Why does chlorine replace a hydrogen in propene instead of adding to the double bond? I read about the following reaction in my book:

$$\ce{CH3-CH=CH2->[Cl2][\pu{773 K}] (Cl)CH2-CH=CH2}$$

It surprises me that the $\ce{Cl}$ doesn't attack the $\pi$ bond, rather it attacks the $\sigma$ bond.

It seems to me that the reaction follows free radical mechanism, for the high temperature of the reaction system. That's why, the $\ce{Cl}$ attacks the $\sigma$ bond.

But, I am not sure, why won't free radical $\ce{Cl}$ attack the $\pi$ bond.

It's not that the chlorine radical doesn't attack the $\pi$-bond, it just doesn't do it as fast. Alkyl radicals are much less stable than allyl radicals, so hydrogen abstraction will occur much faster than radical addition ($k_1 >> k_2$).


Which Phenols give the colours blue, green or violet in the neutral FeCl3 test? I have read about the neutral FeCl3 test for Phenols, where a positive test is indicated by the colour of the solution which ranges from green to blue to violet. But which specific derivatives of phenol give a specific colour? Or is it random?

The colour may vary from blue, green or even red depending upon the
  nature of the phenol.(Source)

o,m,p-cresols and resorcinol gives violet or blue coloration. (source)
beta-naphthol gives green coloration. (Source)
alpha-naphthol gives pink coloration. (")


How is dimerization done? Nitrogen dioxide is dimerized to form dinitrogen tetroxide according to the equation
$\ce{2NO2 -> N2O4}$. How do two molecules of one compound fuse to form another?

The electronic structure of NO2 leaves an unpaired electron on the nitrogen. This is one of the reasons it is quite reactive. But it is also the reason why it can easily dimerise to N2O4 as the two unpaired electrons in two molecules of the gas can pair up creating a weak bond. The dimer is favoured at low temperatures when the thermal energy isn't large enough to shake the dimer apart (I'm simplifying a bit). This is why the colour of the gas changes depending on temperature as the dimer is colourless but the NO2 is brown.


Stability of carbanion I am confused about how one can determine the stability of carbanions using pKa-values. I know that the stability of carbanions can be determined by the inductive effect, hybridization of the charged-bearing atom and resonance, but when searching about how pKa can do so I don't find anything. The lower the pKa-value, the stronger the acid and hence, the weaker the conjugate base, does this have anything to do with the stability since weak bases are unreactive and thus more stable?

As commented by SteffX, less the $\mathrm{p}K_\mathrm{a}$, more stronger the acid. More stronger acid means that the conjugate base or the anion formed after donating the proton is more stable compared to the carbanion of an acid of a  higher $\mathrm{p}K_\mathrm{a}$ value. So, if the anion formed is a carbanion, then lower the $\mathrm{p}K_\mathrm{a}$, the stable the carbanion is.


When does copper leach to water? Conflicting faqs I have heard conflicting reports:

Copper leaches to water only when the water is high in acid or the Copper is corroded.
Copper leaches to water any time, especially when it is new copper or the water is hot.

Which one would be correct?

The second answer would be correct in an analytical context as the auto-ionization of water decreases in pH as temperature increases. And new copper does not have the surface of reacted material to shield the metal underneath, as such, scenario 2 always occurs, scenario one occurs faster, however, it's conditional statement that the water be acidic and the copper be corroded are invalid.
Cheers,


Will we ever run out of gold, silver, copper and other important conductors? Conductors are very important in technology. At the current rate of extraction how long until the world's reserves of copper are depleted?

How long will these reserves last?

Short answer: longer than you think and will outlast me, you, our children and our grandchildren. Possibly last forever.

Long answer: This is a fairly complex issue. There was a book published about this topic several years ago claiming that we are running out of metals. Turns out that the book had a major flaw in its reasoning, so the same publisher followed up with a second book refuting the first one. Both books are open access and freely available so I recommend having a look.

The problem lies in the definition of the word reserve.

What are reserves?

A reserve is an ore deposit (that is, a concentration of a certain metal) that we know exactly how much is in there to a good degree, and we have detailed plans on how to extract it. This means, we have plans to build a mine, we know the cost of running the mine, etc. In some countries (for example, Canada and Australia) you are not allowed to call something a reserve unless you have the plans. This is a legal definition.

What are resources?

On the other hand, if you know how much metal is in the deposit, but you haven't made the plans to extract it yet, it is a resource. You are not legally allowed to call it a reserve.

The work required to develop a resource into a reserve costs money. You have to employ mine engineers, you need to get environmental approval, you have to do a lot of lengthy and costly things. Therefore, you don't "declare" a reserve until you have to. If you are a mining company and you have reserves of 40 years, why bother converting more of your resources into reserves? And this is exactly the problem: alarmists and the media consider the reserves, without the resources. Of course we have limited reserves! Because it's pointless to turn the resources to the reserves if we still have reserves!

Quoting Antonio Arribas, an exploration geologist, from an enlightening article published in the SEG Newsletter:

  One was published in issue 2605 of the international science magazine
  New Scientist in May 2007 (Fig. 5). In it, the author claims that,
  based on data by researchers from Yale University and the University
  of Augsburg, “if predicted new technologies appear and the population
  grows, some key resources will be exhausted” within what is
  graphically shown as an alarmingly short time frame (Cohen, 2007).
  Among the mineral resources covered, platinum, indium, and copper are
  particularly illustrative of the errors of interpretation contained in
  the article and the overall fallacy that this report represents.

  According to the article, humankind should be running out of platinum
  within four years (15 years from 2007). However, despite production of
  2,000 t of platinum between 2007 and 2017 and a metal price drop of
  26%, the global reserves of platinum group metals have decreased only
  by 2.8%3 (data from USGS, https://minerals.usgs.gov/
  minerals/pubs/mcs/). Likewise, according to this report we should have
  run out of indium sometime between 2012 and 2017. Never mind that, as
  a key component of indium tin oxide (ITO), this metal has been present
  as a transparent thin film conductor in the touch screen displays of
  every smartphone and tablet manufactured since then—including the 1.2+
  billion iPhones sold by Apple alone since the product’s release in
  June 2007, barely a month after publication of the New Scientist
  article.

  The second article and infographic selected were titled
  “Global resources stock check” and appeared in a June 18, 2012,
  article from BBC Future (www. bbc.com/future/story/
  20120618-global-resources-stock-check). In this report, the world
  supplies of indium, antimony, and copper, just to choose three, were
  projected to last through 2024, 2020, and 2044, respectively. I would
  argue that nothing but common sense is needed to believe that by the
  time of the Paris Olympiad in 2024, ITO-based transparent thin film
  conductors (or a new, and likely superior, substitute) will continue
  to be present in digital displays and thin-film photovoltaic cells,
  among many other applications. As for antimony, how do we reconcile
  the prediction that we should have only two years left of the metal
  now when, after production of
  1.1 Mt between 2011 and 2017, global reserves remain unchanged within the
  1.5- to 2.1-Mt range that has existed every year since 1997, when the U.S. Geological Survey started to record reserve estimates? In other
  words, for the time being and for the foreseeable future, the reserves
  of this metal are controlled by economics, technology, geopolitics,
  etc., and not by absolute limitations of availability in nature.


Isolation of salt from solution I just want to clarify my understanding of whether metathesis can be used to form other products by combining an acid and a base (do special rules apply whether its strong/weak acid/base)?

Say I want to form some $\ce{Na2CrO4}$. There's many ways to do this but the most obvious one I can think of is:

$$\ce{2NaOH(aq) + H2CrO4(aq) ->Na2CrO4(aq) + 2H2O}$$

Can you then just filter out the $\ce{H2O}$ to get your desired product? How would you do so? I'm used to looking at these in terms of net ionic equations where two aqueous reactants for a solid precipitate but since everything here would cancel out (except water) is this even a viable possibility?

You obviously cannot ‘filter water out’ as water is a liquid at STP and the solvent in which your reaction occurs in. However, you can evaporate your solution whereby the solvent gets removed and the lower vapour pressure contents (mainly solids or oils) will remain. While water can be more difficult to evaporate at a standard rotavap, it is by no means impossible.


Relationship between Hartree-Fock and Hückel coefficients I am new in computational chemistry. I'm studying the Hartree–Fock–Roothaan (HFR) method and have some of questions about the HFR basis function coefficients.

Do we have the condition on HFR coefficients like in the Hückel method?

In the Hückel method, we have a normalization condition on the coefficients $\{c_i\}$:

$$c_1^2 + c_2^2 + \dots + c_i^2 = 1 \tag{1}\label{1}$$

In the HF literature I didn't see that condition, so my guess is that we don't have it.

Do we have similar to Hückel method condition on HFR coefficients of bf?

I'm wondering because going deeper in understanding HF I find out that we have certain conditions that looks like $\eqref{1}$. For example, a normalization condition for the Gaussian primitives that are not normalized:

$$
S = \left< \Phi_{\textrm{1s}}^{\text{STO-3G}} | \Phi_{\textrm{1s}}^{\text{STO-3G}} \right> = 1
$$
$$
\Phi_{\textrm{1s}}^{\text{STO-3G}} = N_1C_1e^{-a_1r^2} + N_2C_2e^{-a_3r^2} + N_3C_3e^{-a_3r^2}
$$

This is not the same as $\eqref{1}$ but we can represent it as

$$f_1(c_1^2) + f_2(c_2^2) + \dots + f_i(c_i^2) = 1 \tag{2}\label{2}$$

Because HF is more complex than Hückel, $\eqref{2}$ can make a sense. So can I think about $\eqref{2}$ as complex version of $\eqref{1}$? I was hoping to find $\eqref{2}$ or similar equation in HF literature intro part, but those sources that I have do not mention $\eqref{2}$ in intro part. Maybe my attention was scattered or it is just a mess in my head, anyway I will be grateful for any help.

What are the conditions imposed on the coefficients of the basis functions that use in the calculation of the HFR equations?

For example, the electronic charge distribution is

$$
\sum_{mn} P_{mn}S_{mn} = N =\sum_{mn} 2\sum_{a}^{N/2} c_{ma}c_{na}^*\int{f_m(r)f_n^*(r)\mathop{dr}},
$$

where

$$
S_{mn} =\int{f_m(r)f_n^*(r)\mathop{dr}} = \delta_{mn},
$$

and N is the total number of electrons. Can we consider this as a condition?

Let start by definition the Hartree-Fock-Roothaan equation:

$$\sum_\nu F_{\mu\nu}C_{\nu i}=\varepsilon_i\sum_\nu S_{\mu\nu}C_{\nu i}$$

Where the Fock matrix elements are giving as:

$$F_{\mu \nu} = \left< \phi_\mu \left| \hat{f} \right|\phi_\nu\right>$$

And the overlap matrix is given as:

$$S_{\mu \nu} = \left<  \left. \phi_\mu \right|\phi_\nu\right>$$

Here $\phi_\nu$ is our atomic orbitals (i.e. what you get from your basis set).

