You can also use it when you know the $\mathrm{pH}$ and the $\mathrm pK_\mathrm a$ and want to know the ratio of protonated and deprotonated species (for example for relating color and $\mathrm{pH}$ when working with pH indicators). Written as a formula, the following is always correct for a weak acid and conjugate base at equilibrium:

$$ \mathrm{pH} = \mathrm pK_\mathrm a + \log\frac{[\text{weak base}]_{\text{eq}}}{[\text{weak acid}]_{\text{eq}}}$$

or

$$ \log\frac{[\text{weak base}]_{\text{eq}}}{[\text{weak acid}]_{\text{eq}}} = \mathrm{pH} - \mathrm pK_\mathrm a $$

The equation is especially convenient for a buffer because when you mix high concentrations of weak acid and its conjugate base, the concentrations don't change much when the system reaches equilibrium, so you can approximate the $\mathrm{pH}$ by plugging in the initial concentrations. So for buffers,

$$ \mathrm{pH} = \mathrm pK_\mathrm a + \log\frac{[\text{weak base}]_{\text{eq}}}{[\text{weak acid}]_{\text{eq}}} \approx \mathrm pK_\mathrm a + \log\frac{[\text{weak base}]_{\text{initial}}}{[\text{weak acid}]_{\text{initial}}}$$

For a solution of a weak acid only or weak base only, that does not work (one of the initial concentrations is zero, and you get a nonsense result). The approximation is best for high and similar concentrations of acid and base when the $\mathrm pK_\mathrm a$ is close to $7$.


The diference between covalent molecule lattice and covalent network lattice Can you tell me the difference between covalent molecular lattice and covalent network lattice?
Thanks for your help.

Covalent molecular compounds are actual molecules. Examples include $\ce{H_2O, NH_3, CO_2, N_2, I_2, SO_3, N_2O_4}$ etc. They are made of molecules which are made of covalently bonded atoms. Covalent network compounds form a network of bonds, and the whole object can almost be considered a "molecule," such as diamond (C network), silicon (Si network), graphene (C network), and black phosphorous (P network).

It's not a coincidence that all but one of the covalent molecules that I listed are gasses at room tenperature and pressure. The intermolcular bonds between molecules are relatively weak. On the other hand, every molecule in a network solid is surrounded in a network of strong covalent bonds.


Why is this molecule named 1,1-chloroiodoethane, rather than 1-chloro-1-iodoethane? $\ce{CH3CHClI}$

If there is more than one substituent on a carbon, shouldn't they be named separately like 1-chloro-1-iodoethane?

The correct name is 1-chloro-1-iodoethane since the locants are placed immediately in front of the prefixes that represent the substituent groups.

Furthermore, since the release of the IUPAC recommendations of 1993, generally all locants are placed immediately in front of the part of the name to which they refer (e.g. cyclohex-2-en-1-one). The corresponding rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows.

  P-14.3.2 Position of locants

  Locants (numerals and/or letters) are placed immediately before that part of the name to which they relate, except in the case of the traditional contracted names when locants are placed at the front of their names.

  P-15.1.7.1.4 The longest chain is numbered from one end to the other by arabic numbers, the direction being chosen so as to give the lower locants to the substituent groups (side chains) [see P-14.4(f)]. The lower set of locants is defined as the set that, when compared term by term with other locant sets cited in order of increasing magnitude, has the lower term at the first point of difference (see P-14.3.5). The locants are placed immediately in front of the part of the name to which they refer. Identical simple substituent groups are indicated by multiplicative prefixes, such as ‘di’, ‘tri’, etc. [P-16.3.3(b)]. For compound or complex substituent groups (see P-29.4 and P-29.5), the multiplicative prefixes ‘his’, ‘tris’, ‘tetrakis-‘, etc. (P-14.2.2) are used as described in P-16.3.5(a).


Number of stereoisomers of 3-ethylpent-1-ene-1,4-diol I am trying to figure out how many stereoisomers this compound has. I know that there are two chiral carbon atoms, which allow $2\times2 = 4$ possible isomers. There is also a double bond which, as far as I know, doubles the number of stereoisomers, since there are now four possibilities for the cis-isomer, and four for the trans-isomer. So I thought the total number of stereoisomers was eight, but my teacher told me the answer was six. I have thought about it for a while, but I can't figure out why it should be six. What did I do wrong?

The answer lies in the double bond.  The alcohol on the double bond undergoes keto-enol tautomerism. The greatly more stable aldehyde effectively makes the vinylic alcohol non-existent. This removes a factor for stereoisomers.  six is an odd number of stereo isomers to have since 6 is factored to 2 and 3.  This implies a stereoisomer with 3 forms.  I think your teacher may be thinking of the trans, cis and aldehyde state of the alcohol to come up with six, but this would neglect a carbon stereo center. in this case 12 would be the correct answer. However you should ignore the tautomers and just assume aldehyde which means the answer should be 4.


What will be the reaction between tin (II) chloride and copper (II) chloride? I just tried conducting an experiment during lunch today. Now, I calculated this reaction 4 days ago dealing with electrochemistry,

$$\ce{SnCl2 + CuCl2 -> Cu(s) + SnCl4}$$

Theoretically, I got these 1/2 cell reactions from a couple different sources (one of which are posted below):

$$\ce{Sn^2+ -> Sn^4+ + 2e-} \quad E° = -0.15~\mathrm{V}$$

$$\ce{Cu_{(s)} -> Cu^2+ + 2e-} \quad E° = -0.34~\mathrm{V}$$

Using basic electrochemistry, we know that Copper prefers to be copper metal. Due to its overbearing power over the tin(II) ion, the copper(II) ion forces the tin(II) ion to become tin(IV).

I did not check the reduction or oxidation potential of the chloride ion (since the chloride ion is the most stable form of chlorine due to its (relative to the other compounds in this experiment) high ionization potential. I concluded that chloride would be a spectator ion.) Finally resulting in this net ionic reaction.

$$\ce{Sn^2+ + Cu^2+ -> Cu_{(s)} + Sn^4+} \quad E^\circ_{\rm cell} = 0.19~\mathrm{V}$$

I mixed the compounds in $150~\mathrm{ml}$ of deionised water. Later combining them only to end up with a milky-green colored solution. I saw some brown-red precipitate at the bottom of the beaker. however, only a small amount of what I think is copper precipitated. Tomorrow I will place the beaker with the solution onto a hot plate in hopes of speeding up the reaction.

I was wondering if a smaller difference in electrical potential between the two half cells results in a slower reaction. The other thing I was wondering concerns any other ideas or misconceptions I may have faced while conducting this experiment.

This is indeed a redox reaction between copper and tin, but you should instead use reduction potentials for the $+2$ ions to calculate the potential.

This might be quite an exothermic reaction. http://www.periodicvideos.com/videos/mv_copper_chloride.htm


How to compare size of subshells? For the same principal quantum number, on increasing the value of the azimuthal quantum number does the average radius of the subshell increase or decrease?
In other words, which out of, say, 3s, 3p, 3d would have the largest average radius and why?

For a hydrogen atom, according to Hertel and Schulz [1]:
$$\langle r\rangle_{n,\,l}=\frac{a_0 n^2 \left(\frac{1}{2} \left(1-\frac{l (l+1)}{n^2}\right)+1\right)}{Z}$$
and
$$\langle r^2\rangle_{n,\,l}=\frac{a_0^2 n^4 \left(\frac{3}{2} \left(1-\frac{l
   (l+1)-\frac{1}{3}}{n^2}\right)+1\right)}{Z^2},$$
where $\langle r\rangle_{n,\,l}$ is the average distance of the electron from the nucleus as a function of n and l, and  $\langle r^2\rangle_{n,\,l}$ is the average distance squared of the electron from the nucleus as a function of n and l, $a_0$ is the Bohr radius, and Z is the nuclear charge (Z = 1 for hydrogen).  These are known as expectation values, because these are the expected values based on the probabilistic distribution of electron density.
Using these, we can compare $\langle r\rangle_{n,\,l}$ and $\langle r^2\rangle_{n,\,l}$ for the $3s \,(n = 3, l = 0), 3p \,(n = 3, l = 1) \text{ and } 3d\, (n = 3, l = 2)$ subshells:
$$\langle r\rangle_{3,\,0}= \frac{27 a_0}{2 Z}$$
$$\langle r\rangle_{3,\,1}= \frac{25 a_0}{2 Z}$$
$$\langle r\rangle_{3,\,2}= \frac{21 a_0}{2 Z}$$
$$\langle r^2\rangle_{3,\,0}= \frac{207 a_0^2}{Z^2}$$
$$\langle r^2\rangle_{3,\,1}= \frac{180 a_0^2}{Z^2}$$
$$\langle r^2\rangle_{3,\,2}= \frac{126 a_0^2}{Z^2}$$
You can see that, as the azimuthal quantum number (l) increases, the electron's average distance from the nucleus decreases.  Note, however, that in hydrogen (or any single-electron atom, such as $\ce{He^+}$ or $\ce{Li^{2+}}$, etc.), all orbitals in the same shell have the same energy. Thus the energies of the 3s, 3p, and 3d orbitals are identical.  As a consequence, a hydrogen electron excited to the third energy level is actually in a weighted linear combination of all of these orbitals, and its actual average distance would likewise be a weighted average of these.
But still, why does the average distance decrease with increasing azimuthal quantum number?  Well, quantum is not my field, but I believe this is what's going on (perhaps a quantum expert can let us know if this is correct):
As I noted, in any single-electron atom, the 3s, 3p, and 3d orbitals have the same energy. The electron's energy is a combination of potential energy due to its distance from the nucleus, and the energy due to its angular momentum.  As the azimuthal quantum number increases, the electron's angular momentum increases (it has no angular momentum in 3s, some in 3p, and more in 3d), and thus the energy associated with this angular momentum increases.  But since the total energy is constant, as the energy due to angular momentum increases, the potential energy associated with distance from the nucleus must decrease, and hence the average distance from the nucleus gets lower.
Finally, you can see from the formula that $\langle r\rangle$ is much more sensitive to changes in n than l.  For instance, if we compare the sizes of the s, p, and d subshells in the 3rd shell (n = 1, l = 0 to 2), $\langle r\rangle$ ranges from $\frac{27 a_0}{2 Z}$ to  $\frac{25 a_0}{2 Z}$ to $\frac{21 a_0}{2 Z}$.  By contrast, if we compare the sizes of the s-subshells in the first three shells (l = 0, n = 1 to 3), $\langle r\rangle$ ranges from $\frac{3 a_0}{2 Z}$ to $\frac{12 a_0}{2 Z}$ to $\frac{27 a_0}{2 Z}$.

