How to determine the worst resonance structures out of a given set? So I was "happily" doing organic chemistry homework when I came across this question:

For 1-methoxy-1,3-butadiene, which of the following resonating structure is the least stable?
$$\begin{align}(\mathbf{a})\ &\ \ce{H2\overset{+}{C}-\overset{+}{C}H-CH=CH-O-CH3} & (\mathbf{b})\ &\ \ce{H2\overset{+}{C}-CH=CH-CH=\overset{+}{O}-CH3} \\
(\mathbf{c})\ &\ \ce{H2\overset{+}{C}-\overset{-}{C}H-\overset{+}{C}H-\overset{+}{C}H-O-CH3} & (\mathbf{d})\ &\ \ce{H2C=CH-\overset{+}{C}H-CH=\overset{+}{O}-CH3}\end{align}$$

I decided to take a look through my textbook to see what I was missing. The rules to compare resonance structure energies (and therefore determine relative contributions) were in my textbook.  It said:

The following structures are considered to contribute most:

Structures having filled octet a for second row elements (C, N, O, F) are stable.
Structures having minimum number of formal charges and maximum number of bonds.
Structure in which negative charge appears on the most electronegative atom (C < N < O).
Structure in which there is minimal charge separation while keeping the formal charges closer together.

Instead of answering my initial question, two more questions came to mind. They are (in order):

Why do the octets of only the second row elements have to be filled in for greater relative contribution?
I suppose it has something to do with the greater electronegativity values of the elements, but would still like some clarification.
From option (a) and (c) in my question, how did the two positive charges get next to each other?

Resonance structures, as stated in my textbook, can be represented by two or more Lewis structures that differ only in the positions of the electrons.
The options clearly do not just redistribute the electrons. Or at least I don't think so.
So if I had to guess, the answer would be (b) because the terminal carbocation is worse than an internal carbocation.
But it turns out, the correct answer is (c). And I think that it shouldn't even be considered a resonating structure.

(C) is technically a valid resonance structure. You can derive it from (B) by moving the electron pair from the double bond between the carbons 2 and 3 (numbering from the left: this may not be correct from a nomenclature standpoint, but I'm using it for this answer) onto carbon 2 and moving the electron pair from the double bond between the oxygen and the carbon onto the oxygen.

(C) is least stable because it is, simply put, absolutely disgusting. Charge separation (carbons two and three) is gross, you have two cations in close proximity (carbons three and four), a terminal carbocation (carbon 1), and three atoms with empty valence shells (carbons 1, 3, and 4). So while the resonance structure is valid enough for (C) to be considered a resonance structure, that resonance structure happens to be the spawn of the devil.

That said, the compound in the answers isn't the same as the compound named in the question, like ron said.

As to the first part of your question, the filled valence shells are more strictly necessary for the second row elements because they don't have d orbitals to confound the picture.

According to inductive effect OCH3 is electron withdrawing group and positive charge next to this carbon make it relatively least unstable .So c is correct answer.


Why do (did) they make trans fat instead of using saturated fat? When making chocolate unsaturated fat is used for a good melting point (and perhaps health reasons.)

To make chocolate edible it has to be a bit more solid, so other fats were required. Therefore they use trans fats, which are unsaturated fats made saturated in cis or trans form.

Why do they do that instead of just mix it with normal saturated fat? Why use trans fats?

Catalytic hydrogenation tends to make trans fats... unfortunately catalytic hydrogenation is the easiest (read lowest cost) way of hydrogenating fats so it became the industry preferred way of hardening fats before anyone realized there was a health risk to trans fats. Cheap plant oils can be hardened by hydrogenation... and hydrogenated plant oils are generally cheaper than naturally saturated fats.

Because they are unnatural, trans fats also don't go rancid so quickly... so goods made with trans fats can have longer shelf lives. Which cuts costs. Given so much of mass food production is severely cost conscious, it was inevitable that trans fats became widely used and are now having to be removed as ingredients.

Consider olive oil spread. Olive oil is composed largely of unsaturated fats which do not  "stack" easily onto each other at the molecular level, and thus are a liquid at room temperature. If you want to use it as a spread or in baking it is usually needed to be a solid. This is done in a process called Hydrogenation where the unsaturated molecules become saturated with hydrogen, thus as you stated making trans fat. This process is conducted in order to make often healthier vegetable oils (usually unsaturated) into solids at room temperature.


Resonance stabilized carbanion I am sure that the answer is (b) because I am familiar with the other rings and I think they are showing resonance (and it is a single choice question so it was easy to guess)

I wish to know why is (b) not showing resonance (because I think it can as far as I know) and if I am wrong about the answer then please correct me.

(b) does not show resonance because the second resonance structure (with negative charge at oxygen and double bond in the ring) violates Bredt's rule. The carbon $\ce{sp^3}$ orbital with the negative charge and the $\ce{p_z}$ orbital of the adjacent carbonyl C cannot overlap because they are not in plane, so the charge cannot be delocalized into the carbonyl group.

For any cyclic compound to exhibit resonance, the following rules have to be obeyed:

1) $(4n+2)\pi$ electrons.

2) Conjugation. Conjugation is not possible for either b) or d).

3) Planarity. Compound b) is not planar.

And of course, compound b) violates Bredt's rule as well.

Hence, neither b) nor d) can be resonance stabilised.


Method to know the number of isomers of metal complexes Whenever I try to find the number of geometrical isomers (including optical isomers) of coordination compound, I got confused, and mostly miss few isomers. Is there any standard method to know the number of isomers of compounds such as $\ce{Ma2b2c2}$ or $\ce{[M(A-A)a2b2]}$.

where $\ce{M}$ is metal and $\ce{a,b}$ are monodentate ligands while $\ce{A-A}$ is a bidentate ligand.

Do it many times over, so as not to get confused.

Alternatively, there is a thing called Polya's formula, but you won't be able to use it anyway. In trivial cases like this, the said formula is about 100 times more complicated than counting isomers by hand. It is not before polysubstituted fullerenes that its use in chemistry starts making any sense.

For Ma2b2c2, you could go by:

All cis (aCa) (bCb) (cCc)
All trans (aTa) (bTb) (cTc)
Only 1 trans (aTa)
Only 1 trans (bTb)
Only 1 trans (cTc)

Similar cases could be done for M(AA)b2c2, except of the fact that, (AA) will remain always cis [all cases EXCEPT where the 2 donor atoms are connected by NOT-A-VERY-LONG-CHAIN]


Redox reaction with tin and sulfuric acid solution I need to write the redox reaction from tin and a sulfuric acid solution. Here is the solution given:

  The first half-reaction: $\ce{Sn->Sn^2+\+2e^-}$
  The second half-reaction: $\ce{2H+\+2e^-->H2}$
  Adding both reactions will give you: $\ce{Sn\+2H+->Sn^2+\+H2}$

Why is the sulfuric acid solution replaced with $\ce{H^+}$?

The sulfuric acid dissociates in water: $$\ce{H2SO4 -> H+ + HSO4-}$$$$\ce{HSO4- -> H+ + SO4^{2-}}$$

When writing redox equations the spectator ions are left out because they do not participate in the redox reaction.

In this case it is the hydrogen ions which are oxidising the tin and the sulfate and hydrogensulfate ions play no part in the reaction so they are left out. Including them in the hydrogen half equation would give (with sulfate ions): $$\ce{2H+ + SO4^{2-} +2e- -> H2 + SO4^{2-}}$$

Clearly the sulfate ions are unnecessary and so they are omitted from the equation.

I doubt the answer because I'm sure you'll get $\ce{Sn_{(s)} + 2H2SO4_{(aq)} → SnSO4_{(aq)} + 2H2O_{(l)} + SO2_{(g)}}$ as your Products

anyways , The Arrhenius concept of acids and bases is as follows: an acid is a substance that when dissolved in water increases the concentration of the hydrogen ion,  $$\ce{H+}$$
When Sulphuric acid is dissolved in water it gives you $$\ce{H+}$$and $$\ce{SO4^{2-}}$$ ions.
Sulphuric acid is a VERY Strong Oxidising agent, that means that it will oxidise  your Tin
therefore we get  $\text{Sn}\rightarrow \text{Sn}^{2+}+2\text{ e}^-$
in the second part I think you have reacted your 2 electrons formed in the previous reaction with Hydrogen ion, But I doubt this part


How much CO2 could you remove from the atmosphere with 2 trillion USD? I know it's possible to capture $\ce{CO2}$ with various chemical reactions. For example NASA's space shuttle had some kind of regenerative $\ce{CO2}$ scrubber. But how expensive is it? Could a huge number of these devices, or something like it, significantly reduce the atmosphere's $\ce{CO2}$ level?

I realize this is probably not practical, but it's an interesting thought experiment. To make my question more specific: can anyone show some rough calculations of how much $\ce{CO2}$ you could remove if you had, say, 2 trillion USD to spend? To simplify, assume that energy is at current prices, but clean. (i.e., if the device requires electricity, assume it's coming from a nuclear power plant, solar farm, etc.).

This is very hard to answer precisely, as there are many different carbon capture strategies, and economics at the scale required is quite different from our normal understanding. However, I'd love to see some attempts to at least get order of magnitude estimates, or sources with more in-depth analyses.

Here is an implementation of carbon capture and storage which is useful to consider for its simplicity rather than its real-world applicability, to give a sense of scale. One way to remove anthropogenic $\ce{CO2}$ is to do "inverse combustion", more specifically:

$$\ce{CO2(g) -> C(s) + O2(g)}\ \ \ \ \ \ \ \mathrm{\Delta H=+390\ kJ/mol}$$

Assume this process can be done with perfect efficiency, and that the only energy expense in the process is to drive the reaction forwards (that is, zero energy consumed in transportation, collection, construction, etc). According to this source, the amount of anthropogenic $\ce{CO2}$ emissions from 1750 to 2008 has totalled about $1250\times 10^9\ \mathrm{t_{CO_2}}$. Suppose you wish to remove all this carbon dioxide (only about half is in the atmosphere, the rest is trapped in the ocean or land) using the above process. This would require about $\mathrm{10^{22}\ J}$ of energy, which Wolfram Alpha suggests is about 50% more energy than can be retrieved from combustion of all global proven oil reserves in 2003. Put another way, this is about 20 times the world energy consumption in 2012, or 150 times world electrical energy production. This would put the cost of this process in the range of hundreds of trillions of US dollars, meaning two trillion USD barely makes a dent.

Is there some other process where two trillion USD is close to enough? I seriously doubt it.