When we do a Hartree-Fock calculation we want to find a set of orthonormal orbitals, also known as molecular orbitals. The molecular orbitals are constructed as:

$$\psi_i=\sum_{\mu=1}^KC_{\nu i}\phi_\nu$$

For the molecular orbitals to be orthonormal we can write the following condition:

$$\left< \left. \psi_i \right|\psi_j\right>=\delta_{ij}$$

This means that the orbitals have zero overlap with one-another and an overlap of $1$ with them self. If we insert the expansion of the molecular orbitals in atomic orbitals in the above equation we get:

$$\sum_{\mu=1}^K\sum_{\nu=1}^LC_{\nu i}^\dagger C_{\mu j}\left< \left.\phi_\nu \right|\phi_\mu\right>=\delta_{ij}$$

I.e. we can see that when we do a Hartree-Fock calculation we have the constraint on our molecular orbital coefficients that have to transform our atomic orbital overlap to a matrix with ones on the diagonal and zero elsewhere. This can also be written more compact in matrix form:

$$\mathbb{C^\dagger SC=1}$$

Which is a condition equivalent to the one you know from Hückel theory. At last let me make a breif comment about basis set, since you also talked about those in your question. In general we use contracted Gaussian type basis sets, just like STO3G, that you mentioned. Our atomic orbitals can thus be written as:

$$\phi_\nu = \sum_{k=1}^M N_k d_k P(r)\cdot e^{-\alpha_k\cdot r^2}$$

Here $P(r)$ is a function related to the angular moment quantum number.The way the contraction coefficients $d_k$ are choose implies that we normalize our Gaussian functions, i.e:

$$N_k^2 \int_{-\infty}^\infty \left( P(r)\cdot e^{-\alpha_k\cdot r^2} \right)^2 dr = 1$$

We can now write $\left< \left.\phi_\nu \right|\phi_\nu\right> = 1$ as:

$$\overline{N}\sum_{k=1}^M N_k d_k\sum_{l=1}^T N_l d_l\left< \left.P(r)\cdot e^{-\alpha_k\cdot r^2} \right| P(r)\cdot e^{-\alpha_l\cdot r^2}\right> = 1$$

Where $\overline{N}$ is the normalization we need. As we can see there is no easy condition we can put on the contraction coefficients, in order to avoid evaluation $\overline{N}$. It can be shown (I will not show it here) that $\overline{N}$ only depends on the total angular moment and on the Gaussian exponents, thus it is possible to choose the contraction coefficients in such a way, that we do not need to evaluate $\overline{N}$.


Is "ice point" the same as "freezing point"? I am using an old book and they almost use both terms interchangeably sometimes. Do they mean the same thing? Similiarly for "steam point" and "boiling point"

Ice point is the freezing point of water.

Freezing point is the point where liquid turns to solid.

The difference is that in the case of ice point is substance is always water.

(Note that there is a slight discrepancy of 0.01 K between the triple point of water, melting point, freezing point and/or ice point. And technically, Ice point is the point where water and Ice are in equillibrium.)

See e.g. http://www.thefreedictionary.com/freezing+point, http://www.thefreedictionary.com/ice+point.


Home-made way to detect lye in solution I believe that electrolysis of salt water creates lye ($\ce{NaOH}$ - sodium hydroxide). In salt water, there's $\ce{Na+}$, $\ce{Cl-}$, $\ce{H+}$ and $\ce{OH-}$ as of ionts.

So at cathode, you get $\ce{Na2}$ and that should react with water into $\ce{NaOH}$.

I want to test, if this is true - how do I do it with household items? The solution didn't react with aluminium.

The easiest way to measure the pH of a solution is to use an acid-base indicator. Limited to household supplies only, I would suggest a solution made from red cabbage, which should turn blue-greenish in the alkaline region.

The ions you actually get from your reaction is a higher concentration of $\ce{Na+}$ and $\ce{OH-}$, due to the fact that you are removing $\ce{H2}$ and $\ce{Cl2}$ (from $\ce{H+}$ and $\ce{Cl-}$ ions respectively) via the performed electrolysis. Under no circumstances would you generate a $\ce{Na2}$ species, you would rather write it as $\ce{2Na}$. Also, in aqueous solutions the $\ce{NaOH}$ most certainly dissociates. So speaking of the formation of solid $\ce{NaOH}$ from $\ce{Na}$ in water is quite a stretch.

Bottom line: Since the salt is already dissolved when you start the electrolysis, all you get is a change in concentration of the ions involved. The thereby induced change in pH can be measured observed using an acid-base indicator such as the juice of the red cabbage.


Do all metal nitrates give nitrogen dioxide on decomposition? Some metal nitrates decompose to give nitrites: $\ce{2MNO3 -> 2MNO2 + O2}$

whereas some metal nitrates decompose to give $\ce{NO2}$. A possible reaction is $\ce{M(NO3)2 -> MO + NO2 + O2}$
Based on my observation my question is that, is it true that all metals (including transition metal) nitrates decompose to $\ce{NO2}$ apart from $\ce{[Na , K, Rb, Cs]}$ (alkali metals which give nitrites instead) ?

TL;DR: Decomposition of nitrites always afford $\ce{NO2}$ if you apply the correct reaction condition to it.
Long answer
To generalize the decomposition of metal nitrates (I concluded from the article above, and hereby using a generic monovalent metal M for the equations), we can say this:

The first stage of decomposition is always the decomposition to nitrites: $\ce{MNO3 -> MNO2 + 1/2 O2}$

Note that nitrites are almost always (if any) less thermally stable than the corresponding nitrates. Thus, the formed nitrites from the reaction above quickly decomposes. I think that (and from the article), nitrites decompose to form the oxide and $\ce{N2O3}$, the latter disproportionates into $\ce{NO}$ and $\ce{NO2}$: $\ce{2 MNO2 -> M2O + NO + NO2}$.

Everything becomes interesting from here. It is due to the fact that the nitrite formed from the first step and not decomposed (kinetics matter) can be oxidized by either the $\ce{NO2}$ or $\ce{O2}$ back to the nitrate. This is especially significant when the nitrite is stable enough ($\ce{KNO2, NaNO2}$ etc). In fact, the decomposition of $\ce{KNO3}$ is reversible (equilibrium) at about 650 - 750 °C in the atmosphere. In the other hand, nitrites of transition metals (which are mostly covalent in character) are so unstable that they decompose rapidly and thus only appear as the so-called reactive intermediates.

Finally, the oxide can decompose into the metal and $\ce{O2}$. This is a characteristic of metals that lie in the less-reactive end of the reactivity series. If the decomposition temperatures of the oxide and the nitrate are different enough, you can control the temperature to get the oxide or the metal ($\ce{HgO}$ from $\ce{Hg(NO3)2}$ is an example.)

The re-oxidation of stable nitrites can be avoided by two ways: either do the reaction under a flow of inert gas to displace the gaseous products away, or crank up the heat more to increase the reaction temperature, which means faster decomposition of nitrites. For example, decomposition of $\ce{KNO3}$ goes to completion at about 800 °C.
Reference

High Temperature Properties and Decomposition of Inorganic Salts Part 3, Nitrates and Nitrites, Kurt H. Stern, Journal of Physical and Chemical Reference Data 1, 747 (1972); DOI: 10.1063/1.3253104


How to know the aluminium sulfate solubility as function of the pH? I'm trying to find a diagram or a reference for to know the solubility of aluminium sulfate $\mathrm{Al_{2}(SO_{4})_{3}}$ in differents pH values. Something like:

If the question only is about the diagram and the reference:

  Rubin, A. J., & Hayden, P. L. (1973). Studies on the Hydrolysis and Precipitation of Aluminum (III). page 57.

But the interpretation is not clear for me, so here is a new question


Isopropyl Alcohol leaves a white haze on plastic, is there anything I can use to clean it off? Recently I wanted to clean up some lubricant on some keyboard switches. On some forums, an Isopropyl alcohol bath was recommended so I decided to give that a try and went to purchase some 91% Isopropyl from Target(purest I could find near me and I admit, I am a little impatient).

Good news - it worked in getting the lubricant off the switch! Bad news, on some of the switches, it left this white haze.

I'm not sure what it is, though from what I've been reading, it sounds like this is a common issue when using Isopropyl, my best guess is that it's some mix of the lubricant mixing with the alcohol and reacting poorly with the switch housing plastic given that I'm only seeing this haze on the bottom half of the switches which were the only parts of the housing that were lubed, the other parts are clean. Another reason for my guess is that this haze isn't appearing in the same amounts on all the switches if at all, some switch bottoms are basically clear.

It does appear to scrape off but looks like it will require a fairly strong brush which I don't have at the moment and I'm not sure if there's one available for purchase that could be used in this situation.

Is there any plastic-friendly household cleaner available that might be able to help speed up the process?

Even the common alcohol ( ethanol, ethylalcohol) is known to corrupt surface of some plastics, and rubbing alcohol ( isopropanol, isopropylalcohol) is even better in that, as it contains bigger organic chain and is less polar.

These alcohols do not dissolve the plastics in large extent, like acetone often does, but they may corrupt surfaces, especially if they are shiny or polished.

The typical case is compact polystyren or polymethylmethacrylate, that are often clear and shiny, where the white milky haze is very well visible. The opposite case is filled black shiny plastic where the haze and lack of shining  is notable.

For similar reason, lukewarm water with detergents is preferred for plastics cleaning.

The only way I see to get rid of the haze is polishing, or applying a selfpolishing  coating, if applicable and worthy. Perhaps polishing of wax coating.


Why do we use glacial acetic acid in lidocaine synthesis? The lab sheet says the following :

  This reaction is carried out in glacial (anhydrous) acetic acid as the
  solvent. The acetic acid reacts with 2, 6-dimethylaniline to establish
  an equilibrium between it and its acetate salt.

What is the purpose of establishing an equilibrium between 2,6-dimethylaniline and its salt here?

EDIT:
This is the source of the quoted text & the procedure to synthesize lidocaine : https://tigerweb.towson.edu/jdiscord/www/332_lab_info/summer2016chem332labs/webexpt_lidocaine_a.pdf

I think acetic acid is used because it allows you to play funny games with the varying solubility of reactants, products, and by-products. Note that the final product is not soluble in acetic acid/water mixture, but the reactants and salt by-products are. This means that instead of performing an extract in a separatory funnel (extra step, plus you'll need to evaporate solvent and purify), you can just filter the solid product directly from the rest of liquid.

This is the kind of simplifying "magic" you can get when the process chemists get a stab at designing how to run the reaction.


Propane Combustion in a Gas Grill with Irregular Flameouts I'm troubleshooting a problem with a brand-new propane gas grill that is experiencing flameouts and making puffing/popping sounds.  I'm asking here because I want to understand the basic chemistry that would cause this problem.

The problem is demonstrated in this video.

I've already been through all the expected troubleshooting steps:

Replaced the gas regulator
Replaced the burner
Tried three different propane tanks

The grill was purchased about 3 weeks ago. The valve/regulator assembly was replaced last week, and the burner was replaced yesterday, so both parts are essentially new and unused.

It's hard to tell from the video if the puff causes the flameout or the flameout occurs first, followed by accumulation and detonation of a small amount of unburned propane.

I would like to understand the chemistry principles that govern the combustion of propane in a grill, and what design flaw(s) could cause this behavior.