Hertel, Ingolf V., and Schulz, Claus-Peter. Atoms, molecules and optical physics. Berlin: Springer, 2015


Acidic strength comparison between sulphuric acid and nitric acid Why is $\ce{H_2SO_4}$ more acidic than $\ce{HNO_3}$. Shouldn't it be the other way around because $\ce{N}$ is more electronegative than $\ce{S}$?

My reason: $\ce{O}$ can take electrons more easily from $\ce{S}$ than from $\ce{N}$. Hence, it will be more difficult to remove $\ce{H}$ from $\ce{H2SO4}$ than from $\ce{HNO3}$. Hence, $\ce{HNO3}$ should be more acidic.

The important thing about acidity is the stability of the conjugate base, which is enhanced by resonance effect and inductive effect. The more stable the conjugate base, the easier deprotonation becomes, and thus the stronger the acid.

In a $\ce{NO3^-}$ ion, the negative charge can delocalise among three O atoms, but in a $\ce{HSO4^-}$ ion, in addition to delocalisation among three O atoms, there is an additional OH group, which further stabilises the anion by inductive effect. So $\ce{HSO4^-}$ is the more stable conjugate base, and $\ce{H2SO4}$ is a stronger acid than $\ce{HNO3}$.


Ligand Binding Paradox? I have a ligand $\ce{L}$ and a receptor $\ce{R}$. As can easily be shown, with dissociation constant $K_\mathrm{D} = \frac{[\ce{L}][\ce{R}]}{[\ce{LR}]}$ the fraction of receptors occupied by ligand is

$$\theta = \frac{[\ce{LR}]}{[\ce{LR}] + [\ce{R}]} = \frac{[\ce{L}]}{[\ce{L}] + K_\mathrm{D}}$$

It's easy to see that the fractional saturation of receptors is independent of receptor concentration. Now here's where I get confused. Let's say we have a constant volume of $\pu{1 L}$ with $\pu{5 mol}$ ligand and a $K_\mathrm{D}$ of 5:

$$L = 5, \quad K_\mathrm{D} = 5 \quad \to \quad \theta = 0.5$$

If we have $\pu{10 mol}$ of receptor total ($[\ce{LR}] + [\ce{R}] = 10$), then $\pu{5 mol}$ will be bound to ligand. Now let's say we increase the number of receptors by $10\times$. Our formula for $\theta$ is independent of $[\ce{R}]$ or $[\ce{LR}]$. We now have $\pu{50 mol}$ receptor bound to ligand.

That being said, we only had $\pu{5 mol}$ ligand to start with, which is itself a paradox. Can someone explain to me where I went wrong?

I see where I made my mistake. I set $[\ce{L}] = 5$, but this "$L$" is referring to the equilibrium concentration of free ligand -- not the total ligand concentration. In order to do that I should have constrained the system to $[\ce{L}] + [\ce{LR}] = 5$. Working through what this would do currently.

So in the first case, my assumption actually dictated that there be $\pu{10 mol}$ ligand total instead of the $\pu{5 mol}$ I erroneously stated. I incorrectly equated total ligand and free ligand the first time I did this.

So I see where I messed up, but now I'm curious if the values of $[\ce{L}]$, $[\ce{R}]$, and $[\ce{LR}]$ can be solved as functions of $[\ce{L_{total}}]$, $[\ce{R_{total}}]$ and $K_\mathrm{D}$.


Mechanism of radical cyclisation to form hydroazulene core Recent studies on radical-induced cyclisation reactions have led to a simple, one-step method for the preparation of hydroazulenes from appropriately substituted cyclopentanone precursors. Treatment of 1, for example, with triphenyltin hydride and AIBN in refluxing toluene gave 2 in 79% yield. Give a mechanism for this transformation. (TES = triethylsilyl, SiEt3)

  (taken from: McKillop, A. Advanced Problems in Organic Reaction Mechanisms; Tetrahedron Organic Chemistry Series, Vol. 16; Pergamon Press: Oxford, U.K., 1997; p 1.)

I know that triphenyltin hydride, Ph3SnH, can act as a reducing agent but I have no idea of its reduction mechanism. And I know that AIBN is a radical initiator:

The given conditions strongly suggest a radical pathway. However, looking at the product I also feel like there is a carbocation intermediate, as there is a ring expansion, but I don't know whether or how that happens.

This reaction is taken from Tetrahedron Lett. 1995, 36 (17), 3015. (DOI:
10.1016/0040-4039(95)00502-4)

The heat (provided by refluxing) is required to initiate the reaction by the decomposition pathway you have drawn. The cyanopropyl radical thus formed abstracts a hydrogen atom from the tin hydride, a consequence of the weak Sn–H bond:

The triphenylstannyl radical is the one that actually does interesting chemistry. Generally, there are a few options for radicals at this stage. One common pathway is to abstract a halide or chalcogenide from the starting material, but there is no such atom here (C–F and C–O bonds are stronger and rarely broken, so the OTES group is out of bounds). The other common pathway is to add to a π-bond. In this case the best place to add is the terminal carbon of the alkyne, which is least sterically hindered.

After this, the remainder is a series of addition-elimination steps. Note that in each case the ring size formed is 3, 5, or 6: these are all kinetically fast cyclisations.

In particular, note that:

Intermediate 4 is stabilised by electron donation from adjacent oxygen (similar to carbocations, radicals - which are also electron-deficient species - can be stabilised by adjacent lone pairs).
You are correct that a ring expansion is needed – in fact, two are needed – but this does not require a Wagner–Meerwein shift in a carbocation. Radicals are perfectly capable of doing this job, usually by a sequence of cyclisation (5 to 6) and retro-cyclisation (6 to 7). See also J. Org. Chem. 2012, 77 (19), 8634–8647 for another example of this motif, which is very common in radical chemistry.
There isn't tin in the final product, so we will eventually have to kick it back out again: this can be achieved via a β-scission mechanism, so we have to work towards an intermediate which has a radical β to SnPh3, i.e. 7.

The original paper supports this mechanism (although they skipped some steps, as is usual in journals):


Is it really possible for a liquid to stay afloat a gas (vapor)? I have a book at home called Chemical Thermodynamics for Process Simulation (I have the first edition) and at the very beginning of chapter 2 of the book one may read the following:

No reference is given. I have never found anything about it so far and I am curious to know whether this statement is true.

Yes! Here's a video of Cody floating NaK (liquid) or ethanol/water (liquid) on compressed xenon (gas):
https://youtu.be/AsP4yMY-a6U


Which is the suitable process for ball milling? Why do we mainly choose the wet ball milling process in spite of dry ball milling process? Both processes are used to make a homogeneous mixture of different compounds.

Wet ball milling is much more effective due to Rehbinder effect producing a disjoining pressure.  Adsorption layers from surfactants that arise in microcracks significantly reduce the energy expenditure necessary to crush solid materials as a result of a decrease in their free surface energy upon contact with a medium containing substances capable of adsorption. Effect is observed in solids and porous materials, e.g. metals, ionic, covalent and molecular mono- and polycrystalline substances, glasses, polymers, both partially crystallized and amorphous.

This reduces milling time and wear of the working surface, at the same time improving homogeneity of the grinding fraction. The same effect is also used at the lab scale when, for example, metal oxides are dispersed in a mortar much more effectively when acetone is added. Sometimes after such treatment the reaction even takes place in a solid phase.


how to separate gamma-butyrolactone from N-methyl-2-pyrrolidone Since these two chemicals have different freezing temps (pyrrolidone is $-11~\mathrm{^\circ F}$ [$-24~\mathrm{^\circ C}$] and $\unicode[Times]{x3b3}$-butyrolactone freezes (also called melting) at $-46~\mathrm{^\circ F}$ [$-43~\mathrm{^\circ C}$]), could a person wanting to separate these just immerse them in liquid nitrogen (or something else?) to get the desired chemical to freeze, then filter off the unfrozen butyrolactone? Other routes seem quite difficult for a non chemist. Any objections or other routes? Anything you can add? Even a step by process will be much appreciated.

Cud [sic] a person wanting to separate these just emerse [sic] them into liquid nitro [sic] or any [sic] to get the desired chemical to freeze then filter off the unfrozen butyrolacton [sic]?

With liquid nitrogen this would not be possible because liquid nitrogen condenses/boils at -195 °C.  It would freeze both the pyrrolidone and the lactone.