Edit: Several comments and answers mention biological sequestration, which is a legitimate carbon capture strategy. I did not consider it, however, because its costs are far more complicated to calculate. My intention with this answer was to find a quick and comparatively simple way to attach an energy cost to carbon sequestration. Whether the monetary cost even makes sense at this scale (how do you define "monetary cost" when it's larger than world GDP?), I don't know.

But here's another amusing comparison, which should somewhat temper hopes that biological sequestration will be a magic bullet. The amount of anthropogenic carbon released between 1750 and 2008, ~350 billion metric tons of carbon, is comparable to a significant amount of the biomass on Earth (all eukaryotic life contains approximately 560 billion metric tons of carbon, multicellular life is a fraction of this). Thus, biological sequestration of the majority of anthropogenic carbon would be broadly equivalent to sacrificing all eukaryotic life on Earth (or ~10-30% of all living organisms by mass) in order to collect and bury carbon, then seeding the Earth back to its current biological state.

Over 250 years of constant stimulation to produce as much energy/goods from fossil fuels, we have released a lot of carbon.

the $64,000 Question is ... Is it economically feasible to use a 'synthetic' process over a naturally occurring one. In this case, No. Zooplankton are the prime movers of CO2 in our atmosphere. They basically turn gaseous CO2 into Calcium Carbonate a lab phrase for sea shells. Increasing the number of Zooplankton is key to the process. UV and oil slicks and the Sargasso seas of plastic are lowering those numbers.

so the answer to the question is what process would one employ that was the most cost effective solution to the problem, clearly A. clean up the oceans and promote natural zooplankton growth or B. mimic the metabolic pathway that results in the production of calcium carbonate, as a GM/organic solution. Either A or B is a far better and sustainable solution that does not reintroduce/produce CO2 for the energy to power an artificial non-organic/biologic method.


Hyperconjugation in ethane conformers I refer you to this set of presentation slides. On slide 8, the author explains the preference for the staggered conformation of ethane by saying that there is a favourable hyperconjugative interaction between $\sigma^{}_{\ce{C-H}}$ of the $\ce{C-H}$ bond in one methyl group and $\sigma^{*}_{\ce{C-H}}$ of another $\ce{C-H}$ bond in the other methyl group.

However, I am puzzled by the diagram. Since both $\ce{C-H}$ bonds are similar in terms of bonding, their bonding and antibonding $\sigma$ molecular orbitals should have the same energies. Instead, the diagram shows that one of the $\sigma_{\ce{C-H}}$ bonds is lower than the other in energy. How can this be right? I have shown a screenshot of the diagram below.

Also, how can a $\sigma_{\ce{C-H}}$ donate electron density to a $\sigma^*_{\ce{C-H}}$ effectively? After all, the energy difference between them is rather significant since the orbital interaction between the atomic orbitals of $\ce{C}$ and $\ce{H}$ is large.

I am not entirely sure what purpose the lower sigma bonding orbital in the MO-diagram serves, because an ethane molecular orbital would, as you have surmised correctly, have six degenerate $\sigma_{\ce{C-H}}$ MOs and a $\sigma_{\ce{C-C}}$ MO, lower in energy.

My best guess would be the position of the $\sigma_{\ce{C-H}}$ on the right hand is quite arbitrary and the basic purpose of the diagram is to show the possible overlap between the $\sigma_{\ce{C-H}}$ of one $\ce{C-H}$ bond and the $\sigma^{*}_{\ce{C-H}}$ of $\ce{C-H}$ bond anti to it - you don't really need to be worried about the $\sigma_{\ce{C-H}}$ of that anti bond because it cannot overlap with the other $\sigma^{}_{\ce{C-H}}$.

That brings us to the second part of your query. A $\sigma_{\ce{C-H}}$ can donate electron density to an empty $\sigma^{*}_{\ce{C-H}}$ anti to it, because the latter is oriented correctly (anti) to accept the electron density from it. It's a question of orbital symmetry and not energy difference.

Just to build on the very good answer from Sagnik, the purpose of the σC-H bond on the right side of the diagram is to show that the C-H bond is destabilized by electron density in its antibonding orbital. But the energy of the overall system is reduced.

The red orbital structure isn't in the Angew. paper either.


Relative rate of solvolysis I underestimated this question and just compared the carbocation stability of all the compounds (after removal of chloride via $S_N1$) and got the order D>C>B>A... but my answer is incorrect, what am I missing out?

The rate of solvolysis is highest in (C) as a 6 membered ring is formed which is more stable than 4 & 5 membered rings.

And (B) more readily formed than (D) because the NGP ability of oxygen is more than that of an alkene(due to high charge density of oxygen).

So the correct order is C>B>D>A.


Why are lithides not known? In the last few decades, many alkalides - anions of alkali metals - have been synthesised. The most famous is undoubtedly that of sodium: $\ce{[Na(\text{2.2.2-cryptand})]+Na-}$, but the alkalides $\ce{K-}$, $\ce{Rb-}$, and $\ce{Cs-}$ are all known. However, $\ce{Li-}$ is not.

James Dye writes in a 1984 review1 that for the product of the reaction

$$\ce{M(s) + N(s) + L(s) -> [ML]+N-(s)}$$

to be thermodynamically stabilised ($\ce{M}$, $\ce{N}$ are metals, and $\ce{L}$ is the macrocyclic ligand), several criteria must be met:

  Small lattice energies for $\ce{L(s)}$, $\ce{M(s)}$, and $\ce{N(s)}$ so that the enthalpies of sublimation will not be too large.
  Low ionization energy of $\ce{M}$.
  High electron affinity of $\ce{N}$.
  Large complexation energy of $\ce{M+}$ by $\ce{L}$.
  Large lattice energy of $\ce{[ML+]N-}$ (which depends mainly on the interionic separation).

From these given factors, I suppose the only possible explanation is the larger heat of sublimation of lithium. However, lithium is still known to form electrides in solution - compounds of the form $\ce{[ML+]e-}$. In order for these to be formed, the sublimation energy of $\ce{Li}$ still has to be overcome.

Is there a thermodynamic reason why lithides, $\ce{Li-}$, have not yet been made?

Reference

Dye, J. L. Electrides, negatively charged metal ions, and related phenomena. Prog. Inorg. Chem. 1984, 32, 327–441. DOI: 10.1002/9780470166338.ch4.

One barrier is the greater tendency of lithium to form partially covalent bonds than heavier alkali metals.  This is a good thing for organic chemists, since this stabilizes organolithium compounds and enhances the solubility of lithium chloride as a chloride ion source (see this answer).  But it's not so hot for forming lithides.  Covalent bonding to more electronegative elements, meaning to practically all other elements in this case, oxidizes the intended lithide ion.  In effect the cryptand in an intended lithide is converted into one of those organolithium compounds.

Linear Christmas, in the comments to the question, hints at a way to beat this tendency.  Two of the references given there, https://doi.org/10.1021/ja056314+ and https://doi.org/10.1002/cphc.201200805, involve theoretical investigations of lithides built with pyrrole-based cryptands.  Such cryptands resist covalent bonding to lithide (or other nucleophilic) anions because this would destroy the aromatic of the ring and require electrons to overlap into a high-lying, strongly antibonding orbital.  These compounds, or counterparts with heavier alkalide ions, draw interest for their expected nonlinear optical (NLO) properties, making them synthetic targets.  So, stay tuned.

Not too complicated of an answer - don't make a mountain out of a mole-hill. This is simple nuclear stability. You can just google "Nuclear Band of Stability" and it explains the necessary balance between protons, neutrons and electrons. Li-1 would have 3 protons, 4 neutrons, and 4 electrons. Crowded around such a small nucleus, the electrons would experience too much repulsion which would cause at least one to be ejected from the cloud. Additionally, 3 measly protons does not suffice for another electron, at a greater distance, experiencing electron repulsion, to experience an even greater force by effective nuclear charge.


Salt bridge in galvanic cell When we make a galvanic cell using zinc rod (in zinc sulphate solution) and copper rod (in copper sulphate solution), there is a flow of sulphate ions from the copper compartment to the zinc compartment through the salt bridge. Once the sulphate ions reach the zinc compartment, what happens to them?

According to my opinion i think you are wrong because the sulphate ion does not enter the other compartment. This is because the function of the salt bridge is to maintain electrical neutrality by either allowing the sulphate ion to reside in it or by allowing the cations inside the salt bridge to enter the compartment so as to balance the ions.


Improving Elephant Toothpaste experiment we are trying to perform the Elephant Toothpaste experiment involving the reaction of yeast and Hydrogen Peroxide for the students at school.

However, the results have been very underwhelming, likely because we have to use the 3% version of Hydrogen Peroxide.

My question is: What are some other ways to help create a stronger propulsion or thicker foam?

Some things I have been experimenting with are reducing the water mixed with the yeast, and swirling the bottle to mix the yeast and Hydrogen Peroxide. Any advice is much appreciated.

You might try horseradish as a catalyst, rather than yeast. Use the fresh root, not the prepared sauce, and test a few different plants - some are more effective than others.
Cosmetic supply stores stock 6% (20 volume) $\ce{H2O2}$, but that concentration is more likely to cause skin irritation or bleach fabrics.
The most effective catalyst for decomposing  $\ce{H2O2}$ that I've used is potassium permanganate, $\ce{KMnO4}$, but it has a few liabilities:

It's a strong oxidant and can start a fire when mixed with organic chemicals.
It stains skin and fabric anything from pinkish-purple to black. The stain can be removed with sodium bisulfite, $\ce{NaHSO3}$, sold as an iron-stain remover. That can be a fun demonstration but is not suitable for young students to use.