The brilliance of the orange in flares suggests suggests incomplete combustion of a carbon residue.  This can happen with a poor oxygen mix, but the blue in the core of the flame indicates that mix is fine.

Also, on the base of the grill there is some charred residue.

Based on the color, and the residues I expect somewhere in the flow you have some carbon deposits that are occasionally getting dislodged, causing the sputtering.


What is the difference, if there is, between this two distillation methods? Reading about water vapor distillation, I found two differents approaches:

Direct one
Indirect one.

The first has an arrange like:

Here the chamber contains water and the plant. It is cooled and as the water vapor flows away it carries the low volatile compound vapor of the plant/vegetable. Both condense and the oils can be obtained.

The vegetal is on the same pot as water.

The second one:

Here the water vapor is generated on a first po (the blue one) and when flowing through the balloon, it carries the oil vapor. I should remark: the vegetal is on a different pot than water.

Question

Do you see any fundamental difference on the process or something related?

It might has to do with the aggregation state of the material we need to extract, but I'm not sure.

There is no fundamental difference between the two methods.
In both cases steam is formed and it is passed through the plant material.
The intention is to extract the non volatile water insoluble essential oils from the plant.
The method is called steam distillation.

On the other hand, normally it is not done in this way.
Since generally only the essential oil is of interest, the integrity of the plant does not matter, it is placed in water and the whole is heated and the water is distilled.
The students do this in the lab with cinnamon and orange peels, for instance.
It is much easier because you only have to heat one vessel.
In these two step setups, you must isolate the part containing the plant very well so that the steam does not condense, once the whole system is hot enough.
Preferably, you should also heat the part containing the plant, especially in your second example.


Calculation of silver nanoparticle size using Mie Theory I have synthesized silver nano-particles and I wish to determine the approximate particle size using the wavelength that was most absorbed according to its absorption spectrum.

I find out that Mie theory provides a good approximation for this, (Eq 4 in this paper). However, I am clueless as to how to use the equation. In other words, how should I use the $\lambda_{max}$ obtained through the experiment in the provided equation?

Plug it in to the math equation
Ok, so we have this equation: $γ(R)=γ_0+(Av_f)/R$
First, we substitute the values into the equation that we know.
$γ(R) = (5\cdot10^{12}) + (3/4\cdot1.39\cdot10^6)/R$. Simplified, this is:
$γ(R) = (5\cdot10^{12}) + (1.0425 \cdot 10^6)/R$
$γ(R)$ can be measured in your experiment. Then we can solve your equation for $R...$
$\large{R = \fbox{$\frac{1.0425 \cdot 10^6}{γ(R) - 5\cdot10^{12}}$}}$
Plug in your value of $γ(R)$, and you're there!*

*In other words, how should I use the λmax obtained through the experiment in the provided equation?
The equation only mentions $\gamma (R)$, I don't know how to help you there**

**Zhe: @JavaScriptCoder For these types of questions, the first thing to ask is "do you work in a lab?"
If not, why are you doing experiments outside of a lab, and if so, why is it that no one in your lab has any expertise on this?


Why are there multiple lines in the hydrogen line spectrum? Hydrogen has only one electron, yet it exhibits multiple lines in a spectral series, why is this?

As the electrons fall from higher levels to lower levels, they release photons.  Different "falls" create different colors of light.  A larger transition releases higher energy (short wavelength) light, while smaller transitions release lower energies (longer wavelength).

The visible wavelengths are caused a by single electron making the different transitions shown below.  There are even more transitions that release invisible wavelengths.

 Wavelength   Transition  Color
 (nm)
---------------------------------------------
 383.5384      9 -> 2    Ultra Violet
 388.9049      8 -> 2    Ultra Violet
 397.0072      7 -> 2    Ultra Violet
 410.174       6 -> 2    Violet
 434.047       5 -> 2    Violet
 486.133       4 -> 2    Bluegreen (cyan)
 656.272       3 -> 2    Red
 656.2852      3 -> 2    Red

*Values taken from Hyperphysics: Hydrogen Energies and Spectrum

(Why are there two different 3-> 2 transitions?  See here: Hydrogen Fine Structure )


Balancing Ionic and Net Ionic equations I am currently taking a chemistry course about acids and bases. One of the things I am a bit confused about, is the acid and base neutralization reaction, and how to balance it. For Example:

$$\ce{NaOH + HCl -> NaCl + H2O}$$ How do you write the ionic and net ionic equations for that?

Note: My main problem here is that my teacher requires us to first write the ionization of the $\ce{HCl}$ into $\ce{H3O+}$ and $\ce{Cl-}$. This is when I get confused, because I don't know how to balance the equations with the $\ce{H3O+}$

Also, I had another question in my exam, which I got wrong because of the balancing. Can you please tell me the correction of it:

  Consider the neutralization reaction between nitric acid and calcium hydroxide. Write the balanced neutralization reaction and the ionic and net ionic reactions of it:

As $\ce{NaOH}$ and  $\ce{HCl}$ are strong base and acid respectively, they are completely dissociated in water. The salt $\ce{NaCl}$ is a  strong electrolyte and completely dissociated in water. So, the ionic equation is:

$\ce{(Na^+ +OH^- )}+\ce{(H^+ +Cl^- )}\rightarrow \ce{(Na^+ +Cl^- )}+\ce{H2O}  $

As the ions $\ce{Na^+}$ and$\ce{Cl^-}$ do not combine and remain in the solution as spectator ions. Their role is to keep the solution electrically neutral, the net ionic equation is:
$\ce{OH^- }+\ce{H^+ }\rightarrow \ce{H2O}  $.

On the other hand, we have to notice that the ion $\ce{H^+}$ is not stable in water, it combines with $\ce{H2O}  $ to form the ion hydronium $\ce{H_3O^+}$ . We usually write the complete dissociation of $\ce{HCl }$ as follows:

$\ce{HCl} +\ce{H_2O }\rightarrow$ $\ce{Cl^-} +\ce{H_3O^+}$

and the ionic equation is

$\ce{(Na^+ +OH^- )}+( \ce{H_3O^+}$$ +\ce{Cl^- })\rightarrow \ce{(Na^+ +Cl^- )}+2\ce{H2O}  $

And the net ionic equation is

$\ce{OH^- }+\ce{H_3O^+ }\rightarrow 2\ce{H2O}  $.

As for your exam questions:
$2\ce{HNO_3} + \ce{Ca(OH)_2 }\rightarrow \ce{Ca(NO_3)_2} + 2\ce{H_2O}$
 Now you can try to write the ionic and the net ionic equation to see if you understand what I have explained.


What is the significance of heating value? Say I burned for example methane and propane to boil some water.
Methane has the heat of combustion of $\pu{890kJ/mol}$ and propane has the heat of combustion of $\pu{2200kJ/mol}$
What does the difference in heat of combustion tell us? In other words, propane's heat of combustion is bigger. Does that mean that methane emits less energy and therefore is less "environmentally harmful"? Is a higher heat of combustion better or a lower heat of combustion?

Note that the mentioned values are molar heats of combustion. Propane is a larger molecule than methane and it is therefore obvious that more energy per mole is released.
If we replace the molar amount by molar mass, the resulting specific heat of combustion has the opposite relation:
$$\begin{align}\text{Methane:}&&\frac{\pu{890kJ/mol}}{\pu{16g/mol}} &=  \pu{55.6kJ/g}\\[0.5em]
\text{Propane:}&&\frac{\pu{2200kJ/mol}}{\pu{44g/mol}}&=\pu{50kJ/g}\end{align}$$
For the enviromental aspect of $\ce{CO2}$ emissions, the relevant value is the heat of combustion per mole of carbon atoms. It is $\pu{890 kJ/mol}$ for methane, but $\pu{2200 kJ/mol}$ for propane, which is roughly $\pu{730 kJ/mol}$ for carbon. For methane, both numbers are the same, as a propane molecule contains 3 carbon atoms, while a methane molecule just 1.
So methane has a an advantage as energy source, as it has higher energy production per the same $\ce{CO2}$ emission.
PLA is partially already oxidized, therefore it has significantly lower energy content per 1 kg of produced $\ce{CO2}$, compared to  PP.
PLA has an energy content of $\pu{16.3MJ/kg}$
PLA has structure [-O-CH(CH3)-CO-]n, with basic unit containing 3 C + 4 H + 2 O
$\pu{1 kg}$ of PLA contains 36/72 kg of carbon
PLA has an energy content of $\pu{16.3 * 72 / 36 MJ/kg} \simeq \pu{32.6 MJ/kg of C}$ .
PP has an energy content of $\pu{46.4MJ/kg}$
PP has structure [-CH2-CH(CH3)- ]n, with basic unit containing 3 C + 6 H
$\pu{1 kg}$ of PP contains 36/42 kg of carbon
PP has an energy content of $\pu{46.4 * 42 / 36 MJ/kg} \simeq \pu{54.1 MJ/kg of C}$ .
To get the same energy by the burning of the stuff,
PP releases $54.1/32.6 \simeq 1.66$ times less $\ce{CO2}$ than PLA.
Or, with the same emission of $\ce{CO2}$, PP releases 1.66 times more energy than PLA.


Demonstrating decomposition of hydrogen peroxide using iron(III) nitrate catalyst I need a way to prove/show that hydrogen peroxide was decomposed through use of catalyst.
I want to ensure that my catalyst: $\ce{Fe(NO3)3}$ or iron(III) nitrate is a catalyst, not a reactant/ consumed during the reaction.

$$\ce{2 H2O2 (aq) ->[Fe(NO3)3 (s)] 2 H2O (l) + O2 (g)}$$

When the reaction is happening, I will introduce a wooden glowing splint over the bubbling reaction and the wooden splint glowing brighter or reigniting will show that oxygen is being produced.

Question:

Now, how do I show that water is produced? Would I just boil the product I get after reaction (image above) and put cobalt chloride paper at the water vapour?

Also, if I were to use the orange solution above again as a catalyst (since it still contains iron(III) nitrate), would hydrogen peroxide decompose again? If so, is there a way to put iron(III) nitrate back to its solid state? or any way to reuse as catalyst?

Edit:
If it is easier to answer, it doesn't have to be with iron(III) nitrate. I have an option to use manganese dioxide, which is another catalyst that I can substitute for iron(III) nitrate. I think it should be OK since it does the exact same reaction.

Detection of oxygen: Detection of $\ce{O2}$ by a glowing splint is a good way to detect the oxygen. Also you could capture the gas by a simple fixture e.g.  and demonstrate the volume change in the receiver. This way you can actually measure moles $\ce{O2}$ produced (by $PV=nRT$) then moles $\ce{H2O2}$ decomposed stoichiometrically by the formula you've written.

Choice of catalyst: Using $\ce{MnO2}$ would be better if you want to make sure you have a catalyst. $\ce{MnO2}$ will not be consumed during the decomposition; I'm not sure about iron(III) nitrate.

Or, detecting change in $\ce{Fe(NO3)3}$ concentration: Addition of a very small concentration of potassium thiocyanate, $\ce{KSCN}$ (say 1/100 of your $\ce{Fe^{+3}}$ concentration) will yield a deep red product, iron thiocyanate ($\ce{Fe(SCN)^{+2}}$). If you have access to a spectrometer you can measure absorbance of the initial solution's product and the product after decomposing $\ce{H2O2}$.