Bending of water and methanol towards a charged objects I just had a question about why does water, methanol or any other polar substances tend to bend towards the charged object.

I think it is because of a slight dipole dipole movement between the charged object and the polar substance, but not sure if that's the reason.

As you state, in an electric field, a polar molecule such as methanol or water, having an uneven charge distribution, is attracted (or repelled) by an electric field. This orients the molecule, but is not primarily responsible for the attraction of the bulk substance.

An electric field causes a force on even a nonpolar molecule such as methane or sulfur, though. You can pick up powdered sulfur with a charged piece of plastic because of electrostatic induction, which moves electrons in the bulk material.

This induced charge is similar to the effect of moon (and sun) on tides, pulling the water underneath the moon more strongly than it pulls on the more distant core of the earth, and on the core more strongly than on the ocean on the opposite side of the planet, causing uneven distribution of the water. In a similar way, an external field causes an uneven distribution of charge.


What do "degrees of freedom" refer to? Can you explain simply what are the degrees of freedom in chemistry? I don't fathom it very intuitively from Wikipedia.

I have almost no background in physics. I understand that it refers to all the variables that can describe a system, e.g. when talking about the motion of the molecule, the degrees of freedom refer to the ways in which the molecule can move. I understand that an atom can only have translational motion, whereas a molecule can rotate, vibrate and translate. But how this gives the molecule more "freedom" is beyond me.

To find the number of degrees of freedom of a system, ask yourself the question, "in how many ways can I describe the motion of this system?"
Let us begin with monoatomic gaseous molecules. Visualise them as minute spheres.
Image source: http://vle.du.ac.in/mod/book/print.php?id=13264&chapterid=29093
Do you think there are any other ways to describe its motion? No, there aren't, but you might argue that it can also rotate about its own axis, so why do we exclude that? If that's the case, refer to this answer (tl;dr: The moment of inertia of an atom about its own axis is negligible due to the extremely small size and can be safely ignored.)
Now, let's switch to diatomic gases:

Source: https://physics.stackexchange.com/a/210955/168273
The diatomic molecule too is free to translate in any three coordinates of space. Also, it can rotate about the two axes perpendicular to the axis joining the two constituent atoms.
Again if you are wondering about the rotation about the axis joining the two atoms, apply the moment of inertia argument here too.
Alternative definition: The number of terms appearing in the energy equation of the molecule is called its degree of freedom.
$$E_\text{monoatomic}= \frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2
+ \frac{1}{2}mv_z^2$$
$$E_\text{diatomic}= \frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2
+ \frac{1}{2}mv_z^2 + \frac{1}{2}I_x\omega_x^2+\frac{1}{2}I_y\omega_y^2$$
where $I$ denotes the moment of inertia about an axis and $\omega$ the angular velocity about the same axis.
Do you see how this agrees with our qualitative analysis? There are $5$ terms in the energy equation of the diatomic molecule and $3$ in the case of the monoatomic one.
But there's a problem, at higher temperatures, under the provision of sufficient amount of energy the molecules can vibrate too, leading to an increment in the number of terms in our energy equation.
From Wikipedia:

The number of vibrational degrees of freedom, or vibrational modes, of a molecule is determined by examining the number of unique ways the atoms within the molecule may move relative to one another, such as in bond stretches or bends. This can be determined mathematically using the $3N-6=n_\text{DOF}$ rule, where $N$  is the number of atoms in the molecule and $n_\text{DOF}$ is the number of vibrational degrees of freedom. Note that for linear molecules, the rule instead becomes $3N-5=n_\text{DOF}$, meaning a linear polyatomic made up of $N$ atoms will have one less vibrational degree of freedom than a non-linear polyatomic with $N$ atoms.

Why are degrees of freedom important in thermodynamics?
According to the equipartition theorem an ideal gas distributes its energy equally in all degrees of freedom.  The average energy of a molecule in a gas associated with each degree of freedom is $kT/2$ where $k$ is the Boltzmann constant and T the temperature.
Task: Use the equipartition theorem to derive the concise internal energy equations for a diatomic and monoatomic gas molecule at low temperatures (i.e. ignore the vibrational degrees of freedom.)
Besides, knowing the degrees of freedom of the gas under investigation, we can find out its approximate molar specific heat capacities $C_p$ and $C_V$ which we often use in our thermodynamics calculations.


Heat of hydrogenation of 1,3-substituted cyclohexanes From these two molecules in question, (C) should be more unstable as it is in a cis-configuration. The t-butyl and methyl groups being on the same side of the ring would exert a larger steric force on each other as compared to the trans configuration. However, (D) is mentioned as the correct answer.

How is a trans-1,3 disubstituted cyclohexane more unstable than the corresponding cis-isomer? I drew the chair projections of both compounds and the cis one has both groups on the same side, which would cause 1,3-diaxial interaction as mentioned here

First of all, as Jan pointed out on the comment, better nomenclature for thes compounds is syn- and anti- instead of cis- and trans-, respectively. However, OP is more familiar with cis- and trans- nomenclature I'd continue it with my answer.

Each substituted cyclohexane ring (e.g., di-substituted ones here) has two chair conformers as depicted in OP's bottom diagram of methylcyclohexane (a mono-substituted one). In one conformer, the methyl group is in equatorial position (left hand structure) and the methyl group in the other one is in axial position (right hand structure). Because of two 1,3-interactions cost additional approximately $2 \times \pu{3.8 kJ/mol}$ steric strain energy, the predominant conformer is equatorial one.

Now, if you put tert-butyl group in $\ce{C3}$-equotorial position on equatorial-methylcyclohexane (left hand structure, vide supra), you get cis-1,3-substituted compound (C). On the other hand, if you put tert-butyl group in $\ce{C3}$-equatorial position of axial-methylcyclohexane (right hand structure), you get trans-1,3-substituted compound (D). Because it is too bulky, the tert-butyl group is always in equotorial position in cyclohexane ring (1,3-interaction with one hydrogen would cost additional approximately $\pu{11.4 kJ/mol}$ steric strain energy), as depicted in following diagram (The larger the substituent, the more the equatorial substituted conformer will be favored):

The most predominant cis-isomer (C) has both substituents in equatorial-positions (the left hand side structure of (C)). Other conformer (get by the ring flip; the right hand side structure of (C)) has both substituents in axial-positions. Thus, as a rule of thumb, the conformer with both substituents in equatorial positions is more stable.

On the other hand, the both trans-isomers of (D) have one substituent in an equatorial-position and the other in an axial-position. Yet, the conformer with the tert-butyl group in the equatorial-position (left hand side structure of (D)) is more stable between two, because tert-butyl group is comparatively much larger than methyl group. However, because most predominant cis-isomer (C) has both groups in the equatorial-positions, it is relatively more stable than (D), which has methyl on axial-position, making additional steric strain energy.


How does baking soda works as an desiccant? The thing is that I did a project from a website which is about the science of mummification and I can't find a way to explain the process of desiccation of the hot dog with the baking soda.

By this I mean to answer questions like: Does the baking soda absorb or adsorb water? If it adsorbs then which type of adsorption is it? What's happening in details between the hot dog and the surrounding baking soda?

Here is the link to the project

My results did show a decrease in mass and size which is explained in the lost of water from the hot dog but how?

Frankly speaking, dehydration occurs because of osmosis. In fact, cellophane is used for "peel-before eating" hot dogs and also for osmosis experiments. Natural sausage casing (from intestinal submucosa) are also semipermeable membranes.
You might try an experiment demonstrating dialysis using sausage casing: in a cellophane tube, put plain tap (or distilled) water and tie the ends of the tubing. Weigh the watery "weiner". Immerse the tube in a concentrated solution of baking soda for a day or so, then weigh the sausage again -- some of the water should have leaked out into the outer solution, faster than larger sodium and carbonate ions can infiltrate, so the sausage should shrink.


Why is the nitroso group a deactivating group for electrophilic aromatic substitution? Does the lone pair of nitrogen in nitrosobenzene take part in resonance? If not, then why? If it takes part in resonance then the hybridisation of nitrogen is sp. Can a sp hybridised atom be attached to the ring?

Let's look at nitrosobenzene.

In order to be delocalized (participate in resonance), the lone pair on the nitrogen needs to be in a $\ce{p}$ orbital. As I have draw it above, the lone pair is in a $\ce{sp^2}$ orbital, and the $\ce{N=O}$ pi bond is conjugated with the ring:

If the lone pair on the nitrogen atom is delocalized, then yes the nitrogen atom is $\ce{sp}$ hybridized. As a consequence, the $\ce{N=O}$ pi bond is perpendicular to the pi system. It is no longer conjugated:

Either the lone pair or the $\ce{N=O}$ pi bond is conjugated, not both.

If the lone pair is conjugated, the nitroso group can be electron donating by resonance:

If the $\ce{N=O}$ bond is conjugated, the nitroso group is electron withdrawing by resonance:

Finally, because the nitrogen is more electronegative than carbon, the nitroso group is electron withdrawing by induction:

It is more likely that the $\ce{N=O}$ bond is conjugated because that gives a higher degree of conjugation: 8 atoms versus 7 atoms.


What exactly is peptone (tryptone?) and why does it not get precipitated? According to Wikipedia Peptone (also called Tryptone) is a form of amino acid that has been treated by stomach acids. According to Google it also stays in solution when treated with acids or heat. Why is that so?

Sigma-Aldrich has listed long list of their Basic Ingredients of Microbiology, which are called Protein Sources (Peptones):

Peptones and extracts are excellent natural sources of amino acids, peptides, and proteins in growth media. They are most often obtained by enzymatic digestion or acid hydrolysis of natural products such as animal tissues, milk, and plants or microbial cultures. The number of available peptones and extract is enormous, and can promote and sustain the growth of most organisms.