H2O2 can cause eye damage. When used with reasonable care, it is unlikely to cause anything but temporary skin bleaching - if kept away from the eyes. That and maybe some ruined clothing (or other surfaces). I'd not use 30% around children, although as a controlled demonstration -physically separated from the audience, it can be safely used. First thing you could do, depending on your resources, is to determine if your 3% is actually near 3%. There are tables which list its density, if you can weigh some out into a volumetric flask, say. If you have a chemical refrigerator, it should be stored cold (and of course, away from sunlight). I guess if it were me, I'd buy some fresh 30% and dilute it for the kids to use (prior to the session). I'd use potassium iodide rather than yeast (what is it that you're trying to teach?) and possibly consider FeCl3 (if the lesson is about catalysts) - although staining would be more of an issue with the iron salt. Make sure that the peroxide isn't phosphate stabilized (if possible). If you really want to use yeast, make sure it's fresh, and I'd give it at least a minute or two in 35°C (95°F, which is just warm, not hot) water. Try different brands. Also consider your liquid dishwashing soap - they're not all the same. Some sources suggest adding glycerol to stabilize the bubbles - I've not tried that, but it seems reasonable (my old bubble recipe calls for 0.5 cups DW soap and 2 Tablespoons glycerine...). No reason you shouldn't add the KI to the soap (technically, it's a detergent) - you'd have to futz around to see if any water was needed and if so how much. At 3%, that means (do the math) that 100 ml will generate ~2 liters of O2 (at room temperature) - so a 20X total expansion maximum (ignoring volume increase due to steam formation from exothermicity). Figure the reaction is only partially complete and that figure drops. Seems to me the important thing is mixing - how the two solutions are added to one another. If it were me, I'd add the soap to the peroxide...although if I was leading a bunch of kids maybe I'd use them to try out different techniques. Proper eye protection is, imho, eye goggles - not safety glasses. This experiment is energetic and uncontrolled, use caution (with anything higher than 3%-6% peroxide).


How do I calculate how much sodium hydroxide I need to add to get a new pH? A 100 ml $\ce{HCl}$ solution has a pH of $3.7$. You want the solution to be of pH 4.5. You have a solution of $10\ \mathrm M$ $\ce{NaOH}$. How much $\ce{NaOH}$ do you need to add to to the $100\ \mathrm{ml}$ solution of $\ce{HCl}$ to get a pH of 4.5?

I firstly calculated the concentration $\ce{HCl}$ in $100\ \mathrm{ml}$ solution = $10^{-3.7}$ = $2.0\times10^{-5}\ \mathrm{mol}$ $\ce{HCl}$ per $100\ \mathrm{ml}$. After this I calculated the $\left[\ce{HCl}\right]$ of the new solution: $10^{-4.5}=3.16\times10^{-6}\ \mathrm{mol}$ $\ce{HCl}$ per $100\ \mathrm{ml}$ of new solution. Now I'm stuck.

At the beginning, the amount of substance of hydronium ions is:
$$n(\ce{H3O+})=10^{-3.7}  \times 100 \times10^{-3}=10^{-4.7} \mathrm {mol}$$
After adding sodium hydroxide, the amount of substance of hydronium ions is:
$$n(\ce{H3O+})=10^{-4.5}  \times 100 \times10^{-3}=10^{-5.5} \mathrm {mol}$$

The difference represents the amount of substance of hydronium ions that have reacted with hydroxide ions.
$$n(\ce{H3O+})=10^{-4.7}-10^{-5.5}= n(\ce{OH-})$$
The volume of sodium hydroxide to add is:
$$v(\ce{OH-})=\frac {10^{-4.7}-10^{-5.5}}{10}= 1.68\times10^{-6}\mathrm{L}=1.68\,\mu \mathrm{L} $$

Assuming the volume of a droplet to be about 60 micro liter, it is much less than a droplet!

$$\pu{pH}_{(\text{before adding}~\ce{NaOH})}  =-\log[\ce{H3O+}]_{(\text{before adding}~\ce{NaOH})}=3.7$$
$$[\ce{H3O+}]_{(\text{before adding}~\ce{NaOH})} =\pu{1\times{10^{-3.7}}~M}= [\ce{HCl}]_\text{(initial)}$$
$$\pu{pH}_{(\text{after adding}~\ce{NaOH})} =-\log[\ce{H3O+}]_{(\text{after adding}~\ce{NaOH})}=4.5$$
$$[\ce{H3O+}]_{(\text{after adding}~\ce{NaOH})} =\pu{1\times{10^{-4.5}}~M}$$
$$\mathrm{V}_\ce{(HCl)} = \pu{100ml}=\pu{0.1L}$$
$$\color{red}{ [\ce{H3O+}]_{(\text{after adding}~\ce{NaOH})} =\frac{\mathrm{V}_\ce{(HCl)}\times[\ce{HCl}]_\text{(initial)}-\mathrm{V}_\ce{(NaOH)_\text{added}}\times[\ce{NaOH}]}{\mathrm{V}_\ce{(HCl)}+\mathrm{V}_\ce{(NaOH)_\text{added}}}}$$
$$1\times{10^{-4.5}} =\frac{0.1\times{10^{-3.7}}-\mathrm{V}_\ce{(NaOH)_\text{added}}\times{10}}{0.1+\mathrm{V}_\ce{(NaOH)_\text{added}}}$$
$$\mathrm{V}_\ce{(NaOH)_\text{added}}=\pu{1.679\times{10^{-6}}~L}$$


Determining the stereochemistry of the product(s) when an alkyl shift in a carbocation generates a chiral carbon Consider this carbocation:

We know that this carbocation would immediately rearrange - by a methyl shift - to form the more stable carbocation.

Now, notice that the migration will produce a chiral center at carbon 1. Assuming that the leaving group was a chloride, will the configuration at 1 - after the methyl shift - be retained, inverted, or racemized?

Interestingly, assuming the nucleophile is $\ce{EtOH}$, in the final product - 2-d-3-ethoxy-3-methylpentane, carbon 2 is also a chiral center. I assume that it will be racemized, as in usual SN1 reactions, but I am not sure. Am I correct?

At a very detailed level, I was also taught that in SN1 reactions, 100% racemization does not occur. The inverted product is slightly more than the retained product. This is because - while the leaving group is still leaving - backside attack is preferred. So, does this phenomenon also have an effect on the molecule above?

To conclude, I wish to know the stereochemistry at both chiral carbon 1 and 2. Are they retained, inverted, or racemized? If racemized, then is the racemization 100%, or is the inverted product formed more than the retained one?

Since this problem is about the stereochemistry, I'd try to address it according to sophomore undergraduate Organic Chemistry text books used in the USA.

Briefly, any nucleophilic substitution reaction at primary carbon predominantly undergoes with SN2 mechanism while those reactions at tertiary carbon undergo SN1 mechanism exclusively. Meanwhile, reactions at secondary carbon would undergo either mechanism, based on reaction conditions, solvents, etc.

The "carbocation" given is a primary carbon even though it is a chiral carbon due to the presence of deuterium atom in place of hydrogen. Note that the stereo-effect caused by deuterium can be considered similar to that of hydrogen. Nonetheless, the bond energy difference (the $\ce {C–D}$ bond dissociation energy listed as $\pu{341.4 kJ/mol}$ while that of $\ce {C–H}$ as $\pu {338.4 kJ/mol}$) may cause the isotope-effects in kinetic of reaction mechanisms (that is a different avenue than what in discussion). Since $\ce{-CH2-Cl}$ and $\ce {-CHD-Cl}$ are similar and $\ce {C}$ atom in both cases is primary, the mechanism of the reaction should be SN2 exclusively. Therefore, the stereochemistry will be $\ce {100\%}$ inverted.

For reading for substitution reactions involving alkyl halides: article. Please also note that, one of example given on that website, the compound $\ce {Ph-CH(CH3)CH2-Cl}$ does not undergo methyl group migration to give benzylic carbocation during its nucleophilic substitution reaction.


How to compare dipole moment in thiophene, furan, and pyrrole? This is what I found (source):

I wonder about several things here.

First, why is tetrahydrothiophene's dipole moment bigger than that of tetrahydrofuran, when considering that tetrahydrofuran has an atom of stronger electronegativity?

Secondly, why thiophene has weaker dipole moment when the S atom has larger p obitals, and therefore, has weaker conjugation ability than O to aromatic ring. The result seems that S in thiophene has better conjugation effect. (that's why it has less stronger dipole.) and I don't seem to explain why.

So what are the reasons for these dipole moment orders?

As another (downvoted) answer points out, the key to this comparison involves the heteroatom donating an electron pair to the ring in the aromatic compound.  This shifts the negative charge away from the heteroatom's corner into the ring.  Thus we either do not get as much of a negative end at the electronegative heteroatom's corner, or with a stronger interaction we can reverse the dipole.  In the examples shown, pay attention to the directions as well as the numbers of the given dipole moments!

Clearly nitrogen has this stronger interaction, while oxygen and sulfur give the weaker effect, with a reduced but not reversed dipole moment.  Oxygen is so electronegative that even in forming the aromatic ring it donates an electron pair only weakly.  The large sulfur atom has lower electronegativity, but overlaps poorly with the smaller carbon atoms. (In the nonaromatic tetrahydrothiophene, the larger size of the sulfur atom makes the dipole longer and thus may give a larger dipole moment with less charge separation.)  Nitrogen in pyrrole has the moderate electronegativity and strong overlap which, together, enable the stronger interaction.

It depends on the extent of which the heteroaton can donate its lonepair into the ring and as N does it the most the dipole moment is upward and similar way it can be determined for the other two


How do I write orders of magnitude in scientific notation when margin of error is involved? My TA has me confused about how to use scientific notation when margin of error and orders of magnitude are involved.

The number I wrote: 3.78 ± 0.02 * 10^-3

My TA told me that this means: 3.78 ± 0.00002

I thought that this meant: 0.00378 ± 0.00002

Which one is correct? I've been looking online to find something that explains the proper notation when writing out margin of error with orders of magnitude but I haven't been able to find anything (I'm at a loss of what terms to search for). If what I wrote isn't the correct way to say 0.00378 ± 0.00002, then what is? Writing it as 3.78 * 10^-3 ± 0.02 or  3.78 * 10^-3 ± 0.02 * 10^-3 seems kind of foolish.

Sometimes you need to go the extra mile to be entirely unambiguous. Your original formulation (3.78 ± 0.02 * 10^-3) could be interpreted in two ways, the way you intended and the way your TA suggested. The majority of people, I expect, would use context to interpret it correct as you intended, but some (as your TA demonstrates) will not.

There is no clear cut widely agreed standard that will resolve this. But you can apply standard mathematical notation (which will look superfluous to some) to make what you write entirely unambiguous. As Ivan suggested in a comment the simple addition of brackets giving (3.78 ± 0.02) * 10^-3 makes your statement unambiguous in its interpretation. This is more compact than 3.78 * 10^-3 ± 0.02 * 10^-3 but just as unambiguous.

If you were using properly typeset formulae, you might well be able to avoid some of the apparent redundancy of the last formulation, but that would take more effort. Better to accept that redundancy is sometimes required to make things clear.

I think your TA is wrong, based on the following argument.