Prove water was produced: The best way I can think of is to measure the (subtle) change of density of your solution before and after decomposition. Using $\ce{MnO2}$ would make this easy because you can remove/filter it as a solid after decomposition and thus measure mass/volume of your solution before and after. $\ce{H2O2}$ and $\ce{H2O}$ have small albeit detectably different densities at RT. $\ce{H2O2}$ is more dense than water so your density should decrease.


Is it correct to say that the pKa is the pH at equilibrium? I've heard this said when looking at the charge of certain groups on a molecule (with different $\mathrm{p}K_\mathrm{a}$ values) in a solution of certain $\mathrm{pH}.$ But is that really correct to say?

Based on speciation diagrams, I would have thought that the $\mathrm{p}K_\mathrm{a}$ is always constant and that $\mathrm{p}K_\mathrm{a}$ is simply the $\mathrm{pH}$ at the point where the concentrations of the acid and its conjugate based are equal.

$\mathrm{p}K_\mathrm{a}$ and $\mathrm{pH}$ are apples and oranges.* We can't equate them on a graph. The correct wording should rather be that the $\mathrm{pH}$ at which the concentration ratio of an acid and its conjugate base is equal to $1,$ then the $\mathrm{pH}$ is numerically equal to $\mathrm{p}K_\mathrm{a}.$

* It will be another painful story if one begins to argue on the units of $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}.$ They are dimensionless. Just a subtle point $\mathrm{p}K_\mathrm{a}$ is constant as long as the temperature/pressure is constant. It is not always a constant.


How can an electron shield another electron of the same subshell? While I was preparing for my upcoming exams, I stumbled upon this sentence which is bothering me quite a bit:

  The contraction of the lanthanoids is due to the imperfect shielding of one electron by another in the same subshell.

Though I am well aware of lanthanoids contraction and the poor shielding effect of the f- and d-orbitals, but still according to what I have studied and understood so far is the shielding effect is the shielding of valence electrons by the electrons between the nucleus and the valence electrons from the nuclear charge.

Then how can an electron shield another electron of the same subshell?

Shielding is simply the cancellation of some of the effective charge of the nucleus.

It's a lot easier to envision an electron to shield another electron that, by time average, is likely to be farther from nucleus, e.g., $1s$ shielding $3p$.

But for 2 electrons in the same subshell, on average, you can expect one to be closer to the nucleus than the other roughly 50% of the time. While it's true that a good deal of the time that other electron may be on the other side of the nucleus, for d- and f- orbitals, we're looking at more than 1 electron, so there's some partial shielding. Partial, because the shielding electrons don't actually shield a good portion of the time due to relative distances from the nucleus.


Software for visualizing molecular force vectors I would like to visualize molecular normal modes as vectors sitting on each atom, and not as actual vibrations. I am able to do this, in part, with Molden by copying in the vector into a file readable by Molden (extension is .geo.molden, and is generated from a Molcas calculation). The vector is represented as red lines, but they do not look very nice (see image). Also the background is black, and Molden has a limited number of ways to visualize the molecule.

Is there some other software (free) that can do this? Perhaps Avogadro, VMD, or something along these lines? Specifically I would like to be able to change to different molecular visualizations, and show the vectors as arrows with a easily distinguished color.

Figure 1. This is what Molden can do... not very pretty!

Or possibly, after changing the background color to white with GIMP:

Figure 2. Not really better, since the Hydrogen atoms are represented with white by Molden, as is the label text.

As long as the output from your harmonic frequency calculation can be read directly using Avogadro, it will do exactly what you want.

Notice that:

The normal mode of interest is selected on the right-hand side.
"Force" is selected on the left-hand side; this box is present when "Display Settings..." is highlighted in the top toolbar.


Why triiodide ion does not form ionic bond with cation having +2 charge? In general, I found that triiodide ion generally forms ionic bond with cations having +1 charge like  $\ce{NH4+I3-}$, $\ce{K+I3-}$, $\ce{Tl+I3-}$. But, I have never seen triiodide ion forming ionic bond with cations having +2 or +3 charge like  $\ce{Ba^2+(I3^{-})2}$, $\ce{Pb^2+(I3^{-})2}$, $\ce{Bi^3+(I3^{-})3}$. If properly written, it would  look like $\ce{Ba(I3)2}$, $\ce{Pb(I3)2}$, $\ce{Bi(I3)3}$. These compound has never existed and would not likely to exist. But what is the reason behind this?

Recently I went through this question:-

  Why isn't thallium triiodide stable?

One of the  commenter  said:-

  Thallium Tri-Iodide $\ce{Tl+(I3)-}$ does exist.  Thallium (III) Iodide, $\ce{Tl^3+(I^{-})3}$ to my knowledge, does not.

I was quite surprised at first seeing that $\ce{TlI3}$ exist because I knew that due to inert pair effect, thallium will form Thallium(I) compounds like $\ce{TlCl}$, $\ce{TlBr}$,  $\ce{TlI}$ etc. but I  realized that the compound contains  $\ce{Tl+}$ and not $\ce{Tl^3+}$. But what about the corresponding $\ce{Pb^2+}$ and $\ce{Bi^3+}$ ions? Why does not they form triiodide compounds?

Why triiodide ion does not form ionic bond with cation having +2 charge

It does. Specifically, quick googling for "bis-triiodide" gives out of the blue links like this: poke, poke, so the question is not 'why no bivalent triiodides', but 'why only super-large bications can form triiodides.'

The answer, I suspect, lies in charge density of the cations and their ability to form strong ionic bonds.

It is well known, that energy of ionic crystal lattice strongly increases with decrease of the size of ions involved. Triiodide ion, being quite large, is not very good for forming strong ionic lattices, forming only very weak ionic bonds. Iodide ion, on the other hand, is much better for that.

The energy of formation of triiodide anion from iodine molecul and iodide anion is very low. Thus, triiodie will break to form a crystal lattice with higher formation energy at first chance. So, for a solid triiodide to form, an ion with very low charge density is required, so the energy of crystal lattice would not benefit significantly from triiodide anione dissociation. This rules out any 'naked' cation with charge +2 or more and actually most 'naked' cations with charge +1. On the other hand, by covering dication in a shell of organic molecules strongly bound to the cation, we gain a very large cation, and a solid triiodide can form.

A very similar effect guards temperature of decomposition of solid carbonates, for example, as the decomposition involves breaking of a larger carbonate dianion into a much smaller oxide dianion.


half-reaction in a battery A half-reaction on one of the electrodes in a battery produces free electrons (for example) and consumes anions (or produces cations). I understand why, if there's an external wire for electrons to travel on, there needs also to be a salt bridge for ions to travel on inside the battery, and if there isn't, the electrons will soon stop flowing due to separation of charge building up.

However, since each half-reaction on its own is charge-balanced, what stops the half-reaction from just continuing to happen even when the circuit is open? Why for example in an alkaline battery Zn won't continue to react with OH- abundantly coming from the electrolyte?

The half-reaction stops primarily because it's still producing/consuming electrons, and there's only so many free electrons an electrode (the local environment) can accept/supply.

Just like chemical species, electrons have an equivalent to concentration. (Whereas the generalization of the concentration is the "chemical potential" of the species, the "concentration" of electrons is related to the electrical potential or voltage). If the "concentration" of electrons gets too high or drops too low, the reaction rate of the half reaction will drop off - just like it does if the concentration of a chemical product/reactant gets too high/drops too low. In normal operation, the external conductor allows electrons to move, renormalizing the "concentration" of electrons and allowing the reaction to proceed further.

Note that if you have a poorly made half-cell, you may indeed have a pathway for the electrons to travel to the other half cell, or to react in some other (half-)reaction. If this occurs, then your battery will indeed continue to run even when disconnected, draining its stored chemical energy. But decently made batteries control for this, and the only way for the half cells to receive/dispose of electrons is through the external circuit.


Predicting color of a TMB derivative affected by peroxidase The ELISA test on HIV works by verifying the presence of peroxidase within samples. The verification process is done through the enzymatic reaction that said peroxidase has on TMB (3,3',5,5'-tetramethylbenzidine) in an environment that contains $\ce{H2O2}$. This reaction leaves the enzyme untouched and transforms the $\ce{H2O2}$ into $\ce{2H2O}$ by adding two hydrogen atoms taken from the TMB. This also transforms the TMB into another compound, 4-(4-imino-3,5-dimethylcyclohexa-2,5-dien-1-yl)-2,6-dimethylcyclohexa-2,5-dien-1-imine, which has a blue color in that specific case.

Therefore my question is, how could one predict from the structure of the one product that it is going to have a blue color in that situation? I know that it probably comes from the conjugated double bonds but how do their count relate to the absorbed wavelength(s)?

Your question: How could one predict from the structure of the one product that it is going to have a blue color in that situation?

This is a very broad question. OP has to go to chemistry fundamentals to learn, how one can predict the absorbance wavelength of a compound by analyzing the structure of the given compound. One easy subject for this study for OP is basic fundamentals of UV-visible spectroscopy in organic chemistry. For example, how absorbance wavelength increase with each additional conjugated double bond (Absorption in the ultraviolet and visible regions is related to the transition of electrons). Look following table:

$$
\begin{array}{c|lcr}
\text{Substance} & \text{Absorption Peak }(\lambda_{max}) & \text{Molar Absorption Coefficient }(\epsilon) \\\hline
\ce{CH2=CH2} & \pu{180nm} & 10000  \\
\ce{CH2=CH-CH=CH2} & \pu{217nm} & 21000  \\
\text{Vitamin A (5 conjugate double bonds)} & \pu{328nm} & 51000  \\
\text{Benzene} & \pu{255nm} & 180  \\
\text{Naphthalene} & \pu{286nm} & 360  \\
\text{Anthracene} & \pu{325nm} & 7100  \\
\hline
\end{array}
$$

Thus, one can suggest diimine, 4-(4-imino-3,5-dimethylcyclohexa-2,5-dien-1-yl)-2,6-dimethylcyclohexa-2,5-dien-1-imine, which contains 5 directly conjugated double bonds (two of them are imines), may absorb wavelengths of lights way larger than $\pu{328  nm}$ given for Vitamin A. That's because it involves transition of $n \rightarrow \pi^*$ in addition to $\pi \rightarrow \pi^*$ in Vitamin A (e.g., $\pi \rightarrow \pi^*$ in acetone is $\pu{195 nm}$ in addition to $n \rightarrow \pi^*$ transition at $\pu{274 nm}$, when compared to $\pi \rightarrow \pi^*$ transition of ethylene at $\pu{180 nm}$). Experimental evidence has shown that 4-(4-imino-3,5-dimethylcyclohexa-2,5-dien-1-yl)-2,6-dimethylcyclohexa-2,5-dien-1-imine absorbs light at $\pu{450 nm}$, giving its characteristic complimentary color $\color{blue}{\text{Blue}}$ (Ref.1).

References:

P. D. Josephy, T. Eling, R. P. Mason, “The horseradish peroxidase-catalyzed oxidation of 3,5,3',5'-tetramethylbenzidine. Free radical and charge-transfer complex intermediates,” The Journal of Biological Chemistry 1982, 257, 3669-3675 (http://www.jbc.org/content/257/7/3669).
For basic learning, read: Master Organic Chemistry.