Some protones are animal-based and some are plant-based. For instance, peptone water is a microbial growth medium composed of peptic digest of animal tissue and sodium chloride. The $\mathrm{pH}$ of the medium is $7.2\pm 0.2$ at $\pu{25 ^\circ C}$ and is rich in tryptophan. Peptone water is also a non-selective broth medium which can be used as a primary enrichment medium for the growth of bacteria (Ref.1).
References:

J.E.L. Corry, G.D.W. Curtis, R.M. Baird, Eds, Culture Media for Food Microbiology, Volume 34; 1st Edition, Elsevier Science: London, UK, 1995 (ISBN: 9780444814982).


Why does plastic tear easier when you pull it quickly? If you've ever taken the plastic from a plastic bag and pulled on it, you might know what I mean. If you pull quickly, the plastic tears relatively easily, but if you pull it slowly, it gets stretched out into an indestructible plastic string. Once stretched, pulling the plastic as hard as you did to tear it before yields no results.

Similarly, adhesive bandages are pulled off the skin (almost) painlessly if pulled quickly, but seem to stick even under the same level of force if it is applied more gradually.

What explains this phenomenon?

Polymers have a special property called viscoelasticity, which means they may strain elastically (elasticity) then flow (visco). The models below are mechanical representations of this phenomenon using springs and dashpots (dampers).

$\qquad\qquad\qquad$ Maxwell Model   $\qquad\qquad\qquad\qquad\qquad\qquad$ Kelvin–Voigt Model

$\qquad\qquad\qquad$Time Stress Deformation Plot

This property of viscoelasticity allows plastics, when pulled slowly (low strain rate), time for the polymer chains to rearrange and the plastic to flow which means it is less rigid but more elastic (think stretching silly putty). When the plastic is pulled quickly the plastic is more rigid and breaks (like breaking silly putty). AS I have alluded you can observe this with silly putty which flows when strain is low and will stretch to long lengths while snap into two pieces when pulled rapidly.

So far I have only explained why plastics will stretch or snap, but not why they are hard to tear after stretching. This is caused because as the plastic is stretched the polymer chains align as shown below. As it turns out polymers are hardest to tear perpendicular to the direction of the molecules because you have to break the bonds of the chain rather than some bond breaking and some chains separating. As a result, the plastic becomes hard to tear across the stretched part, but it is now easier to tear along the stretch (parallel) since in is mostly chain separating to do so, and indeed a tear in a stretched bag will follow the direction of a stretch.


Hybridization molecule Why is SO32- sp3 hybridized even though it forms a double bond with the oxygen (in resonance)?

I thought double bonds needed a p orbital so I would think it would be sp2d so it can still have an unhybridized p orbital for the double bond. Thanks!

This is a point of continuing drama.  Theoretical calculations as long ago as the 60s (cf. J. Am. Chem. Soc., 1994, 116 (10), pp 4414–4426) strongly suggest there is no d orbital participation in hybridized orbitals for Period 3 "hypervalent" compounds like the sulfite anion.  In fact, these calculations appear to suggest the observed bond shortening is probably explained quite simply by a significant ionic component to the bond, which means the best Lewis structures may well be the simple octet-rule following one, with 4 sp3 bonds around the central S, occupied by 3 bonding pairs (giving 3 sigma bonds) and 1 lone pair. No pi bond at all. This gives you higher formal charges than the resonance structure with a single double bond -- but that's what's meant by saying the calculations suggest the best VB description simply has more ionic content to the bonding between the S and O atoms (represented by the higher formal charges), rather than any d mixing.


Simultaneous solubility of CaF2 and SrF2 Solubility of two electrolytes having common ion when they are dissolved in solution is called simultaneous solubility.

How do I calculate the simultaneous solubility of the above mentioned salts given that:

  $\mathrm {K_{sp}(\ce CaF_2)= 3.9 \times 10^{-11}}$ and $\mathrm {K_{sp}(\ce SrF_2)= 2.9 \times 10^{-9}}$

My attempt:

Let the solubility of $\ce {CaF_2}$ be $x$ and that of $\ce {SrF_2}$ be $y$.

Two equations are obtained:

$3.9\times10^{-11} = x(2(x+y))^2$
$2.9 \times 10^{-9} = y(2(x+y))^2$

Dividing the two we get, $x=0.013y$
Substituting this value of $x$ in equation $1$, I got $y= 5.2\times 10^{-3}$ which should be the solubility of $\ce {SrF_2}$. However, answer given is $\pu{9\times 10^{-4}M}$.

How do I solve this problem then?

Calculate the simultaneous solubility of $\ce{CaF_2}$ and $\ce{SrF_2}$ given that:

$\mathrm {K_{sp}(\ce CaF_2)= 3.9 \times 10^{-11}}$ and $\mathrm {K_{sp}(\ce SrF_2)= 2.9 \times 10^{-9}}$

Let $\ce{[Ca^{2+}]} = x$ and $\ce{[Sr^{2+}]} = y$
So the dissolving $\ce{CaF_2}$ contributes $2x$ of $\ce{F^-}$ and the dissolving $\ce{SrF_2}$ contributes $2y$ of $\ce{F^-}$. So the total $\ce{[F^-]} = 2x + 2y = 2(x+y)$
Substituting into the two $\text{K}_{sp}$ equations leads to "messy" cubic equations which could be solved numerically since there are two equations with two unknowns.
$3.9\times10^{-11} = x(2(x+y))^2 = 4x(x+y)^2$
$2.9 \times 10^{-9} = y(2(x+y))^2 = 4y(x+y)^2$
But going back to the original $\text{K}_{sp}$ equations we have:
$\ce{[Ca^{2+}][F^-]^2 = 3.9 \times 10^{-11}}$
$\ce{[Sr^{2+}][F^-]^2 = 2.9 \times 10^{-9}}$
and by dividing the two we get $\ce{[Ca^{2+}] = 0.0134 [Sr^{2+}]}$ or $x = 0.0134y$. Substituting this into the second equation we get:
$2.9 \times 10^{-9} = 4y(x+y)^2 = 4y(1.0134y)^2 = 4.108y^3$ or $y=8.9\times10^{-4}$
So:
$\ce{[Sr^{2+}]} = 8.9\times10^{-4}$
$\ce{[Ca^{2+}]} = (0.0134)(8.9\times10^{-4}) = 1.2\times10^{-5} $
$\ce{[F^-]} = 2(8.9\times10^{-4} + 1.2\times10^{-5}) = 1.8\times10^{-3}$
Note that this can also be simplified using significant figures a different way. Since the two $\mathrm {K_{sp}}$ values only have two significant figures and  $\ce{[Ca^{2+}] = 0.0134 [Sr^{2+}]}$, we can assume that for all practical purposes that all the $\ce{F^-}$ comes from the dissolution of the $\ce{SrF_2}$. So:
$\ce{4[Sr^{2+}]^3} = 2.9\times10^{-9}$
Thus
$\ce{[Sr^{2+}]} = 9.0 \times10^{-4}$
$\ce{[F^-]} = \sqrt{\frac{2.9\times10^{-9}}{9.0 \times10^{-4}}} = 1.8\times10^{-3}$
$\ce{[Ca^{2+}]} = \dfrac{3.9\times10^{-11}}{(1.8\times10^{-3})^2}=1.2\times10^{-5}$


Is the volume of a protein substrate equal before and after an enzymatic reaction? (pepsin degrading egg white suspension) I am trying to determine the reaction rate of pepsin at different pH values by performing an experiment.

I am using egg-white suspension as the substrate and am adding differently concentrated $\ce{HCl}$ to the substrate to vary the pH of the reaction. Eventually, I add the enzyme. However, for the reaction rate I require the time and the substrate disappeared and product formed.

Is the volume of the egg-white suspension different before and after the enzymatic reaction so that I could use it to calculate the reaction rate or does the protein's volume remain the same after the reaction?

The volume of the protein will probably change (although not by much) because it will be denatured as it is degraded and as the pH decreases. As it denatures, the hydrophilic surface area and hydrophobic interior volume will change, leading to a change in the local water liquid structure and density. Denatured, short polypeptides also pack into smaller volumes than native proteins do, and so that would also cause the volume to decrease in most cases. The amount of water in the system will affect the magnitude of the change. I would predict that the volume would decrease slightly, probably on the order of what happens with a few degrees of temperature increase of water (around 0.1%).

So, the short answer is "Yes, the volume will change, but probably not consistently or by enough to be used to reliably measure the extent of the reaction."

This is not my area of expertise, so I don't know what the current standard protocols are for determining reaction rates for pepsin reactions. However, I was able to find a couple of research papers that talk about using ninhydrin to measure reaction extent:

The pH-Dependence of Pepsin-Catalyzed Reactions

Kinetics of Hydrolysis of Egg White Protein by Pepsin

In the second study, they quenched the reaction at a specific time by adding NaOH to bring the pH up, then centrifuged the products before measuring the reaction extent.

If this is not a professional scientific study (meaning it's a school project, or something like that) then you might look at the method used in this report:

The Effect of pH on the Reaction between Pepsin and Egg – White

Basically, they just added pepsin and watched to see how long it took for the solution to become "clear". This won't give you very good accuracy or precision, but it might work for a science project or school report.