When you write a number derived from calculation or experiment, and following standard protocols for the propagation of uncertainties, you regard the last reported integer in the number as uncertain. Therefore, it simply does not make sense to report an uncertainty of ±0.00002 for 3.78, since the uncertainty in 3.78 must be at least ±0.01. I would say that although there may be cases in which explicit notation should be followed, it is generally safe to write

  $3.78 ± 0.02 \times 10^{-3}$

as meaning

  $0.00378 ± 0.00002$

A safe alternative (consult a standard analytical chemistry textbook such as Skoog, West and Holler's) is to write

  $3.78 (±0.02) \times 10^{-3}$

The use of parentheses emphasizes that the uncertainty defines a range of possible values.

Postscript: Another similar argument can be made for the placement of units, for instance should one write

  $3.78 ± 0.02 ~kcal/mol$

or

  $3.78 ~kcal/mol ± 0.02 ~kcal/mol$

Clearly the second case borders on the absurd.
This is a recurrent problem during drafting of scientific text. Another similar situation is the following: should one write

  A and B were determined to be 1.0 and 2.0 kcal/mol, respectively.

or

  A and B were determined to be 1.0 kcal/mol and 2.0 kcal/mol, respectively.

or

  We determined values of A=1.0 kcal/mol and B=2.0 kcal/mol.

Clearly the first case is acceptable, even if it could be construed as ambiguous, and is not necessarily the most concise. Scientific writing is subject to different rules than standard prose.

Clarity is very important, conciseness too, and there are, in general, recommended conventions. See this answer for instance, which suggests that using the "±" symbol is not recommended unless you mean to use the uncertainty to estimate the distribution of errors.


Gibbs Free Energy and Entropy I understand Enthalpy, the exchange of energy between products and reactants. But what is Gibbs Free Energy and Entropy? I know that Gibbs free Energy is the difference between the change in Enthalpy - [(the change in entropy)*the Temperature]. However I am not able to understand the physical meaning. What does this really mean? On the molecular scale where is it stored? If you can make it less abstract than it is right now, that would help a lot.

I am assuming once I understand Entropy, I should be able to understand Gibbs Free Energy. So, my question on Entropy is, where is entropy stored? I know it is the total disorder in the ENTIRE system, not just one molecule, the units for Entropy are J/K, and by disorder I mean the preference of the system to go to a chaotic/disordered state.

So, when solving for Gibbs Free Energy, what are we finally getting by subtracting the energy from the system (dS) from the reactions exchange in energy (dH).

Like you said Gibbs free energy is the enthalpy minus TdS so basically it's the energy available (free energy) after you take the thermal energy out of the system. The TdS term is the energy due to random thermal fluctuations so sometimes it's more useful to remove that from the enthalpy and deal with the free energy especially when trying to determine spontaneity of reaction.

Entropy is a measurement of disorder in a sense mostly related to probability and statistics. A system with greater entropy contains a greater amount of possible microstates. For instance, two gas chambers with the same amount of gas molecules will have different entropy values if one of them contains different types of gases. The molecules in the chamber with more gas types can be displayed in many more ways.

In fact, by understanding entropy one can truly understand the statistical nature of all chemical processes and understand why Gibbs free energy can be seen simply as a measurement of the likeliness of a state existing or not and ΔG can be seen as the likeliness of a process happening or not. This can be shown by an analysis of the famous equation:

ΔG = ΔH – TΔS

We can begin by remembering that ΔH represents the heat exchange involved in a reaction at constant pressure, if it’s negative, the system releases energy and, if it’s positive, the system absorbs energy. Now, enthalpy is “stored” in the form of chemical bonds or interactions that are present in the system and, with the progress of a reaction or other processes like solvation, interactions are formed or broken and this energy is released or absorbed.

And what does –TΔS represents? As an endothermic reaction progresses, heat is transferred to the system from the surroundings, but why does that happen? Why is that spontaneous and not, say, the opposite reaction, which would be exothermic and release heat to the surroundings (supposing the reverse reaction is possible)?

Well, chemists and physicists know that the entropy of the universe always increases (second law of thermodynamics), that is just a way of saying mathematically that all that is most likely to happen will happen. So –TΔS is the term that accounts for the effect that the exchanged heat will have in the system or surroundings when it is transferred.

In the case of an endothermic reversible reaction, we have  ΔH > 0, so the reaction can only be spontaneous if TΔS is big enough. That means that the reaction can only be spontaneous if it leads to an increase of disorder in the system (increase in the number of possible microstates of the system) that is big enough to compensate the decrease of disorder in the surroundings, associated to the heat loss by the surroundings, which is accounted for by ΔH.

ΔG will only ever be 0 if the system is in equilibrium and if ΔSu (entropy variation of the universe) is also 0, meaning that the variations of entropy both in the system and in the surroundings caused by the two opposite reactions occurring at the same time are exactly compensated given the constant reaction rates achieved in equilibrium. The system is not only in equilibrium with itself, but also with the surroundings.

The key is to realize that you don’t need to measure the ΔS of the surroundings and yet you obtain information that ultimately depends on it. This is one of the great merits of this equation.

So ΔG can be seen as just a convenient way for chemists to evaluate the balance of factors that “decide” if a process is spontaneous or not, mathematically equivalent to simply checking which processes lead to an increase in the entropy of the universe.


How do I prepare citric acid solution (from anhydrous powder) of accurate concentration, given that the solution volume expands when water is added? I attempted to prepare saturated citric acid solution from anhydrous powder. According to online data, the solubility limit was 59.2 w/w at 20 °C, i.e. 59.2 g citric acid in 100 mL water. To ensure full saturation (as in the case of preparing brine solution), I ended up weighing 67 g of the powder and tried adding 100 mL of water to it.
However, when I had added about 67 mL of water, the 100 mL Schott bottle was already full, and all of the powder dissolved (estimated 5.22 M solution). Since there is an apparent volume expansion, how then should this solution (or a citric acid solution of any other desired concentration) be prepared?

The "expanding solution" problem is irrelevant if you use standard techniques
First, you need to realise that a standard solution is a known amount of solid in a known volume of solution not a known amount of solid and a known amount of the solvent. You don't need to know how much solvent you use. So you need to target getting a result that occupies, say, 100mL not a solution made from 100mL of water.
There is an easy way to achieve this. Use a volumetric flask with an accurate volume marking. Add the solid to the flask. Fill the flask slowly (while shaking enough to ensure the solid dissolves before the flask is full). Keep adding small increments of the solvent until the total volume reaches the graduation. You now have an exact volume of a solution containing an exact amount of the solid. At no point do you need to worry about any volume expansion or contraction.

One thing is usual solubility in g/100 mL, other thing is solubility as percentual mass fraction.
(Improperly used) 59.2 w/w obviously means 59.2 weight percent, as this solubility data – noted properly – is listed in wikipedia: Citric acid.
You need 59.2 g of the acid and 40.8 g of water ( about 40.8 mL or approx. 41 mL ), in total 100 g. The volume will be less than 100 mL, due the higher than water density.
Recalculated to solubility it is approximately $\pu{145 g/100 mL}$
In the case an acid monohydrate is used instead, the stoichiometric mass of crystal water has to be obviously subtracted from the acid mass and added to the water mass.
That can be recalculated for any target total mass and particular volume can be taken from this stock solution.

comments to comments:
@matt_black Sure, there are multiple ways to prepare saturated solutions.

One is the preparation of arbitrary mass of w/w% solution according to provided data.
Other is recalculation of w/w% to g/100 mL and dissolving in this nominal volume.
Other is reaching equilibrium of stirred solution with an excess of acid.
Other is to obtain density of saturated solution, calculating the needed acid mass as percentage of the total volume mass. And then not fully filling up into given volume. After near full dissolution, mixing/filling up in few steps until the nominal volume.
Other is to calculate the acid mass equivalent to the acid volume with the filling volume as a complement. But this is not recommended, being tricky, not applicable to well soluble and quickly dissolving compounds. As it implies filling is done before significant dissolution occurs, because there is volume change during dissolution.


Which one is more acidic – 3,5-dinitrophenol or 4-nitrophenol I’m confused because in 3,5-dinitrophenol, the inductive effect is dominating whereas in 4-nitrophenol the mesomeric effect dominates. So which one should we consider for acidity?

As Zhe's comment points out, intuition does not count for much here, unless you have enough actual experience working with these systems to know what is more acidic. There are some things which cannot be predicted a priori using simplified models.

The only way to find out is to look up the data. The acidity constants of some of these phenols are tabulated in J. Org. Chem. 1998, 63 (19), 6680–6688. 4-Nitrophenol has a $\mathrm pK_\mathrm a$ of $7.14$, and 3,5-dinitrophenol a $\mathrm pK_\mathrm a$ of $6.73$, in aqueous solution.

I have used MarvinSketch 17.28 to estimate the acidic strengths of substituted nitrophenols, as measured by their $\mathrm pK_\mathrm a$ values.

$$\begin{array}{cc} \hline
\text{Substitution} & \mathrm pK_\mathrm a \\ \hline
\text{2,4,6-trinitro} & 2.344 \\
\text{2,4,5-trinitro} & 3.224 \\
\text{3,5-dinitro} & 6.238 \\
\text{4-nitro} & 7.075 \\
\text{3-nitro} & 7.892 \\ \hline
\end{array}$$

It is evident that 3,5-dinitrophenol is more acidic than 4-nitrophenol.

The relative acidities of the meta- and para-monosubstituted nitrophenols are in line with predictions based on resonance arguments.


Why the bond strength of HCl is more than HBr? Most people say that it is because the 3p orbital of chlorine when overlapping with the s orbital of hydrogen covers more area than when 4p orbital of bromine overlaps with the s orbital of hydrogen which makes bond between H and Cl more stronger than the bond between H and Br. But how can we say that the area of overlapping is more for 3p and s orbital compared to overlapping of 4p and s orbital?

You can think of the overlapping volume between the electron clouds, but a simpler form to show why this happens is when you consider the electrical potential between the atoms. Even tough this isn't an ionic compound, it has many characteristics that are similar.

If I remember correctly, this is the expression for the potential energy between charged particles: $E_p = k\frac{Q_1 \cdot Q_2}{d}$

Since the radius of $\ce{Br}$ is greater than the radius of $\ce{Cl}$, the potential energy involved in $\ce{H-Br}$ bond will be smaller than $\ce{H-Cl}$ bond.

Another thing that came to my mind, and I'm not sure, is that as the 4p cloud of $\ce{Br}$ is bigger than that of $\ce{Cl}$, the relative amount of the volume that is overlapped with the 1s cloud of $\ce{H}$ gets smaller with $\ce{Br}$ than with $\ce{Cl}$, which means the connection is weaker.