Find the structural compound from carbon and proton NMR spectrum The proton and carbon NMR spectrum datas for compound W (201 g/mol) are
shown below. Based on Diagram 1 and Diagram 2, draw the structural formula of compound W according to the type of hydrogen and carbon.

I have extracted the information from the NMR spectrum into a table.

Interesting Problem and quite a nailbiter.
I will try to sort some of my ideas, even thought this won't be a full answer.

This looks like a question from an exam. Did you have any reactions during study with hydrochloric acid and propanoyl chloride (acid chloride of propionic acid). So maybe look into different reactions of acid chlorides.
As @MethewMahindaratne mentioned, it is hard without integration but for carbon we dont need integration. We have 7 Carbons, some hydrogen, pretty sure some oxygen and for sure some nitrogen. Why? Because only nitrogen is able to produce odd numbered masses. If we have CH compound, there will be always an even number of hydrogens and carbon has a mass of 12  so overall its even. When we create double bonds or add oxygen, the amount of hydrogen is decreased by 2 so it remains uneven. Only with nitrogen, we can add 1 hydrogen per nitrogen, therefore we have 1 or 3 or 5 etc. nitrogens in it.
The proton g is also quite specific. This high shift in proton NMR usally comes from a carboxylic group. So we have $C_7O_2H_xN_y$.The idea of the carboxylic group is also supported by the C atom at 175 which is a really high shift.
With a molecular mass of 201 g/mol this leaves quite some but not endless options. We can take of 7 x 12 (for the seven carbons) and 2 x 16 (for oxygen carboxylic group) and 1 (for hydrogen carboxylic group) leaving us with a mass of 84 split along oxygen, hydrogen and nitrogen.
This is a lot. On the bottom line it is mentioned, that W consists of certain things. halogenides which usally make compounds heavy are not included there.
Appart from the carboxylic group we are left with 6 carbons which could give the idea of a benzene. But I think it is not benzene because we could only attach $NH_2$ or $OH$ groups onto it and even with 5 $NH_2$ groups we get not enought mass. In addition, aromatic C atoms have high shifts usally 120-150. But we only have 3 others (next to our carbon from the carboxylic group) so not enought for a six membered ring.
But also not enought for just one alkene which I would prefer. Alkene shift ist around 110-140 and fits well.
Now that is all I could provide without crazy speculations.

I can throw in some suggestions but they are realy wild and you need to look in deeper.

The big peek a looks a bit like some alkohols I worked with so maybe a hydroxylic group?
Maybe Carbon Atom u contains 2 or even 3 peaks? (the line looks thicker than the others)= that would contribute to the idea of a substituted benzene ring and provide more carbons so we could add other groups.
Could $H_f$ be two dupletts instead of a quartett so that could come from a double bond?
With integrals you could look at the singuletts and determine wheather it is an amine (singulett with an integrall of 2)
Maybe my whole carbon acid idea is wrong and it is an ester or something.
If we are allowed to include hallogens, then it is pretty sure a benzene with a carboxylic group maybe hydroxylic and a hallogen atom.
If you have more information, maybe you can use the sdbs data base https://sdbs.db.aist.go.jp
I was not able to find an entry that fits.

Hope this brings the project forward.I am very happy to hear, what the compound was when you get a result.


Can gold cause tin or other metals to corrode? I stumbled upon the following article while trying to identify some old RAM modules:

https://www.techrepublic.com/article/quickly-identify-ram-chips-with-these-tips/

In the article, it mentions the following about gold contacts:

  A note about contact metals

  Chip contacts are either gold- or tin-plated. I recommend chips with
  gold contacts because they don’t corrode. Also, never insert
  tin-plated chips into gold-plated sockets. Different metals will
  increase the likelihood of corrosion.

I don't understand how gold cannot corrode on it's own, yet cause another dissimilar metal such as tin to corrode.  Is this article correct?

First off, read carefully:

  Different metals will increase the likelihood of corrosion.

It doesn't say:

  Different metals will increase the likelihood of corrosion of gold

With that in mind, it's called galvanic corrosion and occurs when contact between two (or more) dissimilar metals causes the more reactive one to corrode when it normally would not have or faster than it would on its own. This action is similar to a galvanic pile (though its only one cell) that has short circuited thus corroding the more reactive metal.

In your case it is definately the tin that will be corroded, not the gold as tin is more reactive.


Why is 2-ethylpentane is not a valid IUPAC name, but 3-ethylpentane is? I'm currently writing all the names of alkanes having molecular formula $\ce{C7H16}$. I wish to ask why 2-ethylpentane is not a valid IUPAC name, but 3-ethylpentane is?

Due to IUPAC conventions, the main alkane name is based on its longest chain. Here is an image showing different chain lengths for both alkanes. The numbers denote the longest alkane chain:

As you can see, 3-ethylpentane is symmetrical; no matter what chain you count along, the longest chain will always be 5 carbons long (pentane), and the ethyl group will always be on the third carbon (3-ethyl). Together, this is 3-ethylpentane.

However, if you look at '2-ethylpentane', there's a new longest chain which isn't 5 carbons long anymore. The image shows a new chain which is actually 6 carbons long (hexane), and there's now a methyl group on the third carbon (3-methyl). Together, this is 3-methylhexane. Because of this, '2-ethylpentane' isn't a real structure, since drawing it will leave you with 3-methylhexane as shown in the image.


Why does the degree of dissociation change when we dilute a weak acid even though the equilibrium constant is constant? $K$ represents the ratio of concentrations of molecules in a solution at equilibrium, which means that $Q_\mathrm{r}$ (that ratio at any given point) looks to be identical to $K$. In other words, the molecules in that solution react accordingly so that they reach equilibrium and the ratio of their concentrations is equal to $K$.

If $K$ is large enough (bigger than $10^4$ in my curriculum), this means that the the concentration of the reactants is almost zero. In other words, the  equilibrium position of that solution looks very much like a reaction that went to completion.

The more we dilute an acidic/basic solution, the higher the degree of dissociation, even though $K$ stays the same. So, does that mean that the more we dilute a solution the harder it is for it to reach the point of equilibrium for that specific molecule/solution or what?

For instance, say you found $K$ of solution to be $10^{-5}$. This means that when the reaction happens there are lots of reactants left, and not much products produced, which means that the degree of dissociation is low. But the more we dilute a solution, the closer it gets to a "complete reaction" (if you pour a small amount of weak-acid molecules into a large tank of water, it's certain that all of the weak-acid molecules are going to react with the water, i.e. the degree of dissociation approaches $100\%$).

So, how come $K$ can be independent of the initial reactants concentrations, and tell if a reaction was complete or not, when the "completion" of a reaction (the degree of dissociation) depends on the initial concentrations of reactants?

See my comment above.

As a numerical example, take acetic acid ($\ce{AcOH}$), which has $K_\mathrm{a} = 1.8 \cdot 10^{-5}$.

This means that:

$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]}$$

And the total nominal concentration of acid is:

$$C_\mathrm{a} = [\ce{AcOH}] + [\ce{AcO-}]$$

Combining these two equations, you can see that the % of dissociated acid is:

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} = \frac{K_\mathrm{a}}{K_\mathrm{a} + [\ce{H+}]}$$

i.e. not a constant, but a value that depends on the $\mathrm{pH}$. So no, there is no need for the 'degree of completion' of the reaction to be a constant.

You can even calculate the explicit concentrations of all the species.

Given that:

$$C_\mathrm{a} = \left(\frac{1}{[\ce{H+}]} + \frac{1}{K_\mathrm{a}}\right) \cdot ( [\ce{H+}]^2 - K_\mathrm{w}^2)$$

you can see that, for $\mathrm{pH} = 5$ you need:

$$C_\mathrm{a} = \left(\frac{1}{10^{-5}} + \frac{1}{1.8 \cdot 10^{-5}}\right) \cdot \left((10^{-5})^2 - (10^{-14})^2\right) = 1.5554 \cdot 10^{-5}$$

Then:

$$[\ce{AcOH}] = \frac{C_\mathrm{a} \cdot [\ce{H+}]}{K_\mathrm{a} + [\ce{H+}]} = 5.555 \cdot 10^{-6}$$

$$[\ce{AcO-}] = C_\mathrm{a} - [\ce{AcOH}] = 9.999 \cdot 10^{-6}$$

So the % of dissociated $\ce{AcOH}$ is:

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} \approx 64 \%$$

Now repeat for $\mathrm{pH} = 3$:

$$C_\mathrm{a} = \left(\frac{1}{10^{-3}} + \frac{1}{1.8 \cdot 10^{-5}}\right) \cdot \left((10^{-3})^2 - (10^{-14})^2\right) \approx 0.05656$$

$$[\ce{AcOH}] = \frac{C_\mathrm{a} \cdot [\ce{H+}]}{K_\mathrm{a} + [\ce{H+}]} \approx 0.05556$$

$$[\ce{AcO-}] = C_\mathrm{a} - [\ce{AcOH}] \approx 0.001$$

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} \approx 1.8 \%$$

So you see, in a more dilute mixture the $\mathrm{pH}$ is higher, and the dissociation reaction of acetic acid is 'more complete', whereas at higher concentration the pH is lower, and there is proportionally less acetate anion.

All of these systems are at equilibrium.


How to set up an equation to solve a rate law computationally? I am confused on how to set up my equations to use in the solver. I have an equation
$$\ce{A + 2B -> C}$$
As a given, it is first order in A and second order in B. What I need to do is to solve for the concentrations of all three of $\ce{A}$, $\ce{B}$ and $\ce{C}$ using an ordinary differential equation (computationally in Matlab). The rate constant is given as $\pu{8.5E-4}$. I am given all initial concentrations ($\pu{1.2 mol L-1}$ for both).

To start off I have
$$r = k\ce{[A][B]2}$$
I have no idea where I should take this next, I feel completely lost here.

The second part of this is to do the same thing, but instead of being irreversible, the reaction is given as reversible.
$$\ce{A + 2B <=> C}$$
The concentrations are all given as are the forward and backward rate constants. Its first order in $\ce{A}$, 2nd in $\ce{B}$, and first in $\ce{C}$. $\ce{[A]} = \pu{1.2 mol L-1}$, $\ce{[B]} = \pu{1.8 mol L-1}$, $\ce{[C]} = \pu{0.4 mol L-1}$, $k_\ce{for} = \pu{3.4E-2}$, $k_\ce{back} = \pu{1.2E-2}$.

Again my initial thoughts are
$$r = \frac{k_\ce{for}}{k_{back}} \times \frac{\ce{[A][B]^{2}}}{\ce{[C]}}$$
But again I just don't know where to go next.

I hope I made my question understandable, if not I'll try to clarify anything. Any and all help would be appreciated, I really need it!