When is the Assumption that Enthalpy is Constant over Temperature Change Invalid? In a question, I was asked to integrate the Gibbs-Helmholtz equation to derive some formula. Then the next question asked:

  Identify one assumption that you have made in deriving the above equation and identify one
  scenario where it is a very bad assumption, and explain why the assumption fails

Now the assumption that was made was the $\Delta H_{\mathrm{fus}}$ is constant over the change in temperature. However, could someone please explain when and why this assumption fails.

Thanks.

The $G{-}H$ equation describes the effect of $T$ on $\Delta G$ at constant pressure. But, along the fusion contour of $T$ vs $p$, $T$ is a unique function of $p$. So, along this contour, $p$ is changing. So the $G{-}H$ equation is not valid to use for a change of phase such as fusion. The $G{-}H$ equation was developed to describe the effect of $T$ on $\Delta G$ for chemical reactions at constant pressure, usually for the change from separate pure reactants to separate pure products at the standard pressure of $1\ \mathrm{bar}$.


How do I write a procedure for creating a buffer? I need to write an original procedure with exact numbers for creating a buffer:

Create a buffer using $\ce{CH3COOH}$ and $\ce{CH3COONa}$ that has a $\pu{pH}$ of exactly $3.75$.
A $\pu{50 mL}$ sample of your buffered solution has to be able to withstand the addition of $\pu{25.00 mL}$ of $\pu{0.10 M}$ $\ce{NaOH}$ solution.
"Withstand" here is defined as less "than $0.5$ change in $\mathrm{pH}$."
The buffered solution will break after the addition of no more than $\pu{35.0 mL}$ of the $\pu{0.10 M}$ $\ce{NaOH}$.
We are given $\pu{0.10 M}$ $\ce{NaOH}$ and $\pu{6 M}$ acetic acid.

Where I've got so far:
I took the $\mathrm{p}K_\mathrm{a}$ of acetic acid $(4.75)$ and the desired $\mathrm{pH}$ $(3.75)$ and put it into the Henderson-Hasselbalch equation:
$$3.75 = 4.75 + \log\frac{[\text{salt}]}{[\text{acid}]}$$
$$\log\frac{[\text{salt}]}{[\text{acid}]} = -1$$
So
$$\frac{[\text{salt}]}{[\text{acid}]} = 0.10,$$
which means we want the ratio $[\ce{CH3COONa}]:[\ce{CH3COOH}]$ to be $0.10$.
I think we can prepare a solution containing $\pu{0.10 M}$ $\ce{CH3COONa}$ and $\pu{1.00 M}$ $\ce{CH3COOH}.$ But with what we're given $(\pu{0.10 M}$ $\ce{NaOH}$ and $\pu{6 M}$ acetic acid), I don't know how to proceed and find exact numbers for a procedure.

Initial note

The specific problem with your scenario is, that you need a buffer of buffer capacity progressively decreasing with increasing $\mathrm{pH}$, but capacity of simple buffers is for $\mathrm{pH} \lt \mathrm{p}K_\mathrm{a}$ increasing.

As consequence, its "breaking" capacity is too big, compared to "withstanding capacity".

General analysis

Let there is given the initial $\mathrm{pH_{ini}}$ and the acid dissociation constant
$\mathrm{p}K_\mathrm{a}$

Let there there is given "breaking volume" of $\ce{NaOH}$ $V_\mathrm{br}$, where the buffer losses it's buffering ability.

Let there is given "withstanding volume" $V_\mathrm{ws}$ when $\mathrm{pH}$ changes by value $x$.

Then:

$$\mathrm{pH_{ini}}=\mathrm{p}K_\mathrm{a} + \log {R}$$

where

$$R=\frac{ [\ce{A-}]_\mathrm{init}}
{[\ce{HA}]_\mathrm{init}}=10^{\mathrm{pH_{ini}}-\mathrm{p}K_\mathrm{a}}$$

$$\mathrm{pH_{ini}}+x=\mathrm{p}K_\mathrm{a} +\log  \frac {R+V_\mathrm{ws}/V_\mathrm{br}\cdot (1-R)}{1-V_\mathrm{ws}/V_\mathrm{br}}$$

If we consider
$$R_1=10^{\mathrm{pH_{ini}}+x-\mathrm{p}K_\mathrm{a}} $$
$$V_\mathrm{r}=V_\mathrm{ws}/V_\mathrm{br}$$

then
$$R_1=\frac {R+V_\mathrm{r}\cdot (1-R)}{1-V_\mathrm{r}}$$

$$V_\mathrm{r}=\frac {R_1-R}{R_1-R+1}$$

Goal

The desired $$V_\mathrm{r}\ge \frac{25}{35}=0.714$$

Acetic acid case

If we consider acetic acid with $\mathrm{p}K_\mathrm{a}=4.75$ then $$R=10^{3.75-4.75}=0.1$$

If the change should be $\Delta \mathrm{pH}=0.5$:

$$R_1=10^{3.75+0.5-4.75}=0.316$$

$$V_\mathrm{r} =\frac{0.316-0.1}{0.316-0.1+1}=0.177\lt 0.714$$

If the change should be $\Delta \mathrm{pH}=1.0$:

$$R_1=10^{3.75+1.0-4.75}=1$$

$$V_\mathrm{r} =\frac{1-0.1}{1-0.1+1}=0.475 \lt 0.714 $$

Formic acid case

If we consider as a replacement e.g. formic acid with $\mathrm{p}K_\mathrm{a}=3.75$ then $$R=10^{3.75-3.75}=1$$

If the change should be $\Delta \mathrm{pH}=0.5$:

$$R_1=10^{3.75+0.5-3.75}=3.16$$

$$V_\mathrm{r} =\frac{3.16-1}{3.16-1+1}=0.684\lt 0.714$$

If the change should be $\Delta \mathrm{pH}=1.0$:

$$R_1=10^{3.75+1.0-3.75}=10$$

$$V_\mathrm{r} =\frac{10-1}{10-1+1}=0.9 \gt 0.714$$

Summary

Acetic acid is unuseable for making a buffer to fit your requirements.
Formic acid can be used to make a buffer with "withstanding" parametr $\Delta~ \text{pH} \lt +1.0$, seen in the other question.
Formic acid very tighly missed the "withstanding" parametr $\Delta~ \text{pH} \lt +0.5$, so seems about usable.
Even better results could be achieved with an acid with $\mathrm{p}K_\mathrm{a} \lt 3.75$ .


How to calculate molecular dipole moment from a known wavefunction? Say I have a molecular wavefunction as a set of molecular orbitals and want to calculate the molecule's dipole moment, but don't know how! I searched a lot but couldn't find any practical example.

$$\psi _i=\sum ^N_{i=1}C_i\mathrm e^{-\alpha _ir^2}$$

The necessary formal derivation has already been nicely done by AngusTheMan. I'll start from the last equation:

$$
\langle \mu_{z} \rangle = \langle \Psi | \hat{\mu}_{z} | \Psi \rangle
$$

where $\Psi$ is the variational wavefunction; it can be any molecular state. It's important that it's variational, otherwise the expectation value approach is not exact. So, this works for SCF, CI, and MCSCF wavefunctions, but extra derivatives need to be taken for Moller-Plesset and coupled cluster wavefunctions. More work needs to be done for multideterminental wavefunctions like CI and MCSCF, but the complexity is no different for a single state in each wavefunction. There may be some MO space partitioning I'm neglecting that's required for MCSCF, so I'll restrict my work to a single-determinental wavefunction.

Expand the wavefunction as a linear combination of molecular orbitals (MOs)

$$
\Psi = \sum_{i} \psi_{i},
$$

where each molecular orbital is a linear combination of atomic orbitals (AOs)

$$
\psi_{i} = \sum_{\mu} C_{\mu i} \phi_{\mu},
$$

where $C_{\mu i}$ is the MO coefficient matrix, so our expectation value now looks like this:

$$
\langle \mu_{z} \rangle = \sum_{i}^{\textrm{occ MOs}} \sum_{\mu\nu}^{\textrm{AOs}} C_{\mu i} C_{\nu i} \langle \phi_{\mu} | \hat{\mu}_{z} | \phi_{\nu} \rangle.
$$

The indices $\mu,\nu$ run over all AOs, and the index $i$ runs over the occupied MOs. There's only one index because this is a one-electron operator. I'm also neglecting any complex values here, since we almost always work with real-valued AOs and MO coefficients.

We do one last rearrangement. Replace the MO coefficients with the density matrix

$$
P_{\mu\nu} = \sum_{i}^{\textrm{occ MOs}} C_{\mu i} C_{\nu i}
$$

to give the first explicit "working equation":

$$
\langle \mu_{z} \rangle = \sum_{\mu\nu}^{\textrm{AOs}} P_{\mu\nu} \langle \phi_{\mu} | \hat{\mu}_{z} | \phi_{\nu} \rangle
$$

I say first for two reasons: we usually try and avoid explicit loops like this, and the expression can be broken down further, depending on what molecular properties are of interest; I'll be more clear about this later. Inside the sum there are two terms:

The density matrix $P_{\mu\nu}$, which comes from a converged SCF calculation.
The integral of the dipole operator over two basis functions, $\langle \phi_{\mu} | \hat{\mu}_{z} | \phi_{\nu} \rangle$. Atomic orbitals are represented as atom-centered basis functions. These can be calculated once and at any time, since the quantities here don't change over the course of a calculation.

Each of these terms is represented as a matrix. Since the index $\mu$ runs along the rows and $\nu$ runs along the columns for each matrix, "contraction" involves either a matrix product followed by the trace, or an elementwise product followed by an accumulation sum over all matrix elements. There are other details one needs to be careful about, such as what units the result should be in, which changes the prefactor (programs work internally in atomic units), and what the origin for the dipole operator is, but that's really it.