If you aren't convinced by the overlap area argument,  then think of it in terms of polarization.  The H+ cation will polarize Br- more as compared to Cl-, so HBr will possess a more covalent character as compared to HCl, or conversely, H-Cl bond will be more ionic than the H-Br bond. And as we know, ionic bonds are generally stronger than covalent bonds


The increasing order of reactivity of substituted benz0yl chlorides towards nucleophilic substitution Arrange the following compounds in the order of their increasing reactivity towards nucleophilic substitution:

The answer to this question is $[3<2<1<4]$. We have to compare the positive charge density on the carbons to which leaving group is connected. In $(1)$, the phenyl ring shows negative inductive effect but my teacher did not mention anything for that. In $(3)$ and $(4)$ the connected groups ($\ce{OCH3}$ and $\ce{NO2}$) show +M and -M respectively. But, they also show -I effect. Why are we ignoring the inductive effect?

Electronic effects govern in the order
Resonance/Mesomeric > Hyperconjugation > inductive
+M or +R dominate over inductive. That's because +M involves π electrons which are more reactive than the sigma electrons. + I operates through sigma electrons. So therefore +M effect has priority over +I.
Therefore the order of reactivity towards nucleophilic substitution is 4>1>2>3


Why does adipic acid give a cyclic ketone on heating while glutaric acid and succinic acid give a cyclic anhydride? Why does adipic acid give cyclic ketone on heating while glutaric and succinic acid give cyclic anhydride?

According to Blanc Rule for Dibasic Carboxylic acids, A cyclic Ketone is formed unless a five or six membered cyclic acid anhydride can be formed. This rule is valid for 1,4; 1,5; 1,6. dicarboxylic acids

Because cyclic anhydride formed will be a 7 Member ring compound which is less stable than 5 membered cyclic ketone


Does copper react with HCl when current is present? Copper does not react with hydrochloric acid, according to many sources.  However, I found an experiment posted in a video, which I will try to repeat, that shows a copper pipe dissolving into HCl and plating a nail when the copper pipe and the nail are connected to a 12-volt battery, becoming electrodes.  No other chemicals, according to the video, are present.  How does the electrical current cause the copper to react with the HCl when otherwise it would not?

A reaction between copper metal and hydrochloric acid would go something like this: $$\ce{Cu + 2HCl -> CuCl2 + H2}$$
...which we can write as: $$\ce{Cu + 2H+ -> Cu^2+ + H2}$$
If we break this up into half reactions, we get: $$\ce{Cu -> 2e- + Cu} ~~~E°=-0.34~V$$ $$\ce{2H+ + 2e- -> H2} ~~~E°=0.00 ~V$$
If we add the voltages together, we get the cell potential of the combined reaction is $-0.34+0.00=-0.34~V$. Since the cell potential is negative, the reaction is not spontaneous, so the reaction is not going to occur without any artificial interference ($\Delta G°=-nFE°_{cell}$).

However, if you connect a battery (galvanic cell) to this reaction, you can supply the necessary electrons for this reaction to occur (electrolytic cell). Since you are connecting a 12 V battery, there is more than enough voltage to cause this reaction to occur.

The short answer is because the electric circuit provides the energy needed to do those reactions. You should know that you could try the same experiment but instead of using HCl, try table salt dissolved in water. That should work (12 V is a whole lot of voltage), but I can't say I actually looked it up to verify it. If that doesn't work, add some vinegar (acid) to the salt solution. Be sure to use the right connections (the copper will only dissolve near one wire). Waiting 5 or 10 minutes should be long enough to see if its working or not.
Dr. Who said "Time makes liars of us all." and he was right. It is just not true that HCl is unreactive towards Cu. I found dozens of YouTube videos and testimonies which contradicted that claim. The real world is complicated. Most general statements of the form "A doesn't react with B" are likely to be true in some circumstances and false in others. "Water doesn't react with sugar" for instance. Well, that seems to be true; I can dissolve sugar in water and then evaporate the water and get back the sugar. Well, wait a minute: it turns out that sucrose reacts (very slowly with water at room temperature, it takes years) and converts to a mixture of glucose and fructose. Not only that, but if I heated the water to 2000° I know that dumping some sugar into it would decompose the sugar. So, we need to qualify statements of fact (either implicitly or explicitly). In the case of your question, we need to make sure that the copper and the acid are pure. We need to prevent contamination with any oxidants (such as O2 from the atmosphere) and we need to specify the conditions of temperature and pressure (and protect the system from other interferences, such as radiation, gravity, etc. etc. (admittedly, Earth's surface gravity isn't strong enough to have much effect on reactions at lab scale.). Most copper is coated with various oxides, so cleaning a penny with HCl (which makes it shiny) doesn't prove that the HCl is reacting with the metal, but demonstrates that the HCl is either reacting with the oxide, with the metal, or with the copper alloy. (Copper wire is quite pure, 99.99%, pennies not so much). So, let's say we condition the HCL solution to eliminate all oxygen and other impurities, abrade the surface of a (pure) copper wire, and place the wire into the solution and leave it there at room temperature, 25°C. Will the copper react? Well, let's say that it won't. We can assume this because supposedly the "standard potential" tables indicate it won't. (Be careful here!, what concentration of acid is used to determine "standard potential"??) 1M HCl is ~4% but you can easily buy 30% concentrations. (The devil is in the details.) So, assuming your "sources" are trustworthy, then they are referring to specific conditions, and not some "universal truth". If you doubt it, you could try (the dangerous experiment of) carefully heating the diluted HCl with a scraped-clean-of-oxide piece of Cu wire in it, and seeing if the copper starts to react. It will but it might not be an experiment you should do. HCl is volatile, and can vaporize. The problem with that is that it is also corrosive and toxic. If you breath it, you could die. If it condenses on various metal surfaces around where you're heating it, they might corrode (and lets not even think about what it will do to electronics!).======== OK. So, we've diluted and cleaned the system and if your sources are right, the copper just sits there. But if you connect the metal to an the right side of a 12V transistor battery, you can indeed drive the reaction. Why? Well the flow of current indicates that energy is being used, just like the flow of water. You can use some of that energy (from the battery) to do work. (Incidentally, you could also use a battery charger and do the same thing, even a dc charger or power source which you probably have dozens of (you know, those little black cubes you have to plug into your AC wiring and then plug into your cell phone to charge it, or into thousands of other things that require DC power)). Anyway, a battery by definition works by ionizing something (usually a metal) and guiding the electrons that flow from one electrode to the other out of the battery (if a circuit is made and closed). The opposite of M → M+ + e- which is what powers most batteries (where M is a metal like lead or lithium) is M+ + e- → M. In other words you can use the electrons to ionize some metal, M, if there's enough energy. We measure energy in a circuit by voltage, the higher the voltage, the 'stronger' the energy (voltage, or potential, is like the height of a tank of water - the higher it is, the more work it can do). A 12V battery has a LOT of potential (relative to what's needed to ionize metals) but generally doesn't have much current. That means it can ionize most (I think, all, actually) metals but not very much of them. Electrochemistry is a bit of a strange area of chemistry, because it deals with two different concepts of "potential": the electrical term, same meaning as the word as voltage, more or less and the chemical term, meaning the ability (or potential) to react. Most metals prefer to ionize to the M+ ion, the cation. (By "prefer" I don't mean the atoms are choosing what to do, they do what the Laws of Physics and Chemistry tell them to do, no exceptions) But metals losing electrons is called "oxidation", metal ions gaining electrons (and so going back to the neutral (uncharged) atom, is called reduction. In order for a + ion (a cation) in solution to gain an electron, it needs to take it from something. The easiest place to take it is from a wire which has a negative charge and is "full" of extra electrons (from the battery). It happens at the wire's surface, and this allows the metal to "plate out" on the wire because uncharged (neutral) metal atoms are generally not very soluble in water. The ions are soluble, they are changed to the neutral metal at the electrode surface and plate out (precipitate out) onto that surface. Oh, one last thing. 12 Volts is more than enough to break apart water. In gentle conditions, (not working with explosive mixtures, using small batteries, etc.) the reaction which will occur is the one which uses the least energy. This means that even if you have enough energy to, say, ionize gold or platinum or silver, that they won't dissolve in a dilute acid if the reaction 2H2O → 2H2 + O2 uses less energy. (You can see this by watching bubbles form at the wires. If bubbles form at one wire, you know that something is going on at the other. The water reaction, above, makes hydrogen at one wire (electrode) and oxygen at the other. One explodes, the other we breathe. (Collecting the H2 is mostly safe unless sparks, flames, or heat (or platinum) is present.)


Calorimetry, finding final temperature I was looking for some help on understanding this practice question from my text book, I know how to figure out enthalpy, but this is my first temperature change question and I am stumped. Any guidance would be greatly appreciated.

Nitric acid is neutralized with potassium hydroxide in the following reaction:
$$\ce{NHO3(aq) + KOH3(aq) -> KNO3(aq) + H2O(l)}$$
$$\Delta_\mathrm rH= -53.4\ \mathrm{kJ/mol}\ \ce{HNO3}$$.
$55.0\ \mathrm{mL}$ of a $1.30\ \mathrm{mol/L}$ solutions of both reactants, at $21.40\ \mathrm{^\circ C}$, are mixed in a calorimeter. What is the final temperature of the mixture? Assume that the density of both solutions is $1.00\ \mathrm{g/mL}$. Also assume that the specific heat capacity of both solutions is the same as the specific heat capacity of water. No heat is lost to the calorimeter itself.

I know the answer is 29.7 degrees Celsius, I'm just not sure how to get there. I tried juggling around my equation "delta t r H= m c delta t" but since I don't know the final temperature i don't think the delta t works in this case. I've tried omitting the delta t but I get no where near 29.7 degrees Celsius.
So I took my formula Q=mc delta t and rearranged it, I now have delta t= Q / mc. Delta t= -53.4 kJ/mol / (55.0g)(4.19 J/g C) I changed the 55mL to g as the question says to assume the density is 1.00 g/mL. I also used the heat capacity as water given in my course data booklet). And I get -0.2317. I tried doubling the mass as there is two solutions but that doesn't work either


Iodine Titration with Free SO2 in wine To determine the free $\ce{SO2}$ in wine, a common method used is called the Ripper Titration Method.

The equation is as follows: $$\ce{SO2 (aq) + I2 (aq) + 2H2O (l) -> 4H+ (aq) + SO4^2- (aq) + 2I- (aq)}$$

This method involves titrating an Iodine solution into a sample of wine. I have found that triiodide is made from reacting $\ce{I-}$ with $\ce{I2}$. However, if it is the triiodide we are titrating, how is the iodine reacting with the wine?