It should be set up as a series of first order ordinary differential equations like so: $$\frac{d[A]}{dt}=-k_f[A][B]^2+k_r[C]$$$$\frac{d[B]}{dt}=-2k_f[A][B]^2+2k_r[C]$$
$$\frac{d[C]}{dt}=k_f[A][B]^2-k_r[C]$$
If we substitute $[C]=[C]_0+x$, $[A]=[A]_0-x$, and $[B]=[B]_0-2x$ into these equations, they all reduce to the single equation:$$\frac{dx}{dt}=k_f([A]_0-x)([B]_0-2x)^2-k_r([C]_0+x)$$


Alkene reaction with acid via SN1 reaction I want to ask a question about alkenes and their reactivity with acids.

From high school studies, I recall that if you had an acid $\ce{H3O+}$ i.e. $H^+$ that it would react with an alkene as shown below:

However, when I was given a lecture today on Allylic alcohols in a $\ce{S_{N}1}$ reaction, the acid does not react with the alkene at all:

As you may understand, attempting to google or find an equivalent mechanism to explain this observation didn't yield a successful result.

The two mechanisms in my opinion appear to contradict each other, but I couldn't work out how the second mechanism for the Allylic Alcohol functioned whereby the alkene did not react with the acid. My assumption was that it was easier to protonate the alcohol group than react with the acid (with maybe a reasoning along the lines of pKa) but I couldn't find anything sufficient to justify my idea.

Why does the acid not react with the alkene group in the allylic alcohol shown above?

My suspicion is that the resonance structures accessible to the allylic carbocation provide a lower energy state than the formation of the C--H single bond and the isolated carbocation that your expected pathway would require.

In order of reactivities, you are right to think that the alcohol is the controlling factor because of its pKa. Under acidic conditions it will behave more as a base than the alkene (It is much more basic than an alkene, or think of the Oxygen in an alcohol as being generally a stronger nucleophile than the alkene). This was one main consideration my O-Chem professor stressed was for us to be aware of how other functionalities in our molecules would behave under the reaction conditions of other steps (Eg, if going into a strongly acidic/basic environment after installing a reactive group). I am not an organic chemist by training, but I suspect that if you wanted to access the alkene reactivity, you would want some kind of alcohol protecting group.

This page has a brief discussion about three classes of nucleophile (and the next page describes what makes a good nucleophile): https://www.masterorganicchemistry.com/2011/03/04/the-three-classes-of-nucleophiles/


Mechanism of three-membered lactone hydrolysis in base I was reading up on the participation of neighboring functional groups and in the reaction below, I was wondering why hydroxide attacks the C–O bond instead of the carbonyl group of the lactone. Lactones tend to be more electrophilic compared to acyclic esters, as the lone pairs on the ether oxygen are not positioned correctly to participate in conjugation, so I would expect hydroxide to attack the carbonyl. Does it have to do with the angle of attack, given the small ring?

According to Organic Chemistry: Second Edition,[1, pg 934](from which I assume that the question was taken):

Enantiomerically pure (R)-2-bromopropanoic acid reacts with concentrated sodium hydroxide to give (S)-lactic acid. The reaction goes with inversion and is a typical SN2 reaction—and a good one too, since the reaction centre is adjacent to a carbonyl group. If, on the other hand, the reaction is run using $\ce{Ag2O}$ and a low concentration of sodium hydroxide, (R)-lactic acid is obtained — there is overall retention of stereochemistry.

Therefore this reaction that has been questioned takes place at a low concentration of $\ce{OH-}$ in the presence of $\ce{Ag2O}$.
How do we decide between competing mechanisms?
The next paragraph clearly outlines what we need to look for in order to proceed.

In the first, we have conditions suited to an SN2 reaction: a very good nucleophile ($\ce{OH-}$) and a good leaving group ($\ce{Br-}$). Improve the leaving group by adding $\ce{Ag+}$ ($\ce{Ag+}$ assists $\ce{Br-}$'s departure much as $\ce{H+}$ assists the departure of $\ce{OH-}$ by allowing it to leave as $\ce{H2O}$) and worsen the nucleophile ($\ce{H2O}$ instead of $\ce{OH-}$, of which there is now only a low concentration), and we have the sorts of conditions that would favour an SN1 reaction. The trouble is, without neighbouring group participation, the cation here would be rather unstable — right next to a carbonyl group. The carboxylate saves the day by participating in the departure of the $\ce{Br-}$ and forming the lactone. The key thing to remember is that a reaction always goes by the mechanism with the fastest rate.

From what I have understood the question lies in the step after the lactone is formed. What reacts, $\ce{C-1}$ or $\ce{C-2}$? Let's assume that both are possible and proceed with the two possible cases.
Firstly, the $\ce{OH-}$ attacks the carboxylic carbon($\ce{C-1}$). We get the possible reaction mechanism to be:

In the second case, let's assume that $\ce{OH-}$ attacks $\ce{C-2}$. The possible reaction mechanism would be:

So, now we are faced with the choice of which reaction could take place, attack on
$\ce{C-1}$ or attack on $\ce{C-2}$.
From the second quotation, we can say that if we are able to decide which is faster, that would be the reaction that takes place and so we reach our next question.
Which reaction is faster?
In pathway 1, we are talking about the base hydrolysis of esters. However, esters are not very reactive and the concentration of base in this case is low as well.
In pathway 2, we are talking about an SN2 reaction taking place on the $\ce{C-2}$ carbon which removes the ring strain that is seen in the lactone. Therefore, this would be faster than base hydrolysis that takes place at $\ce{C-1}$
Another way of checking the speed of the reaction would be to use the principle of least motion, (as pointed out by Yusuf Hassan).
According to Advances in Physical Organic Chemistry, the
principle of least motion is defined as:

The principle of least motion, states that those elementary reactions will be favored that involve the least change in atomic position and electronic configuration.

Using this, we can compare the relative rates of the two reactions.
During the base hydrolysis of esters, the hybridisation of the $\ce{C-1}$ carbon changes from $\mathrm{sp^2}$ to $\ce{sp^3}$. This leads to a big change in the atomic arrangements since you go from a trigonal planar configuration to a tetrahedral arrangement.
In case of the SN2 reaction on the $\ce{C-2}$ carbon, the configuration as a whole doesn't change. It remains $\mathrm{sp^3}$ throughout the reaction, so there is less rearrangement taking place.
This implies that according to the principle of least motion, the SN2 reaction is favoured over the the basic hydrolysis of the lactone formed via the anchimeric effect.
Hence, we can say that the reaction that would take place would be the attack on $\ce{C-2}$ and not the basic hydrolysis of the lactone.
Reference:

Organic Chemistry: Second Edition; Jonathan Clayden, Nick Greeves, Stuart Warren, 2014; ISBN 978-0198728719


Stoichiometry of incomplete combustion Methane is burned with air in a continuous steady-state combustion reactor to yield a mixture of carbon monoxide, carbon dioxide and water.
The feed to the reactor contains: $\pu{7.8 mole \% }\ce{CH4}$, $\pu{19.4 mole \% } \ce{O2}$, $\pu{72.8 mole \% } \ce{N2}$. The conversion of methane is $90\%$ and the gas leaving the reactor contains a ratio of $\pu{8 mol}\,\ce{CO2}/\pu{1 mol}\,\ce{CO}$.

My attempt:

$\ce{2 CH4 + 3 O2 -> 2 CO + 4 H2O}$

$\ce{CH4 + 2 O2 —> CO2 + 2 H2O}$

Both of these reactions have $90\%$ conversion.
From the ratio, can I say that approximately $88.88\%$ of the methane in the feed produces $\ce{CO2}$ while $11.11\%$ produces $\ce{CO}$?

Complicated little rascal!
What we know:
1) The N2 is inert.
2) The oxygen is in excess.
3) 90% of the 7.8 mole % CH4 reacts; this is 7.02 mole % CH4.
4) This gives 8 moles CO2 per mole of CO, or 88.89% CO2 plus 11.11% CO.

The resultant output contains CO2: 7.02 mole % of the original CH4 x 88.89% = 6.24 mole % CO2.

The resultant output contains CO: 7.02 mole % of the original CH4 x 11.11% = 0.78 mole % CO.

The resultant output also contains 10% unreacted CH4. The 90% conversion refers to the starting material, not to the products. (Yield % would refer to one product.)

So the stoichiometry is 80% of the methane goes to CO2, 10% goes to CO, and 10% is unreacted.


How would one go about making a cheap freeze spray? I'm interested in making a liquid freeze spray, for a fun science project. My only experience with this comes in spraying inverting canned air dusters, which spray a -60F mist/liquid (liquid if you spray while the can is inverted). I'd like to make a spray that I can store and use. And if it can't be stored more than a few hours, then hopefully something easy to make on the spot.

My chemistry understanding is high-school level basic, so I do not have a working knowledge of chemistry. On one end of the spectrum, I know there is liquid nitrogen, but it's expensive and complex to store correctly (high-pressure canisters).

I found this list of cooling bath configurations. Could someone point out which configuration can be stored long term (months), stored cheaply (e.g. a plastic or thin metal thermos; not a heavy high-pressure fire-extinguisher container), and be a liquid when "sprayed" (assuming keeping conditions in our pressure and room temperature)? I'd like a spray temperature of less than -60F, so ice and water won't cut it.

I'm thinking of going with dry ice and > 90% concentration ethanol. I have two questions regarding this approach and a random question:

If you have a solution of half dry ice and half 99% ethanol, how can you keep it at a low temperature? Won't the solution slowly return to room temperature? My canned air dusters always spray out at the -60F; I'd like something like that if possible.
Can this solution be stored in a cheap plastic or thin metal container? Will there be some pressure buildup from the dry ice that will eventually crack or explode the container? Assuming a 10 oz container with a solution of half dry ice and half > 90% ethanol.
How are canned air dusters always able to spray their liquid solution (if inverted while sprayed) at -60F, even if they are stored for months at room temperature? Is it because the contents of the can are under pressure and that this pressure makes the liquid cold?

The reason the 'canned air' dusters and other aerosol-based things like anti-perspirant cans produce a cryogenic effect is from the decompression involved. They are stored at room temperature, but the can is pressurised.

There are two effects that happen when you spray out some aerosol.

Firstly, by the ideal gas law, $PV=nRT$, when you release the pressure (i.e. lower $P$), the temperature ($T$) also drops. (The increase in volume $V$ mostly compensates for this, though.)

Secondly, the contents of the can are stored under enough pressure to keep at least one of the parts of the mixture in a liquid state. When you release the pressure by spraying it out, the liquid boils and evaporates, and takes energy in the form of heat to do so. You'll notice that the contents of a canned air duster, if you hold them upside-down when you spray, include more than just air -- there's also a liquid in there which is designed to be mostly inert to electronics and to quickly evaporate just as described above when the pressure inside the can drops, maintaining the pressure needed to produce a gas jet. If you spray for an extended period of time, you'll normally notice a boiling sound coming from inside the can, and the can getting cold, as that liquid evaporates. (As a safety note, these substances are often flammable, so don't use these air dusters near naked flames or ignition sources unless you want fire.)

Most people don't have the equipment needed to produce this sort of thing at home.

None of the cooling baths in that list are designed to be left as-is for months at a time -- they're designed to be made up as and when needed, with occasional addition of liquid nitrogen or dry ice to keep them cool. While you could produce a cold pressurised spray on the spot using dry ice and ethanol, there's a high probability that any sort of robust switched-release mechanism could make the vessel you're using explode under the pressure of sublimating carbon dioxide. Also, unless you relied on a pressure source other than the carbon dioxide itself, your spray probably wouldn't have time to get very cold.