Well, sort of. I'm actually treating some of the program internals as a black box. If you're familiar with Hartree-Fock, it should be clear where $P_{\mu\nu}$ comes from, but what about the integral? For a general expectation value $\langle A \rangle$ with its corresponding operator, where does $\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle$ come from? If it's already available in the code, then you call a wrapper function that then calls the integral engine to do all the nasty work, and you get back a tidy matrix without having to worry about the details. If $\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle$ isn't present, depending on the complexity of $\hat{A}$, there can be a non-trivial amount of derivation required for the working integral equation, followed by the implementation.

Ignoring any possible contraction of primitive basis functions, expand $\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle$ using the definition of $\phi$ in Cartesian coordinates:

$$
\phi(\mathbf{r}; \mathbf{A}, \mathbf{a}, \zeta) = (x-A_x)^{a_x} (y-A_y)^{a_y} (z-A_z)^{a_z} e^{-\zeta |\mathbf{r} - \mathbf{A}|^2}
$$

where $\mathbf{r} = (x, y, z)$ is the electron position, $\mathbf{A} = (A_x, A_y, A_z)$ is the position of the basis function (almost always atom-centered), and $\mathbf{a} = (a_x, a_y, a_z)$ are the angular momenta for each coordinate, with $l_{\textrm{max}} = a_x + a_y + a_z$ total angular momentum of the basis function. $(0,0,0)$ is an s-function, $(1,1,0)$ and $(0,0,2)$ are d-functions, and so on.

Forming the integral more explicitly gives

$$
\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle = \int\int d\mathbf{r}_1 d\mathbf{r}_2 \left[ (x_1-A_x)^{a_x} (y_1-A_y)^{a_y} (z_1-A_z)^{a_z} e^{-\zeta_a |\mathbf{r}_1 - \mathbf{A}|^2} \right] \\ \times\left[ \hat{A} \right]\left[ (x_2-B_x)^{b_x} (y_2-B_y)^{b_y} (z_2-B_z)^{b_z} e^{-\zeta_b |\mathbf{r}_2 - \mathbf{B}|^2} \right]
$$

Before going any further, $\hat{A}$ must be defined. If $\hat{A} = 1$, this becomes an overlap integral. The dipole operator in the z-direction is given by $\hat{A} = \hat{\mu}_{z} = -ez = -e(z_3 - C_z)$, where $z_3$ is the integration coordinate and $C_z$ is the origin of the dipole in the z-direction, usually taken to be zero. Everything is kept in atomic units until after the integral/density contraction, so drop the prefactor $-e$. We can now generalize this to an arbitrary Cartesian multipole moment operator,

$$
\hat{A} = \mathfrak{M}(\mathbf{r}_3) = (x_3 - C_x)^{c_x} (y_3 - C_y)^{c_y} (z_3 - C_z)^{c_z}
$$

where $(c_x, c_y, c_z)$ determine the coordinate of each multipole, and their sum is the total multipole order; for example, $(1,0,0), (0,1,0), (0,0,1)$ are the x-, y-, and z-directions of the dipole operator. The Cartesian moment operator looks just like a Gaussian basis function where $\zeta = 0$.

Once the form of an operator has been derived, it needs to be implemented as part of an integral package, each of which implement one or more algorithms for computing integrals. Each algorithm is named after the authors of the paper in which they were introduced, and are usually abbreviated. For example, the first one I know of is the Taketa, Huzinaga, O-Ohata paper (THO, DOI: 10.1143/JPSJ.21.2313), where explicit working equations are given for 2-center overlap, 2-center kinetic energy, 2-center electron-nuclear attraction, and 4-center electron repulsion integrals. A working implementation can be found in the PyQuante package. I made an IPython notebook translation of the code snippets on the front page of the official documentation.

Other, more complex algorithms are from the Pople-Hehre (PH), McMurchie-Davidson (MD, DOI: 10.1016/0021-9991(78)90092-X), Obara-Saika (OS, DOI: 10.1063/1.450106), Dupuis-Rys-King (DRK), and Head-Gordon-Pople (HGP) papers. I'm sure I'm neglecting some, including the seminal paper by Boys which introduced the use of Gaussian functions as a substitute for Slater-type functions in basis sets. A good review of these algorithms is found in a Peter Gill paper (DOI: 10.1016/S0065-3276(08)60019-2); he is the original author of the integral code in both Gaussian and Q-Chem.

To bring these things full circle, I wrote some code a few months ago to calculate the dipole moment using pyquante2 and a wrapper that calls an implementation of the Obara-Saika recursive integral algorithm. You can find it here with some comparisons to "industrial" quantum programs.


What are the products formed in acidic hydrolysis of a tetrahydropyranyl ester? Find the major product of the following reaction:

My  approach
From my knowledge of reagents, I figured out that that the acid will protonate the oxygen(s) in the reactant and after that $\ce{C-O \sigma}$ bond should break, but I am not able to decide which $\ce{C-O}$ bonds to break in the reaction after protonation by the aqueous acid.
Am I thinking in the right direction or are there some other concepts involved?

The actual mechanism
Waylander has given an accurate reaction since this is a THP ether

The actual mechanism of the de-protection using alcoholysis would be as follows:

The actual mechanism of the de-protection using acidic hydrolysis would be as follows:

A place to start from if you didn't know the mechanism
Seeing that the compound given resembles a carbonyl group that has been protected using ethylene glycol, a mechanism using a reaction similar to that of protection of carbonyl groups was used here to hypothesis what would happen(not knowing the original mechanism) and to decide the products formed.

Hence, taking a reversed approach here, the first question I ask is:

What could have formed this acetal in the first place?

One part of the acetal seems to be connected to the carbonyl group and is part of the main molecule. The other side, however, is a different molecule. So, this is what the acetal formation looks like in the mechanism. Here, I assume $\ce{R}-$ is just an inert part of the molecule and so is equivalent to $\ce{CH3-CH2}-$. This may not be the right mechanism but it would give us a good starting point.
Formation of the hemiacetal takes place as follows:

Formation of acetal using propanol can happen as follows:

Which is the THP ether in the question.

What are the final products?
Therefore, the products of acid catalysed hydrolysis of the acetal would be $\ce{R-CH2-OH}$ and 5-hydroxypentanal which exists in equilibrium with its cyclic form dominating in mildly acidic conditions.

The hydrolysis of the hydroxyderivative of THP would happen as follows. Both alcohols will get protonated but protonating the oxygen in $\ce{-OH}$ results in the same molecule being regenerated.

Therefore, the final products would give $\ce{R-CH2-OH}$ and a hydroxy derivative of THP.
However, if $\ce{HCl}$ or $\ce{NaHSO3}$ is used instead of a weak lewis acid, we see the dominating compound in the equilibrium being 5-hydroxypentanal as shown in this paper here:

[...] As a result of the acid-catalyzed hydration of 1, the 2-hydroxytetrahydropyran (2) was formed primarily which equilibrated with its open chain form, the 5-hydroxypentanal (3). The acid-catalyst (conc. hydrochloric acid), however, promotes not only the addition of water to the C-C double bond of 1, but also that of the OH groups in 1 and 2, thus resulting in the protected compounds 2-(2-tetrahydropyranyloxy) tetrahydropyran 0, and 5-(2-tetrahydropyranyloxy)pentanal (4). The ratio of 4 + 5 to 1 + 2: depends on the amount of acid used.[...]

Reference:

Peter Vinczer , Zoltan Juvancz , Lajos Novak & Csaba Szantay (1989) AN
IMPROVED SYNTHESIS OF 5-HYDROXYPENTANAL, Organic Preparations and Procedures International: The New Journal for Organic Synthesis, 21:3, 344-346, DOI: 10.1080/00304948909356391


Electrochemistry -gold recovery I was wondering how to answer  a question I got from my students about the possible recovery of gold from solution. We just finished the reaction between silver ions and copper metal and they wondered why copper wouldn't displace gold from solution in an electrochemical reaction. The redox potentials suggest that it can be done but there is no documentation of this reaction anywhere. I can't try the reaction because we have no gold compounds. If it could be done you would think that gold recovery but companies would use the technique since copper is a lot cheaper.
Please advise.

Short Answer: Yes, based on the electrochemical/reactivity series, it is possible to percipitate gold from solution using metals higher on the series like copper or zinc. However, it isn't done often because gold is highly unreactive, and when gold salts are formed, they are formed deliberately under atypical conditions like very high temperatures or acidities.
It would be possible to displace gold from it's ionic salt (gold (iii) chloride) using zinc or copper. This is because, as you mentioned copper is higher on the activity series than gold.
The likely reason to why companies don't use single-displacement reactions to retreive gold from solution is be cause gold salts rarely exist. Gold is extremely unreactive (see here) Gold (III) chloride is normally synthesized by reacting gold metal with chlorine gas at 180 degrees Celsius. This temperature is higher than anywhere you'd find on Earth's surface, and chlorine gas is too reactive to exist in that state in the atmosphere. These conditions are typically found only in the lab and not out in nature. Another example of a man-made gold salt is aqua regia (a 3:1 ratio of hydrochloric acid and nitric acid) which can dissolve gold. To recover the gold, ions are displaced using ferrous sulfphate reducing agent (source).
If you are interested in gold recovery, here is an excellent article on a variety of methods.


Why does the reaction of the primary alcohol cyclopentylmethanol with hydrogen bromide follow SN1 mechanism? Example 2: Identify all the products formed in the following reaction and indicate the major product.