Is it triiodide ($\ce{I3-}$) that is being titrated with the wine sample or iodine ($\ce{I2}$)?

Reaction goes along the lines $\ce{3 I2 -> 2 I3- -> 6 I-}$. The beginning of titration is $\ce{I2}$. The end of titration is $\ce{I-}$. $\ce{I3-}$ forms as an intermediate, but it can be ignored in titration.


Can the structure of an azide ion contain a coordinate bond? Can the structure of an azide ion ($\ce{N3-}$) be $\ce{N#N\bond{->}N^-}$?

I know the actual structure but was asking whether the single bond can be shifted to a N to make it a triple bond and a coordinate bond with other Nitrogen.

This is not a duplicate of Hybridization in azide ion? because I am asking whether it can have a coordinate bond or not while the other question asks what the hybridization of centre nitrogen in $\ce{N3-}$ is.

You can’t really tell a dative bond from a ‘normal’ covalent bond until you break the bond. A dative bond will then dissociate heterolyticly while a covalent bond will dissociate homolyticly.

Just using that simplified way of putting it, you could say that there are many reactions of azides wherein $\ce{N2}$ is liberated leaving a nitrene or related structure; just one of many examples would be the Curtius rearrangement. However, that is not what is meant in the definition above; the definition talks about supplying the bond dissociation energy.

The BDE is not typically supplied in chemical reactions except in coordination compound ligand exchange and radical generation. If you were to apply it to $\ce{N3-}$, I predict homolytic dissociation. Thus, the bond between the nitrogens is more of a typical single bond.

Of course, aside from the chemical implications of dative versus covalent bond you may just have decided to use an arrow notation rather than formal charges. Technically nobody can stop you but still you shouldn’t call it a true dative bond.

No.

The issue with using your proposal is very simple: $\ce{N_2}$ sucks at being reactive. It doesn't want to give up or share its electrons with any other atom. In fact, the azide ion is usually explosive because it keeps trying to form nitrogen gas. $\ce{N_2}$ has no real polarity and no particular motive to coordinate to anything, a single N included.


What is the systematic name of a "double benzene"? Consider the molecule representation below. What is the systematic name of that molecule?

Could it possibly be 3,4-hex-2,6-anebenzene?

As @ron already said in his comment, the systematic name of the naphtalene is Bicyclo[4.4.0]deca-1,3,5,7,9-pentaene or Bicyclo[4.4.0]deca-2,4,6,8,10-pentaene since you can start the carbon numerotation from 2 chemically equivalent carbon.

You can find all the references here and here

Chemical Formula:

  $\ce{C10H8}$

Common Name:

  Napthalene (also accepted by IUPAC as per 1993 recommendations)

IUPAC Name: (refer numbering in figure)

  Bicyclo[4.4.0]deca-1,3,5,7,9-pentaene or Bicyclo[4.4.0]deca-2,4,6,8,10-pentaene

For a general guideline:

Start with one of the bridgehead carbons and number it 1.
Proceed round the longest chain of carbons to the second bridgehead.
Number the second bridgehead carbon and continue on round the next longest chain of carbons back towards the first bridgehead carbon.
Pass over the first bridgehead carbon (it already has the number 1) and along the shortest chain of carbons to the second bridgehead carbon again.
The generic name for compounds such as this is

  bicyclo[x.y.z]alkane

where x, y, and z are the numbers of intervening carbons on the three paths between the two bridgehead carbons cited in decreasing numerical order, and alk refers to the total number of carbons in the ring systems.


How should I identify my leaving group in SN1 reactions if there might be more than one option? Before I can even begin to contemplate the products of a given SN1 reaction, I need to be able to determine what my leaving group will be, which should be a species with a strong conjugate acid, and as I understand it the species should be relatively stable once it has left the substrate. Please correct me if I am making any mistakes in my understanding.

A hydroxide ion will be a terrible leaving group, since its conjugate acid is water, and hydroxide is not very stable by itself.

So when I am asked this question "Draw the major organic product(s) for this SN1 reaction:

How do I know what will be the best leaving group? Hydroxide is not very good, so perhaps $\ce{CH_3OH}$ will work better? How do I know it is not $\ce{CH_3CH_2OH}$ instead?

Is this only one decent option, and I am over thinking this?

This case (and similar cases) you need to think about the following:

When a leaving group leaves, only one bond is broken. By this logic, $\ce{OH}$ is a leaving group - it can leaving by the breaking of the $\ce{C} -\ce{O}$ bond. The alcohols you mention, $\ce{CH3OH}$ and $\ce{CH3CH2OH}$ require breaking two bonds to leave: the the $\ce{C} -\ce{O}$ bond and a $\ce{C} -\ce{C}$ bond.
What is that acid $\ce{H-Br}$ doing? When you are given a strong acid as a reagent, very commonly the first step in the reaction's mechanism is a proton transfer (an acid-base reaction). Proton transfers are very rapid; they often happen before anything else has a chance to. In this case, the proton transfer is also exothermic (or exergonic if you prefer) - the equilibrium constant is on the order of $10^8$ in favor of the products. What does the product of the initial proton transfer look like? Is there a better looking leaving group on that structure?

To answer this question, you must first inspect the species that are reacting first. What are these species? What is acid? what is base?...not necessarily focusing on the leaving group.
In general, acid-base reactions have extremely high Keq (Ka) and in the system will occur first. It is so rapid that the exchange of a proton is close to the diffusion-controlled rate limit.
That happens first. The alcohol is protonated and oxygen becomes formally +. It is now the leaving group, water. As it leaves a stable tertiary carbenium ion is formed. Bromide, the conjugate base of the acid used then attacks to form the Sn1-type product.

Someone mentioned TsCl...tosylation will not be a good choice. True it will activate the alcohol to be a weaker base. But since this is a tertiary alcohol, aa back-sided displacement is not going to happen to give you an alkyl bromide, you will get the elimination instead of substitution.


What is the relationship between the stability constant of a copper (II) electrolyte complex ion and the output voltage Ecell? As a part of a school investigation I constructed a voltaic cell experiment where for each run I added a different ligands to the copper (II) sulfate half-cell electrolyte in a daniell cell to see if there was a change in the output voltage Ecell. When plotting a graph of this, it seemed that Ecell increased as the stability of the ligand increased (indicated by the stability constant Kc) however I am struggling to explain why. I presume its because the standard potential at the cathode Erhe increases with ligand stability due to greater oxidation.

The redox potential of Copper does not depend directly on the stability of the complex Cu-ligand. It depends on the concentration of the residual $\ce{Cu^{2+}}$ ion in solution. And in any solution of $\ce{CuSO4}$, adding a ligand produces a decrease in concentration of the free ion $\ce{Cu^{2+}}$. So, according to Nernst's law, the potential $\ce{Cu^{2+}/Cu}$ becomes less positive.
For example, the standard redox potential of $\ce{Cu^{2+}/Cu}$ is + $0.34$ V. Let's consider the ligand chloride $\ce{Cl-}$. If enough chloride ions $\ce{Cl-}$ is added to a $1$ M $\ce{Cu^{2+}}$ solution so that at the end the excess of $\ce{Cl}$- ions is $1$ M, the complex $\ce{CuCl4^{2-}}$ will be formed, with a concentration $1$ M. The measured residual [$\ce{Cu^{2+}}$] concentration is $2.5 ·10^{-6}$ M. In this case, Nernst's law gives the following $\ce{Cu^{2+}/Cu}$ redox potential : $$\ce{E_{Cu^{2+}/Cu} = + 0.34 V + \frac{0.0592}{2}log(2.5·10^{-6}) = + 0.34 V - 0.0296·5.6 = +0.34 V -0.166 V = +0.174 V}$$ This +$0.174$ V is lower than the standard redox potential (+$0.34$ V).
The stability complex of the $\ce{CuCl4^{2-}}$ ion is : $\ce{K = \frac{[CuCl4^{2-}]}{[Cu^{2+}][Cl^-]^4} = \frac{1}{2.5·10^{-6}·1^4} = 4·10^5 M^{-4}}$
If another ligand is used, which makes a stronger complex with copper $\ce{Cu^{2+}}$, the stability constant is higher, and the redox potential will be still lower.

My speculation starts with a comment by ChemEng, who noted here that the Nernst Equation is essentially a function of concentration and temperature.
In this regard, I start with an example of a side-reaction:
$$\ce{ Cu (s) + Cu(II)L <=> 2 Cu(I)L }$$
where the above reaction proceeding, given the general low solubility of Cuprous (absence a ligand) critically depends on the presence/stability of the highly soluble ligand L itself.
Also, a cited half-cell reaction in the Daniell cell is:
$$\ce{Cu(II)L + 2e− → Cu(s) }$$
Or, can it proceed, at least partially also, in steps dependent on solubility provided by the ligand:
$$\ce{Cu(II)L + e− → Cu(I)L }$$
$$\ce{Cu(I)L + e− → Cu }$$
So, again another possible example of the role of the select ligand in facilitating the cuprous (as an intermediate) concentration.
As such, my supposition is that the presence/stability of the soluble ligand impacts copper ion(s) concentration thereby impacting the Nernst equation.
[EDIT] Based on further research, my speculated explanation is likely too simple. Here is a more technical source on an advanced Daniell cell design. In recharging, note the mention of problematic Cu2O formation which is rectified, in part, by an increased Cu(NO3)2 electrolyte concentration.


What is the relationship between the two definitions of the difference between basicity and nucleophilicity? (Clarification: this is a question about terminology, not so much about when to use different concepts.  I think I do understand which characteristics of a species to use when trying to determine something about a reaction in some circumstances.  I just don't know what word to call it.)

When someone on the internet (here or elsewhere) is asked the difference between "basicity" and "nucleophilicity", they typically give one of the following two answers:

Basicity is about how well a species donates electrons to a proton, while nucleophilicity is about how well a species donates electrons to a more general atom or ion (especially carbon).
Basicity is about thermodynamics (i.e. equilibrium constant) while nucleophilicity is about kinetics (i.e. activation energy).

While these two answers may be related, they do not seem identical!  So I am wondering which one is actually correct, or whether there is some way that they are both correct.  In particular, I'm wondering what happens when you have a situation that puts those two descriptions at odds with each other.