So, I don't recommend it.


Reaction of ethylacetoacetate with Iodine A "reaction map" in my lecture handout suggests that the product of this reaction is:

So, basically overall it just merges two ethylacetoacetate molecules together at the central carbon. What is the mechanism for this? My thoughts are to swap the acidic proton for an Iodine atom by making the enol (in acidic conditions to avoid iodoform reaction) and react with $I_2$. Then, the enol of another molecule of EAA substitutes the Iodine to make the product. However, I don't know whether this is $S_N1$ or $S_N2$. Typically alkylation is $S_N2$ (I don't know why, please explain if you can) but in this case its a bulky secondary substituent and the solvent is protic (acidic conditions). Surely it's $S_N1$, right?

Are you certain that the coupling itself isn't a radical reaction?

The bond energy of an ordinary $\ce{C-I}$ bond is in the range of $210\,\mathrm{kJ\cdot\,mol^{-1}}$, which isn't a hell of a lot. Moreover, the resulting radical of an 1,3-dicarbonyl compound is nicely stabilized.


Just how quantized are energy levels? One thing which has bothered me for a while about the idea of quantized energy levels is that it's clear just how quantized they are.

Let me give an example. If I was looking the energy some electromagnetic wave would need to an $\ce{O-H}$ in water from the 0th vibrational energy level to the 1st vibrational energy level, it would be around $\mathrm{3750~cm^{-1}}$. Now, for some electromagnetic wave to cruise by and give enough energy to make that transition, how exact must the energy of that wave be?

Meaning, does it need to have exactly $\mathrm{3750~cm^{-1}}$ of energy? Well, what if the transition was actually $\mathrm{3750.3~cm^{-1}}$. Then that wave doesn't have enough energy anymore, and the likelihood of some wave having exactly $\mathrm{3750.3~cm^{-1}}$ seems a lot less likely than just being very close to that value.

So what happens in actual transitions that the molecule only takes the exact amount of energy it needs? And what happens to whatever leftover energy there might be from a wave that had enough energy to excite this water to it's first vibrational energy level? Is there ever leftover energy like I just mentioned? Also, do two waves of say $\mathrm{1875~cm^{-1}}$ each ever meet up and excite this same transition? This to me also seems so unlikely because that would involve constructive interference exactly of the two waves.

First off, let's think of the light as photons, instead of waves. That makes it clear that yes, there are two photon and multi-photon processes. As you surmise, the selection rules for these are much more complicated than a single-photon process.

Your main question is about "how exact" an absorption needs to be. In reality, it depends on the system.

Atomic Absorption Spectra

For atoms, there are no vibrational or rotational transitions, only electronic (or perhaps nuclear). So atomic spectra are indeed very tight lines, e.g., the Rydberg formula.

Diatomic Molecules

Now you have one vibration and some rotational levels. The latter are important, since you can go up or down some rotational levels with a vibrational transition. (And if you have an electronic transition, you can usually go up vibrational and/or rotational levels.)

So the spectra is a bit more complex than single lines. For example you can do rotational-vibrational spectroscopy.

In this case, the individual transitions are still pretty sharp - not completely lines, but $0.3\:\mathrm{cm^{-1}}$ probably won't match any particular transition.

Larger Molecules

Now you have very small spacing in the rotational lines, so if you take a vibrational absorption spectra (e.g., IR spectra) you'll see that most molecules have some width to the absorption peaks, so $\pm 0.3\:\mathrm{cm^{-1}}$ will still fit within the peak, just not as strongly.


Why is 2-chloropropene achiral? Could someone explain to me why some alkenes containing at least one couple of different substituents at the two double-bonded C atoms (that is, some of those with at least two different substituents for both C atoms) are considered achiral? For instance, for 2-chloropropene or 1-bromo-2-chloro-2-fluoro-1-iodoethene, where (and what) in the structural formula of each of them is the plane of symmetry that makes it achiral or, in other words, what do their superposable mirror images look like? Thanks.

Any molecule that cannot be superimposed on its mirror image is described as being chiral. If the molecule is superimposable on its mirror image the object is called achiral.

For simple alkenes, the carbon atoms of the double bond and the four atoms bonded to them all lie in the same plane, which divides the molecule into two halves, which are mirror images of each other. This plane of symmetry ensures that the mirror image forms of the molecule are superimposable. Therefore, simple alkenes like 2-chloroprop-1-ene and 1-bromo-2-chloro-2-fluoro-1-iodoethene are achiral.

Nevertheless, owing to the barrier to rotation in alkenes, 1-bromo-2-chloro-2-fluoro-1-iodoethene exists as two (achiral) stereoisomers: (E)-1-bromo-2-chloro-2-fluoro-1-iodoethene and (Z)-1-bromo-2-chloro-2-fluoro-1-iodoethene.


Confusion with usage of heat capacity in Adiabatic expansion My book says, that if a system is expanding adiabatically then $dq=0$ and $U= q+w$, so $U=w$.

It also states that $$\delta U=(C_v)\cdot\delta T$$

Here $C_v$, is the molar heat capacity at constant volume. How can they just $C_v$, which is defined for a specific volume, in a system that is expanding, i.e. its volume is changing?

For an ideal gas, U is a function only of T (and not V).  So the derivative of U with respect to T is an ordinary derivative, and is independent of V.  So why do we call $C_v$ the heat capacity at constant volume, if, for an ideal gas, it doesn't even depend on volume?  It has to do with how we measure $C_v$ experimentally. If the volume is held constant, then the amount of work is zero and $$Q=\Delta U=C_v\Delta T$$  So, in the case of constant volume, we can measure $C_v$ by measuring the amount of heat added Q.  If the volume is not held constant, so that some work is done, we can't determine $C_v$ from the amount of heat added.


What does it mean to equilibrate a column in size exclusion chromatography and why is a buffer used? I'm reading a procedure on size exclusion chromatography and it says that the first step is to equilibrate the column with buffer. What does this mean?

You want to make sure that your protein does not get damaged in the separation process. The protein travels faster than the small solutes in its solution, and will encounter the small molecules that are on the column, i.e. it will encounter different ionic strength, pH etc. So before you apply the protein solution to the column, you make sure the column is treated with an appropriate solution.

The solution is often called the buffer because it contains pH-controlling buffer substance. You can use size exclusion chromatography to separate by size, but also to switch from one buffer (or more general: one set of small solutes) to another. This is called buffer exchange, and there are short thick columns for that purpose (for separation, you would use a long thin column).


Why does HNO2 not have resonance? A question on the 1996 AP Chemistry Free Response asks:

  The $\ce{N-O}$ bonds in the $\ce{NO2-}$ ion are equal in length, whereas they are unequal in $\ce{HNO2}$. Explain.

Since the nitrogen dioxide ion has resonance, the $\ce{N-O}$ bonds are equal as resonance is in reality a hybrid of all of the possible structures for a certain molecule. However, the answer then assumes that $\ce{HNO2}$ has no resonance but rather one single and one double bond ($\ce{N-O}$ and $\ce{N=O}$), and therefore not all of the $\ce{N-O}$ bonds are equal.

How is this possible? Why can $\ce{HNO2}$ not have 2 possible structures where the single and double bonds switch between the two oxygens?

Resonance is a remnant of valence bond theory, which is necessary because it is impossible to describe delocalised bonding within a localised bonding scheme. Resonance structures are not real. One Lewis structure describes one electronic configuration with fully localised electrons. For many molecules it is completely sufficient to describe the bonding situation with one such structure in a first order approximation. For many other molecules this is completely insufficient. For $\ce{NO2-}$ this is the case, hence you need a superposition of more than one structure to get closer to a qualitative description. This caveat is not necessary in molecular orbital theory, since this is designed to treat delocalisation.

Nitrite anion

One question therefore arises from the above: Are the $\ce{N-O}$ bond lengths equal because there is resonance, or is there resonance because the $\ce{N-O}$ bond lengths are equal?

This is obviously more of a philosophical question, but it shows the dilemma of the whole concept. The bonding of any given molecule is a result of many different factors; to name but a few: repulsion of the nuclei, repulsion between the electrons, attraction between electrons and nuclei.

For $\ce{NO2-}$ there is one key question, that gives us a very important clue as to why the bond lengths are equal: Why should one oxygen atom be different from the other? In fact the molecule is symmetric (C2v), therefore the bond lengths must be equal.
If you try to write a Lewis structure that respects this symmetry, you will end up with an electron sextet on nitrogen (and a formal charge of +2), which certainly can only be an insufficient approximation. If you write a Lewis structure with fewer formal charges, keeping symmetry, you will notice that nitrogen will have 10 electrons. That would mean that it would have to use d-orbitals, which simply are not available for nitrogen (more on that another time). There is another way to write a Lewis structure, however, this one has to break the symmetry, and you'll notice that there is a second one, which is equal. In order to keep the symmetry you introduce resonance, to state, that the "true" electronic structure is a superposition of (at least) those two configurations.

TL;DR: The $\ce{N-O}$ bonds in the nitrite anion are equal, because the oxygens are equal.

Nitrous acid

It is false to say that in $\ce{HONO}$ is no resonance. And it is also wrong to say that a lack of resonance makes the bonds unequal. If the solution manual claims this, it is wrong.

  Why can $\ce{HNO2}$ not have 2 possible structures where the single and double bonds switch between the two oxygens?

These are valid resonance structures of nitrous acid. You have to keep in mind though, that resonance does not change the geometry of a molecule (or molecular structure). The contribution of the Lewis structure with fewer formal charges to the total wave function will be much higher than the contribution of the other structure in this case. This is not a general statement, there are molecules where the charge separated Lewis structure has a higher contribution to the wave function.

By analogy to the anion case, there is one key question, that gives us a clue about the molecular structure: If you bind a proton to one oxygen, are both oxygen really equal anymore? In this case the symmetry of the molecule is reduced to Cs. Therefore the bonds, cannot be equal.
For this molecule you will have no problem writing a single Lewis structure that satisfies every imposed criteria.
To answer the initial question: The bonds in $\ce{HONO}$ are unequal, because the proton is bound to one of the oxygens. It is a very polar, but covalent bond. The electron density of the $\ce{N-O}$ bond is therefore shifted towards the $\ce{H-O}$ bond, and simultaneously reduced for the $\ce{N-O}$ bond when compared to the anion. Therefore the $\ce{HO-N}$ bond is longer than the $\ce{N=O}$ bond (I implied the most contributing Lewis structure here).

TL;DR: The $\ce{N-O}$ bonds in the nitrous acid molecule are unequal, because the oxygens are unequal.

There frequently is the notion of a most stable resonance structure. While this is very popular, it is completely wrong. I again refer to the question What is resonance, and are resonance structures real? Resonance is used to describe the electronic structure of a molecule, there is no change in the molecular structure allowed. It is necessary to understand, that only the description in terms of all possible resonance structures is in itself correct, i.e. the theoretical limit of valence bond theory, everything else is an approximation. However, not all resonance structures will have the same contribution to the wave function. Whenever someone speaks of the most stable resonance structure, it is likely that they mean the resonance structure with the highest contribution to the wave function. But even if you know what is meant, it is still wrong to speak of a most stable resonance structure, it must be corrected.