Solution: Three products will be formed in the given reaction as shown below:

In this reaction the primary alcohol undergoes SN1 mechanism by forming a carbocation and then undergoing ring expansion, but primary alcohols prefer SN2 mechanism. So why is this primary alcohol undergoing SN1 mechanism?

It does not. The proposed mechanism is wrong.

The conversion of cyclopentylmethanol to 1-bromo-1-methylcyclopentane is not reported anywhere in the literature. If you could somehow generate the primary carbocation, then one would expect a rapid 1,2-hydride shift to generate the more stable tertiary carbocation, giving rise to 1-bromo-1-methylcyclopentane.

The ring-expansion product could be rationalized by a concerted 1,2-alkyl shift displacing $\ce{OH2^+}$ without violating first-principles.


Neutralization of Sodium Azide using Ammonium Cerium Nitrate I'm using Ammonium Cerium Nitrate to neutralize phosphate-buffered saline solutions containing 0.1% Sodium Azide. I add 0.55 grams of Ammonium Cerium Nitrate into 100 mL of this 1XPBS + 0.1% NaN3 solution and am able to neutralize the NaN3.

Will the resulting solution have any negative, harmful, effects if flushed down the drain? My main concern is polluting ground water or corroding our sewerage system.

I think that you are doing fine!
The azide anion can be used for quite some amazing chemistry, but it's rather toxic and harmful for the environment.
Admittedly, I'm too lazy right now to look up the redox potentials or to write the reaction equation, but
CAN should be able to oxidize $\ce{N3-}$ all the way up to $\ce{NO3-}$, which is safe to be disposed.

Update
Judging from memory can result in bad advices!
A closer look at the SDS of both compounds shows that both have

H410 Very toxic to aquatic life with long lasting effects. and

P273 Avoid release to the environment.

Consequently, the combined solutions (with excess of CAN) should not be poured down the drain, but collected (aqueous inorganic waste).


Has the molecular diatomic oxygen trianion been observed or predicted? Has diatomic oxygen ever been observed with a 3− charge, i.e. $\ce{O_2^{3-}}$? Have any ab intio calculations been done on this ion?

Update:

Just to summarize some of the answers below and add my two cents, it seems that no studies have reported observing this ion experimentally, either in isolation or as part of a condensed phase. Ab initio calculations suggest that $\ce{O_2^{3-}}$ is unstable in isolation, meaning that the loss of an electron is exothermic. However, first-principles calculations have predicted that a molecular oxygen trianion can exist as part of a solid; see Phys. Rev. B 85, 035210 (2012).

According to "Small Multiply Charged Anions as Building Blocks in
Chemistry" Acc. Chem. Res. 1996, 29, 497-502

  it is concluded that all isolated diatomic dianions or
  other multiply charged diatomic anions are not stable
  to loss of an extra electron or fragmentation in their
  ground electronic states

So even $\ce{O2^{2-}}$, let alone $\ce{O2^{3-}}$ cannot exist in isolation.


(S)-glycidyl benzyl ether NMR problem Can someone explain the appearance of the resonances at $\pu{4.6 ppm}$ (spectrum taken from here)?

Is it due to long range coupling? But the coupling constants are too high for such long range coupling. I assumed, according to the shift, that this is coming from the two protons nearest to the benzene ring?

The pattern is neither a true quartet nor a doublet of doublets: it is two doublets. Many thanks to long for pointing out that it is also frequently described as an "AB quartet".

The peaks arise due to the benzylic protons, i.e. the $\ce{CH2}$ between the ether oxygen and the phenyl group. Due to the existing stereocentre on the epoxide, these two protons are rendered diastereotopic: due to their different position relative to the stereocentre, they experience a different net magnetic field, and hence have a different chemical shift. Therefore, in the absence of coupling, one expects to see two singlets.

If we add in spin-spin coupling, then we see that both protons are capable of coupling to each other. The magnitude of the geminal coupling $^2J$ is generally somewhere between 10 to 15 Hz. In this case the zoomed-in diagram shows that the coupling is roughly 12 Hz. So, we now expect to see two doublets.

Why, then, are the intensities of the two peaks in each doublet different? This is due to so-called "roofing": when the difference in the resonance frequencies of two protons is comparable to the coupling constant between them, the shape of the multiplets will be such that they have a "slant" towards each other. In this case, the difference in resonance frequencies is roughly 20 Hz (measure the distance between the centre of each doublet), which is on the same order of magnitude as the coupling constant.

This can, to the best of my knowledge, only be explained with quantum mechanics: see What is the origin of 'tenting' or 'roofing' in NMR spectra? (the question has more examples).


Is the hydrogenation of carbon dioxide to methane a solution or a problem? The hydrogenation of $\ce{CO2}$ to $\ce{CH4}$ is thermodynamically a favored reaction at ambient to moderately high temperatures.
$$\ce{CO2 + 4H2 <=> CH4 + 2H2O (g)}\qquad G^\circ_{\pu{298K}}= \pu{-27 kcal/mol}$$
$\ce{CH4}$ (natural gas) is an excellent fuel for a variety of applications. It is also a good C1 feedstock for manufacturing other chemicals. This reaction can be used to recycle $\ce{CO2}$ and reduce its emission to the atmosphere.
So, why not use $\ce{H2}$ from photoelectrolysis of water to produce $\ce{CH4}$ and achieve multiple goals of storing solar energy, creating a useful feedstock and recycle atmospheric $\ce{CO2}$?

I said, $\ce{CH4}$ is 30 times worse than $\ce{CO2}$ to the atmosphere and also the photoelectrolysis of $\ce{H2}$ is still an expensive method that requires more time to achieve maturity.
Is it right? the other student on the panel said my answer was too vague.

The problem is that the carbon dioxide which is released from burning is often mixed with other things which include air. If we were to want to convert carbon dioxide into methane then the oxygen of the air would make it harder.

While one option might be chemical looping combustion to obtain a purer carbon dioxide stream we still have the problem of where would we get hydrogen from. The normal method in industry to get the hydrogen gas needed for things like hydrogen peroxide production is to use steam reforming of methane.

$\ce{H2O + CH4 -> CO + 3H2}$

Using the water gas shift reaction the carbon monoxide is converted into carbon dioxide and hydrogen (using more steam). My worry about an industrial process for the conversion of unwanted carbon dioxide into methane is that unless we use hydropower, solar power or nuclear power to obtain electricity for the production of hydrogen from water then we will never be able to make a net reduction in the amount of carbon dioxide as the process will form as much carbon dioxide as it consumes.

There is some electrochemistry in which carbon dioxide can be converted into other things by reduction at the cathode but I think that this is currently only at a very small scale.


Mechanism for hydroxyethylation on nucleobases (e.g. Thymine) I'm trying to figure out the following reaction mechanism

There's a reference paper by Yanagi and Akiyoshi(1959) showing the hydroxyethylation of imides resulting in the following

I have two questions :

(1) Why would the hydroxylethylation occur in the 1 position instead of the imide at the 3 position? I would imagine that this reaction occurs for the more acidic amine, but I have no idea why there doesn't seem to be any literature about that reaction.

(2) Why would this reaction produce a HydroxyEthyl and not a urethane, as shown here :

Any insights would be appreciated, thanks!

Yanagi and Akiyoshi : http://pubs.acs.org/doi/abs/10.1021/jo01090a601

Reaction in red (from a patent application): http://www.huntsman.com/performance_products/Media%20Library/a_MC348531CFA3EA9A2E040EBCD2B6B7B06/Products_MC348531D0B9FA9A2E040EBCD2B6B7B06/Carbonates_MC348531D1109A9A2E040EBCD2B6B7B06/files/jeffsol_alkylene_carbonates_synthesis_of_hydroxy_alkyl_urethanes.pdf

The second part of the question is easy to answer: the imide attacks at C3 of propylene carbonate instead of C1 (the carbonyl carbon) because carbonates are stabilized (double ester-resonance), whereas carboxylate is a so-so leaving group.

The initial product from nuclephilic attack at C3 is a carbonate half-ester, which subsequently loses CO2 to form the observed N-hydroxyethyl imide.


Confusion about Titration Formula When performing a titration of a monoprotic acid and a monoprotic base, water can not be added to the titrant but water can be added to the analyte. To prove this statement to myself, I decided to analyze the equation for a hypothetical titration where the acid is the analyte and the base is the titrant:

[Acid] = (Volume of Base) $\cdot$ [Base] / (Volume of Acid)

So, if water is added to the base in the burette, the volume of base would increase but the molarity of the base would decrease and the effects seem to cancel out, leading to the same [Base] calculation. However, when water is added to the acid in the flask, the volume of acid increases which would lead to a decrease in the [Acid] calculation.

This contradicts the common protocol and knowledge when adding water to the burette or flask. Many answers I have seen state that the moles of acid and base stay constant if water is added to the analyte; however, wouldn't this also increase the volume of analyte and skew the calculation of [Acid]?

It would be awesome if someone could explain the effects of adding water to the burette and analyte on the equation above.

Your title and the main body of the post are two different questions. Anyway, it is important to understand the difference in each case.

Let us talk about the analyte. Your analyte can come in two forms: (i) either as a solid or (ii) as a solution.

In the case of a solid analyte you would quantitatively transfer a known exact weight of the sample to a flask, let us say an Erlenmeyer flask. You can now add some water to dissolve it and you do not need to worry because you have recorded the actual weight of the sample. [yes the conc. will change, depending on how much water you are adding].