We could have a reversible reaction involving carbon (though I know that's not the usual situation) that does in fact have some equilibrium.  When we're talking about that reaction's equilibrium, do we still use the word "basicity"?
In a reaction with a proton rather than a carbon, when we're talking about kinetics, do we still use the word "nucleophilicity"?

I hope I've gotten my question across OK.

The second set of definitions is the correct one.

Basicity is a thermodynamic property. In essence, the basicity is measured by determining the equilibrium constant of the following (generic) reaction:
$$\ce{B- + H2O <=> B-H + OH-}\tag{1}$$

The proton transfer that is the gist of this is an extremely fast process — faster than diffusion control due to the Grotthuss mechanism. You may have witnessed this during acid-base titrations: equilibrium is reached extremely rapidly. Another way to observe this is by adding carbonate bases to an acidic solution; $\ce{CO2}$ evolution is basically instantaneous. Therefore, there is no way to observe just one step (forward or reverse); you basically always have them both.

Nucleophilicity is a property that is related to the rapidness of the following, again generic, reaction:
$$\ce{Nu- + R-LG -> Nu-R + LG-}\tag{2}$$

As similar as this may seem to the above, the lack of a reverse arrow may already show that this is tied to only one directiong; that it is not an equilibrium process but rather a kinetic one. Most of the time, there will be one species that is clearly a good nucleophile ($\ce{Nu-}$ in the scheme) and one that is only a good leaving group ($\ce{LG-}$ in the scheme). The reaction will be driven to completion and stay there.

To exemplify the difference, consider what would happen if you added hydroxide to the acid-base equilibrium $(1)$ versus additional leaving group to the nucleophilic substitution $(2)$. In $(1)$, the equilibrium would readjust; if the base is less basic than hydroxide it will be deprotonated, i.e. the reaction will be shifted back to the reactants. In $(2)$, typically nothing will happen, e.g. if you add $\ce{OPPh3}$ to a Mitsuobu reaction; showing the kinetic nature of the process.

Another way to look at the problem — and to separate the parameter basicity from the parameter nucleophilicity — is to look at the influence of steric effects. For example, consider ethanolate, propanolate and deprotonated isobutyl alcohol. In all three cases, the negative charge is unambiguously located on the oxygen atom. Thus, we would expect a thermodynamic property to be relatively equal for the threeof them. At the same time, there is a clear increase in steric hindrance from ethanolate to the isobutyl alcohol. This steric effect should manifest itself in reaction kinetics and lead to the latter being less nucleophilic than the former. Indeed, sterically hindered bases are often described as non-nucleophilic while the same structure with less hindrance often works fine as a nucleophile.

Finally, let’s work at debunking the claims laid out in the first set of definitions. First, basicity. After the Arrhenius and Brønsted-Lowry theories of acids and bases came the Lewis theory. In this, the reaction of a Lewis base with a Lewis acid — not necessarily a proton — is an acid-base reaction. One of many possible examples would be the formation of a transition metal complex such as $\ce{[Cu(NH3)4(H2O)2]^2+}$. Here, the ammine ligand is acting as a base, but it is donating to a non-hydrogen atom (namely copper). The claim that basicity only goes towards hydrogen is obviously false.

Unfortunately, it is not possible to fully debunk the second part since a nucleophilic attack on a proton is not really possible (that would indeed be the thermodynamic acid-base reaction). However, we may remember that most of the non-metals can act as the recipient of a nucleophilic attack; for example sulphur in a Swern oxidation, phosphorus in a Horner-Wadsworth-Emmons reaction or iodine in a Dess-Martin oxidation.

I would rather say that nucleophilicity always somehow describes an attack (followed by an addition) whereas basicity describes the ability of taking away positive charge (like a proton). Nucleophilicity and basicity is quite similar with the antagonism of 'adding negative charge' and 'removing positive charge' which de facto is only a difference in regarding the same phenomenon from two different points of view.


why don't carbons use sp3 instead of sp2 orbitals to make double bonds? I don't have a strong chemistry background, but I was thinking why do carbons have to make different kinds of orbital hybridization for double and triple bonds?
Why doesn't a carbon make two sigma bonds into a double bond?

Only one electron can be in an orbital at a time. There are three p orbitals, one in each of the x, y, and z axes. A pi bond is the overlap of one p oribital from each atom. Like the p orbital, a pi bond also can be rotated from the x axis to the y or z to produce a different unique pi bond. Because a sigma bond has rotational symmetry, rotating it does not give a unique bond. Therefore, you cannot have two of them between the same pair of atoms.

Edit: It appears this question has been asked before and there are much better answers than mine (Why can there not be more than one sigma bond in a set of bonds?)


Lithium ion battery - measure C rate How does one determine the C rate of an aged cell?

I would like to find the discharge rate just before any cliff edge of capacity loss. Can the C rate of a cell be estimated by the cells rated heat dissipation and measured internal resistance? Where I could find the current that will use up all the cells heat dissipation capacity.

I have a number of recycled laptop battery 18650 cells.

Are there any negative effects of the cells being below room temperature?

To determine the C-rate, you should charge and discharge two cycles very slowly say at 0.1 mA, between say 1.5 V to 4.3 V for your Li-ion (unless it's a very old laptop battery it should be LiCoO2), and measure how long it'll take. The C-rate would be your capacity divided by one hour.

Don't know of a way to measure it using heat or internal resistance.

Batteries generally have an adverse effect when heating up, so unless it is much below freezing, they should be ok.


Why is the bond order in the SO₃ molecule 1.33 and not 2? $\ce{SO3}$ molecule has three double bonded oxygen to the central sulfur atom.

Sulfur has $\ce{sp^2}$ hybridization and it has 6 outer electrons which make the bonds with the oxygen.
So shouldn't the bond order be 2?

The bonding situation in $\ce{SO3}$ is a tough nut to understand. In a historical context this molecule belongs to the species of hypervalent molecules, that disobey the octet rule. The concept of hypervalence is still very much debated. Recently there was a question raised by ron, seeking for more guidance: Hypervalency and the octet rule.

I stated in Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻?

  The bonding picture is usually trivialised as each $\ce{S−O}$ bond being a double bond. But this is actually far away from the truth, as it does not respect the charge of $q=+2$ at the sulfur atom and the charges at the oxygens with $q=−\frac23$. This is due to the fact, that the sulfur atom actually only contributes to one of the three π bonding orbitals, resulting in electron density deficiency.

There are a couple of molecules that belong roughly to the same class. I call them Y aromatic systems: $\ce{CO3^2-}$, $\ce{NO3^-}$, and $\ce{SO3}$. The description is best carried out in the framework of resonance. (Please also see Why is there a need for resonance?)

I have explained the delocalisation in the nitrate ion, but I am happy to repeat it here for $\ce{SO3}$. The molecule has an overall symmetry of $D_\mathrm{3h}$, hence the central sulfur can be regarded $sp^2$ hybridised. Since oxygen has only one bonding partner, it can sufficiently be described as $sp$ hybridised (compare here). The combination of these orbitals form a $\sigma$ bond each. The remaining $p$ orbitals perpendicular to the molecular plane form the delocalised $\pi$ system. Since delocalisation is not part of the original Lewis description it is impossible to address this with such structures, which ultimately lead to the concept of resonance. It is important, that the bonding situation can only be described by all resonance structures, as there is no predominant. The blue structure tries to emulate delocalisation, but here formal charges and charges are quite messed up.

However, we can take home an approximate bonding order for the $\ce{S-O}$ bond. Since all of the resonance structures are equally dominant, we can see that all of these bonds consist of a $\sigma$ bond. We also see, that there is a $\pi$ bond in the first three structures. Each of them contributes equally, therefore it makes a third of a full contribution per bond, hence the bond order is expected to be about $1\frac13$.

Now the molecular orbital scheme is quite similar, but it suffices with one structure. The explanation of the $\sigma$ system remains unchanged. From the four $p$ orbitals we can form four symmetry adapted linear combinations. Two of the are non-bonding degenerate, the completely symmetric one accounts for the huge stabilisation. The antibonding orbital is not occupied. See below for the schematics of $\ce{NO3^-}$, which is analogous to $\ce{SO3}$. You can find the complete MO scheme here.

Since each of the $\ce{S-O}$ bond contributes equally to the lowest lying $\pi$ orbital, the bond order is found to be about $1.33$. The calculated bond order is a little bit higher, most likely due to attractive electrostatic interactions.

During the reaction of $\ce{SO2}$ and $\ce{O2}$, there's a formation of nascent oxygen which oxidises sulphur. In doing this sulphur co ordinates it's lone pair to the nascent oxygen. Since only 2 electrons take part here in this co-ordinate bond formation it is just a single bond. Now the $\ce{SO3}$ thus formed has 4 bonds shared between 3 places. Hence the bond order.


How to prove that chloroform is more acidic than fluoroform? How to prove that chloroform is more acidic than fluoroform?

Using the concept that the inductive effect is dominant over the resonance (mesomeric effect) in the case of halide substituents: $\require{mediawiki-texvc}$

Since both have single hydrogen, let's compare stability of carbanions $\ce{CF3-}$ and $\ce{CCl3-}$.
Since fluorine shows more negative inductive effect than chlorine, $\ce{CF3-}$ should have been more stable than $\ce{CCl3-}$, so some greater cause must be acting here.
In $\ce{CCl3-}$ carbon has non bonded electrons in 2p orbital and chlorine has an empty 3d orbital, but since the energy difference is very high, $\ce{p\pi-d\pi}$ backbonding is barely possible. Worst case scenario, even if it's possible, then giving preference to backbonding will violate our initial rule (that inductive > mesomeric).
Bond lengths, $\ce{C-Cl}=1.75\AA, \ce{C-F}=1.35\AA$, but since chlorine is bigger than fluorine, they both will have approximately same electronic $\ce{p-\pi}$ repulsion (of lone pair of halogen and at anionic centre).

Which of my reasoning are wrong? How do I prove the fact that chloroform is more acidic than fluoroform?

Data: $\mathrm pK_\mathrm a (\ce{CHBr3})=13.7$ and $\mathrm pK_\mathrm a (\ce{CHCl3})=15.5$ and $\mathrm pK_\mathrm a (\ce{CHF3})=28$ (source - page 2)

My Views (I'm not entirely sure about these reasons)

(1) $\ce{CF3-}$ is highly unstable due to huge repulsions between the negative charge on carbon and the lone pair of fluorine.

(2)While in $\ce{CCl3-}$ chlorine has an empty d orbital thus the negative will be more dispersed into the this empty orbital  hence the repulsions will be quite less as compared to $\ce{CF3-}$.