(I currently don't have the time to provide a full fledged answer, so instead I was trying to give some pointers in the right direction. I might come back to amend it whenever I have more time.)


Effects of Ionizing a Radioactive Isotope If you took your favorite radioactive isotope and ionized it using a strong electric field (or any other ionization method), thereby stripping it of an electron, what would happen when it decayed? For simplicity, assume this process takes place in a vacuum.

Would the resulting material also be ionized? (Main Question)

If the isotope decayed via beta decay, is there any chance the produced electron could take the place of the stripped electron?

There is no effect. Imagine you are demolishing a building with high explosives; does it matter whether the small window on the third floor was open or closed prior to the explosion? No, it doesn't. The building will go crashing down all the same.

I believe this is a correct metaphor of the relation between chemistry and radioactivity. Of course the product will be ionized. Quite probably, it will be multiply ionized via Auger events, and also will certainly ionize quite a few other atoms around. Whether it was ionized before is about as unimportant as that small window on the third floor.


How to name alicyclic compound with hydroxymethyl substituent? What is the IUPAC name of the given compound?

Is it cyclohexane dimethanol (or) 1,2-dihydroxymethyl cyclohexane?

The principal characteristic group of the compound that is given in the question is $\ce{-OH}$, which is expressed as suffix (‘-ol’). Thus, the senior parent structure of this compound is methanol. The problem in this case is that the compound contains two separate methanol structures.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical senior parent structures that are linked by di- or polyvalent groups can be named by multiplicative nomenclature if certain conditions are met (otherwise the usual substitutive nomenclature is used).

  P-45.1 MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES

  P-45.1.1 Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).
  (1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and
  (2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
  (3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.

  When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.

The given structure fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.

  P-15.3.1.3 Multiplicative name formation
  Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.
  When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:

  (a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);

Here: omitted (The locant 1 is the only available locant in methanol.)

  (b) the name of the linking multiplicative substituent atom or group;

Here: cyclohexane-1,2-diyl

  (c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;

Here: di

  (d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).

Here: methanol

Therefore, the preferred IUPAC name is (cyclohexane-1,2-diyl)dimethanol.

By way of comparison, the (correct, but not preferred) name based on usual substitutive nomenclature is [2-(hydroxymethyl)cyclohexyl]methanol


How do I get the molarity of solution 1 given volume 1, molarity 2, and volume 2? For example:

  What is the molarity of a $\ce{NaOH}$ solution if $25\rm~mL$ of the solution is exactly neutralized by $44\rm~mL$ of $0.32\rm~M~HCl$?

I know that the abstract function is

$$\text{Given Volume of Known Molarity} \times \text{Molarity Factor} \times \text{Mole Ratio} \times \text{Per Volume Factor} \times \text{Conversion Factor Per Liter}$$

The reaction that occurs: H+ + OH- --> H2O

When you multiply the volume of HCl by the molarity of HCl you get the total amount of moles of H+ that reacted (mol/L x L = mol).
This means that the same amount of OH- was needed for the reaction (1:1).
Volume of NaOH is known and thus yields the answer.


What, if any, hydration states does sodium bicarbonate have? Sodium carbonate has a few hydrates in addition to the anhydrous form. I haven't found any references online that assert any hydrates of sodium bicarbonate though. Do they simply not exist?

To my knowledge sodium hydrogen carbonate hydrates do not exist.

However, there exists a mixed hydrate of sodium carbonate and sodium hydrogen carbonate with the formula $\ce{Na2CO3 \cdot NaHCO3 \cdot 2H2O}$ (dihydrate of trisodium hydrogendicarbonate, $\ce{Na3H(CO3)2}$) that occurs naturally as the mineral trona.

The region labeled trona in the phase diagram below shows the conditions under which the hydrate can be crystallized from an aqueous solution of $\ce{Na2CO3}$ and $\ce{NaHCO3}$.

The picture is taken from US patent US7638109B2.


How do Precipitation Reactions behave in the Absence of Gravity? How do precipitation reactions behave in the absence of gravity, say on the International Space Station (ISS)? I have seen water taking the shape of a sphere and not that of the container in space due to its surface tension. It takes a spherical shape in order to minimize the energy due to forces of surface tension. In space all directions are equivalent and hence there is no notion of up or down.

In precipitation reactions, generally, the precipitate formed either floats on the surface or settles down. I know the solid particles will be formed in space due to the chemical reaction between the reactants. What will happen to the solid particles (precipitate), will they concentrate at the centre of the sphere of the reaction mixture (solution) or will they come to the surface, or will they remain suspended, or is there any other possibility? Has such an experiment been done on the ISS to date?

For reference, this is how water looks in the absence of gravity:

Please note: Water is coloured to enhance visibility

In gravityless environment, all directions are equivalent save the very vicinity of the surface(*), where just few water molecules are toward the surface.

Therefore, in the bulk volume, the precipitate would stay where it is, unless some currents exist for whatever  reason.

At the very surface, it depends on nature of precipitate.

If it is rather hydrophilic, water molecules would have tendency to surround it, so it would be rather pulled from surface inwards.

If it is rather hydrophobic, water molecules would have tendency to expelled it, so it would be rather pushed outwards to the surface.

If it is far enough ( tens molecules ) from surface, all effects will get spherically symmetric, but one....

(*)  .....  as in a very very long term, precipitate denser than water would be more concentrated near the centre by the own gravity of the "bubble" and vice versa.

I am not aware of such an ISS experiment, but it makes little sense to do it for me. In the microgravity context, it is very very long term experiment.


How do I remove tetrabutylammonium perchlorate from an organic synthesis? I am trying to remove tetrabutylammonium perchlorate (TBAP) from an organic synthesis where TBAP was used as an electrolyte. I am working with compounds containing the amide group. I have used flash column chromatography (FCC) (gradient) but this is laborious.

To give you a better understanding of the experiment this is the path I follow

Once the electrosynthesis of the compounds has completed, the solvents are removed (MeOH & MeCN) in vacuo.
Now for step two, I usually carry out FCC. But I have tried separation using Sodium hydroxide but this did not give me the desired result, the electrolyte was still present after NMR.

Any help would be appreciated.

It's tempting to use phase separation with an aqueous phase, but you're clearly having issues.  One option is to evaporate the volatile solvents off, add toluene, cyclohexane or heptane and try water or another additive to boost the ionic strength. The approach using NaOH suggests this may not be successful.

A less conventional approach may be to partition between two organic solvents e.g. acetonitrile or methanol with heptane or methylcyclohexane. The solubility of TBAP appears good enough in alcohols to dissolve well in that phase, while your amide seems more likely to have less polar solvent solubility. Your methanol-MeCN mix may convolute things,  so maybe try and strip off the methanol (and likely some MeCN will be lost, but you can top up) or if all else fails,  evaporate both solvents and redissolve from scratch.


What exchange timescales are needed to cause decoupling in NMR? I'll try to explain the source of my confusion as clear as possible:

(*) The methyl protons in toluene are decoupled because the C-C bond between the methyl group and benzene allows very fast rotation. Hence, on average, these protons will see the same environment.

It is said that the time scale for the rotation must be faster than the time scale of the instrument. For a 250 MHz apparatus this would be (1/250MHz)=4 nanoseconds.

(*) In impure solvent the methylene protons of ethanol don't couple to -OH,  since fast exchange of the OH protons results in decoupling and the protons will see an average environment.

Here it is said that this occurs if the timescale of exchange must be faster than the coupling constant J for the coupling. This is typically around 5-10 Hz. So in this case, if the exchange is in the 100 millisecond timescale decoupling already occurs.

This utterly confuses me, in the first case averaging requires nanosecond timescales (related to the frequency of the instrument) and in the second case almost second timescales (related to the small difference in frequency between coupled peaks)? Why?

ALSO and maybe related to this, why do we see in the spectrum of ethanol peak broadening for the OH proton, but not for the methylene group?

You need to differentiate what gets averaged, a CS difference or a scalar coupling.

Methyl protons have a different chemical shift depending on the current conformation, but they rotate so fast this gets averaged out perfectly, and they have a very sharp signal. They show no coupling between themselves because it´s an $AB_2$ system with CS difference zero, so $A_3$ actually.

$J$ coupling to a methyl group (e.g. the ethyl group in ethanol) gives sharp lines and clear splitting, because the coupling varies very rapidly (with the methyl rotation) around a stable and finite value, and it only depends on the occupation of the spin states in both groups.

The coupling methylene to OH is inverted every time the proton gets exchanged for one with different spin state, so it averages to zero. The exchange happens so often that the methylene group still gives a sharp quartett from the methyl.

The OH group coalesces with the water peak. Because the exchange happens randomly and differently for each molecule, the coherence in your spin ensemble deteriorates, hence the line broadening. Note: It´s not a splitting that get´s averaged out, but a change in chemical shift.

If the proton exchange was even faster, the line would sharpen again at some point I guess. Just like the methyl protons themselves. If the exchange was even slower (milliseconds), the line would broaden so much it became invisible. If you have a really dry sample, the exchange becomes so slow it does not happen on the NMR time scale, the line becomes sharp, and you even see the $^3J$ coupling to the methylene group.


Creation of ester from acetylene derivative I took a much different approach from the solutions manual. My synthesis:

1) H2 and Lindlar's catalyst to stop at the alkene.

2) Oxidative cleavage to form benzoic acid. Works here because we have a benzylic hydrogen.

3) Addition of excess allyl alcohol with catalytic concentration of acid and heat. A Fischer esterification step.

This is also one fewer step than what my solution's manual has. The solutions manual went up to the acid chloride and down to the ester, which is a nice step (avoids the heat). Does mine also work on paper?

No, your approach doesn't give the right product. There's a carbon missing between the benzene ring and the carbonyl carbon. See the difference in the products below.

I don't know exactly what your solutions manual suggested, but I've included a conservative approach through phenylacetaldehyde. As you suggest, it's not absolutely necessary to go through an acid chloride to make an ester from a carboxylic acid. The Fischer esterification will work. Also, for many carboxylic acid to ester conversions, you can simply react the carboxylic acid with thionyl chloride and the desired alcohol all in the reaction together, rendering it a one-step transformation.


Is there a reaction to get hydrogen from hydrocarbons? Combustion of hydrocarbons reveals hydrogen, but not sure at what efficiency. Is there a reaction to get hydrogen from hydrocarbons?

Hydrocarbons are in fact the main source of hydrogen gas. According to Wikipedia, 95% of all hydrogen is produced by the steam reforming reaction:

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

This is a very important industrial process which has been studied in depth. It is done at very high temperatures (~1000 °C) and under pressure (~20 atm), and can be "up to 75%
efficient" (though efficiency here may not be defined as solely reaction yield).

As an aside, the fact that so much hydrogen is obtained from methane (itself generally sourced from natural gas, a fossil fuel) is one reason why hydrogen fuel cells currently have limited environmental benefit, though there is a huge amount of research into more efficient techniques for obtaining hydrogen from water splitting (especially using light) which could be carbon-neutral.