In the case of a analyte solution you would quantitatively transfer a known volume with the help of a pipette to an Erlenmeyer flask. You can still add some water and need not worry because you know the initial volume.
[again the conc. will change with the addition of water].

*What is important in titration calculations are absolute moles of the analyte not its total volume in which it is being titrated*.

Now why can't we add water to a buret? The reason is that by adding water to buret, it will change the concentration of the titrant. As I said earlier, only moles are important in the calculations, and moles of titrant can only be calculated by

[Volume of titrant][Conc. of titrant] = moles of titrant

if the [Conc. of titrant] is an incorrect number, which has changed by adding water, the moles of titrant will be a wrong number.

Eventually, we calculate the concentrations with the initial volume we added with the help of pipette or a known mass.

*The problem is arising because you are using a titration formula taught by those who shouldn't be teaching science. This formula is not only wrong (prone to error), but it will lead to all type of problems in future with redox calculations. I strongly discourage using the formulae given in your query.

[Acid] = (Volume of Base)*[Base] / (Volume of Acid)

This formula is valid for all acid-base and redox titrations iff (if and only if) concentrations are provided in normality. Normality is an ancient 18th century concept. It has stuck in certain education systems like cancer. You quash it here but it appears somewhere else.*


Why does the two-electron exchange integral in Hartree-Fock theory have positive values? In Section 2.3.6 of Szabo & Ostlund's Modern Quantum Chemistry, the exchange integral has the form of

$$\int \mathrm{d}\mathbf{r}_1\,\mathrm{d}\mathbf{r}_2\, \psi_a^*(\mathbf{r}_1)
\psi_b(\mathbf{r}_1)\frac{1}{r_{12}}
\psi^*_a(\mathbf{r}_2) \psi_b(\mathbf{r}_2)
$$

or $\langle ab|ba\rangle$ in physicists' notation. According to this textbook (and many other books), $\langle ab|ba\rangle$ is positive. The conclusion seems obvious but I just cannot find a proof. Is there any simple reason that $\langle ab|ba\rangle$ must be positive?

The conclusion seems obvious but I just cannot find a proof.

No, I don't think it is obvious. At least I could not think about a simple algebraic proof that the exchange integral $\langle ab \vert ba \rangle$ defined as follows,
$$
\langle ab \vert ba \rangle :=
\iint \overline{\psi}_a(\vec{r}_{1}) \overline{\psi}_b(\vec{r}_{2}) r_{12}^{-1} \psi_b(\vec{r}_{1}) \psi_a(\vec{r}_{2}) \mathrm{d}\vec{r}_{1} \mathrm{d}\vec{r}_{2}
$$
is always positive. But I could think about relatively simple physical argument that proves that at least the sum of all the exchange integrals is positive. The argument is based on the very notion of a variation principle and goes as follows.

We know that exchange integrals arise only if we use an antisymmetric product of orbitals as our trial wave function. If we use just a simple product of them there will be just Coulomb integrals in the energy expression.
Now since the true wave function is antisymmetric, if we use an antisymmetric product of orbitals as our trial wave function instead of a simple product of them, we are guaranteed to get the lower energy since we use the qualitatively right wave function.
Exchange integrals enter the energy expression with the negative sign, thus, they have to be positive, otherwise we won't get the lower energy by using an antisymmetric product of orbitals instead of a simple product.

Again, it does not prove that each and every exchange integral is positive, only that their sum is.


How can we tell whether a phenomena was a chemical reaction or not? I mean in principle, in general.

How can we tell if some phenomena was a chemical reaction(s)? Change in physical properties cannot be since large-scale structural change can change physical properties as well, like heat treatment of steel. And what about changing diamond into graphite by heating. Hence, assuming there are only chemical and physical properties, only a change in chemical properties implies chemical reaction. But how do we know if chemical properties changed? Observing chemical reactions of the result(s). But how do we know what happens is a chemical reaction? A different instance of the same problem, which we started with.

How can anyone sure claim that a chemical reaction happened, or equivalently that some sample is a different substance and not just an allotrope or has larger "grains" for example?

Maybe mass or energy change can rule out those cases?

For some chemical reactions, it would be very difficult to prove without chemical analysis of the materials before and after the reaction. However, with modern analysis techniques, you could show whether a reaction took place. For example, if you tempered steel or converted diamond to graphite, analysis would show that it was a physical change or change in the crystal structure that occurred.

For a low-tech example, consider the combustion of some condensed-phase hydrocarbon in oxygen. If you ignite the hydrocarbon under oxygen in a container sealed except for an oxygen source and an exit to a condenser, you would see the hydrocarbon vanish and water would appear in the condenser. For this we'll just ignore carbon dioxide production. You then easily confirm that the water is not just another form of the hydrocarbon by measuring it's freezing and boiling temperatures. You could then repeat the experiment without ignition of the hydrocarbon and see that you get nothing.
This would confirm beyond any reasonable doubt that you've had a chemical reaction that converted the hydrocarbon to water.


Change in enthalpy and entropy when sugar dissolves in water We know that dissolution of sugar in water is a spontaneous process.
So, change in Gibbs free energy $(ΔG)$ must be negative for the overall process:
$$ΔG = ΔH - TΔS < 0$$
Hence either the enthalpy or entropy must drive the reaction.
Now, after searching for the values, most websites suggest the process is endothermic (i.e. $ΔH > 0).$
But Jan's answer has me convinced that the overall entropy of a system should decrease (i.e. increase in entropy of sugar molecules is outweighed by decrease in entropy of water molecules).
So, what's the real reason for reaction to be spontaneous if neither entropy and enthalpy are driving the reaction?

The entropy change for ideal mixing is $\Delta S=-R(x_A\ln{x_A}+x_B\ln{x_B})$


How does LC50 translate to "real world" exposure for paint thinner? Last year I started oil painting, which means (in part) that I use paint thinner for working with and/or cleaning brushes. I recently became concerned about my exposure to the vapor from the paint thinner I use, so I requested a toxicology report from the manufacturer. In the document it specifies that the summated $\mathrm{LC}_{50}$ for the product is 21400 mg$\cdot$ m$^{-3}$.

My (basic) understanding of this is that 50% of test subjects will die with a single exposure at the above level; however, I've really no idea how that translates to "real life" and wonder if someone can shed some light on it.

I'm painting in a reasonably small room (around 3 m x 5 m) with windows open, so I'm essentially trying to determine if this is safe or not, or whether I should be using breathing apparatus and/or move to a different room.

This is a link to the whole document which I received from the manufacturer which contains the substance and other information.

The $\mathrm{LD_{50}}$ and $\mathrm{LC_{50}}$ (lethal dose and lethal concentration) values are measures for acute toxicity as you noted. Long-term exposure is generally not depicted well by these acute toxicity measures. A.K. pointed out, how improbable it is to have a serious risk of acute poisoning. (Also note that the vapour pressure is given as $2~\mathrm{mmHg}$, which is $2.7~\mathrm{mbar}$ in proper units, meaning that evaporation will be very slow at ambient temperature and pressure.)

Chronic toxicity data is typically given as workplace exposure limits, under the reasoning that if you work with something every day and its concentration is less than the harmfulness limit, it should be okay. Scanning your MSDS, I did not come across any such value, so there should be no serious chronic toxicity if you keep your room well vented and don’t forget to recap the bottle, etc.


PdPS interlayer distance PdPS is being considered as a layered material but when I look at its crystal structure it does not look like that. Fo example I could not figure out its interlayer distance. Does anyone familiar with this structure?

Here is a visualisation that shows the layered structure of the material.

From the side:


Could observer effect cause electrons in one orbital to be found in another? The observer effect states [1] that

when unobserved, absolutely small particles like electrons can simultaneously be in two different states at the same time.

If we look at an atom of any element, all of its electrons are subject to the observer effect. As the electrons are known for wave-particle duality, the Schrödinger equation can be solved to yield a wave function for each electron in the atom. An orbital is a probability distribution map of a wave function squared, and each orbital can accommodate up to two electrons.
For a single electron, it is not hard to imagine that the observer effect operates upon it to the extent of the probability distribution map specified by its orbital. That is, the electron can simultaneously have two positions if unobserved, and these two positions will not go beyond the space occupied by its orbital.
However, I'm wondering if an electron can have two positions when unobserved, where these positions are in different orbitals. For example, an electron falls under both the $\mathrm{2s}$ orbital and the $\mathrm{2p}_x$ orbital, if unobserved, and observation forces it to collapse into either $\mathrm{2s}$ or $\mathrm{2p}_x.$
Reference

Tro, N. J. (2015). Principles of Chemistry: A Molecular Approach, Global Edition. United Kingdom: Pearson Education Limited.

Your intuition is exactly right. Let's consider an atom of neon. It has ten electrons, which we describe as being in 1s, 2s and 2p orbitals. But the electrons are indistinguishable from each other, and accurate wave functions allow all of them to be in all orbitals. Only when we make a measurement (or some other sort of "observation") do the electrons behave as if they are localized to individual orbitals.
For example, if we measure the binding energy of the electrons by photoelectron spectroscopy, the result will always be an eigenvalue of the energy operator (aka Hamiltonian). But that does not mean that any one electron had exactly that energy prior to the measurement. As is always the case in quantum mechanics, systems that have multiple eigenstates exist as a superposition of those states until the measurement, the result of which is always one of the possible eigenvalues rather than, for example, a weighted average of them. So we will measure binding energies consistent with 2p orbitals, 2s orbitals and 1s orbitals with an intensity ratio of 3:1:1.