What might serve as an initial starting photocatalyst for this large water-splitting solar simulator? Question: What might serve as an initial starting photocatalyst for this large water-splitting solar simulator? Surely there must have been some planned experiments!

The Gizmodo article Insane Light System Blasts the Energy of 10,000 Suns and Phys.org's Let there be light: German scientists test 'artificial sun' both describe the large, powerful concentrated light source called Synlight which has been built by the German Aerospace Center (DLR).

This large (~350 kW!) solar simulator - built from 149 high intensity xenon arc lamps, all collimated/focused to a single 20x20cm area - will be used to test and develop materials and catalysts that will allow photochemical splitting of water to yield hydrogen gas, to be scaled up later and used with a solar concentrator instead of electrically powered lamps. Hydrogen could be used as fuel directly, or to synthesize hydrocarbon fuels when combined with carbon dioxide or monoxide.

Water is transparent over most of the sun's useful spectrum, and this is for the obvious reason that the wavelengths reaching the ground are those not already absorbed in the atmosphere. Thus a photocatalyst will be necessary for a high yield of hydrogen gas from concentrated solar light.

From the Gizmodo article:

  When the array is focused onto a single spot, it heats metal to 1,475 degrees Fahrenheit (800 degrees Celsius), which is then sprayed with water vapor. The metal reacts with the oxygen in water, and hydrogen remains. With further heating, the oxygen is once again separated from the metal.

There are certainly easier ways to heat a metal plate than building a giant solar simulator, so I assume the point is to provide an intense source of photons with a somewhat similar spectrum to sunlight as a test platform for various catalysts. But the quote suggests the surface will react with the oxygen released, and need to then be cycled to higher temperature to release it. This does not sound like a very practical photocatalytic process.

So far I have not found anything clearer about what kinds of catalysts are likely to be investigated, but there must be some potential candidates. Based on known photocatalytic water splitting mechanisms, or if someone can find a discussion by DLR, what would be a starting point, or proof-of-concept photocatalyst that might work here?

below: Images from here, photo credit: DLR (top), DPA (bottom). The shape of each reflector is likely to be close to an ellipsoid with the xenon arc at one focus and the sample near the other. This way a ~100 cm diameter reflector can illuminate a much smaller 20x20cm sample.

[

Vitamin B-12 is a photocatalyst, see, for example, https://pubs.rsc.org/en/content/articlelanding/2018/ra/c7ra13037f#!divAbstract and also https://www.ncbi.nlm.nih.gov/pubmed/29271445 , but it has not been applied to water splitting!

What has been used is TiO2, see this source to split H2O into H2 and O2.

Mechanics: The action of light on TiO2 creates electrons (e-) and electron holes (h+). Some reactions:

$\ce{H+ + e- -> .H}$

$\ce{.H + .H -> H2}$

$\ce{OH- + h+ -> .HO}$

$\ce{.HO -> H+ + .O-}$

$\ce{.O- + h+ -> .O}$

$\ce{.O + .O -> O2}$

One of the less desirable reaction is:

$\ce{.H + .HO -> H2O}$


qualitatively: Why are thioesters more reactive than regular esters? In ochem students are taught that shorter bonds (like alkynes) have more energy than longer bonds such as alkanes. However in biochem we are taught that thioesters are very energetic bonds and release alot of energy upon breaking.

Is there a specific mechanism that explains the relationships between bond energy and length?

background: a university student with a chem minor

http://guweb2.gonzaga.edu/faculty/cronk/CHEM440pub/thioesters.html

"thioester cannot make as much of a contribution because the required overlap between carbon's 2p orbital and the sulfur 3p orbital is ineffective. The large difference in energy between these two orbitals means they cannot "mix" as well as the 2p orbitals of oxygen and carbon."


How does expansion of electron shells work? How does it work? I've read that its about the similarity of the energy levels of $3d$ and $3p$, but shouldn't the electron shell $4s$ be filled before $3d$ if filled? So what is done with the $3d$?
I'd also like to know in what cases will atoms of elements expand their valence electrons? Which blocks or groups or periods do this?

You're right! The 4s subshell is filled before 3d. Electrons fill the shell   with least energy first and the 4s shell has a lower energy than the 3d subshell.
Use the Aufbau Principle to find the shell with least energy.

Basically it says that shells with a lower n+l (n and l are the principal and azimuthal quantum numbers) value are filled first.
If two shells have the same n+l value, then the shell with lower n is filled first.
There are a few exceptions to this rule - Copper, Chromium, Palladium, etc.

Reason: n+l gives the number of nodes in an orbital. The more the number of nodes, the less stable the orbital and so the higher the energy.
Valence shells are filled in the s and p blocks. The inner shells are filled in the d and f blocks.

Regarding #2:

Anything below the carbon row can expand their octets.  The question makes little sense in any other context.


How does High Frequency water electrolysis work? I have read that combining the DC current with a high frequency AC current, the electrolysis of water speeds up. Is this true? In that case, how is less energy wasted as heat? Or does it simply catalyze the process?

First of all, have a look at the wikipedia page on electrolysis of water.

I also like this review: Zoulias et al.: A Review on Water Electrolysis, TCJST, 4 (2) (2004) 41-71
Specifically they list a number of actually existing installations (context: renewable energy) and their actually achieved efficiency.
Speed up does not necessarily have anything to do with higher efficiency. In electrolysis it is often the other way round: if you want to squeeze out the maximum free energy, you need to do the reaction infinitely slowly (despite thermodynamics having a dynamic name, it looks at infinitively slow processes).
Thus, speeding up usually means that you find a way to put more power through your system. The big issue is to find a way of doing this without loosing (too much) efficiency.
Pulsed/modulated DC: looking through a few papers I liked this one: Shimizu et al.: A novel method of hydrogen generation by water electrolysis using
an ultra-short-pulse power supply, Journal of Applied Electrochemistry (2006) 36:419–423, DOI 10.1007/s10800-005-9090-

They aim at avoiding the diffusion controlled situation by having the pulses short enough so that no depletion zone occurs.
Look at these diagrams:

So they report one setting where the pulsed electrolysis is actually more efficient than DC in their cell.

There's more research going on on this, however the papers I found reported increased efficiency compared to pure DC electrolysis, but the absolute efficiencies are around 10%.   However, compare their numbers with the 80% efficiency cited by the review for an industrial alkaline electrolysis. Note that one big difference is the voltage that is applied: for the DC it is around 1.85 - 2.05 V, so much less overvoltage. Note also that when they say that higher voltage speeds up ion transport, then this overvoltage is converted to heat (ions face friction in the medium when they travel) and thus basically lost.
Another line that looks real is that if you go to higher temperatures, a (small) part of the energy can be supplied by heat. As heat is cheap, this may help. One point one has to be aware though, that the efficiency calculations may be done with respect to the electric energy only (neglecting the heat input) and thus look artificially nice (like efficiencies of condensing boilers calculated against the lower heating value).
I found a bunch of nonsense claims in the internet, about the resonance frequency of water helping to split bonds.

The first thing to realize here is that there is no one resonance frequency of water. With suitable energy, you can excite rotational, vibrational and electronic states (I left out translation - there transition energies minute). At room temperature you can say as a rule of thumb that most molecules will be in some excited rotational state, but in the vibrational and electronic ground states. Excitation energies for rotation are in the far infrared or microwave energy/frequency region. Widely used e.g. in the microwave oven at 2.45 GHz ($\approx$ 12 cm). Actually, the whole region is full of bands where water absorbs. Note that microwave heating of water does not cause electrolysis. Vibrational transitions are around 2.9 μm = 105 THz = 3500 cm⁻¹ and 6μm = 50 THz = 1635 cm⁻¹ with lots of combinations and overtones throughout the near infrared region. Quite exceptionally, the visible region is basically free of water absorption. Electronic transitions (breaking of bonds) need energies in the UV, and here we meet bands that lead to photodissociation, e.g. at 166nm (taken from Wikipedia). That corresponds to 1.8 PHz = $1.8 \cdot 10^{15}$ Hz. Compare this to the kHz and MHz where your link claims dissociation.

This doesn't mean that the pulsed DC cannot help, nor that impedance spectoscopy won't give important information. But resonance frequencies in the kHz range are electrical LC-circuit resonances depending on cell and electrode geometries and electrical double layers etc. But neiter on vibrations nor breaking of the bonds of the water molecule.
To give the "method" you ask about some real world numbers,

at the very end of the Wiki page the energy efficiency for industrial water electrolysis is cited as usually between 50 and 80 %.
The paper then proposes to burn the gas in an internal combustion machine. As such a stationary process could be adjusted so that the engine is at its maximum efficiency, we may assume 1/3 or 35% efficiency here.
we then need a generator to convert the mechanical energy into electric energy. Fortunately, that step is rather efficient. Say, 95 %.
A fuel cell would be more efficient than the combustion - generator combination: ca. 40 - 60 % according to Wikipedia.
Unfortunately, also battery charging is not 100% efficient. Let's assume 80–90% (taken from Wikipedia on Li-ion batteries) For batteries that are charged with higher current (or current density) efficiency is less. Example would be lead-acid batteries as used in cars. Wiki quotes efficiencies between 50 and 80 %.

Taking these numbers together, I conclude that after going once through the cycle of the proposed "perpetuum mobile", 8 - 24 % of the energy are retained in a "useful state" while 76 - 92 % became heat. With fuel cell, we may be able to "boost" the energy efficiency to 43%.

Useful general knowledge (in addition to the law of energy conservation)

The US patent system is different from e.g. the German patent system in that here in Germany the patent application has to have commercial/industrial usability. This includes a technical argument why it works (according to the physical laws). A perpetuum mobile would be rejected on these grounds (of course the inventor could prove his case with a prototype). US patents to not have this technical check.
generators (mechanic -> electric conversion) are sources of current, while batteries (galvanic cells) are sources of voltage.

I'm a plasma magrav physicist.  and something to try would be to take a copper tube fill it with neodymium magnets that that look like a donut. seal both ends of the copper tubing.  1 and should be permanently sealed.  nano coat the outside by using the vaporization method.  this will take about two weeks to accomplish.  once that has been done unseal the end that is not sealed perminently and remove the magnets. solder a wire to the inside of the nano coated tube however long the lead you may need. make copper oxide gas at nano state and fill the tube 30% of the way with this material, top it off with distilled water.  take a zinc wire or nickel wire and put that into the tube to make however long a lead you may need. now add a copper wire to the centerand keeping the copper wire and the nickel or zinc wire away from the walls of the copper tubeand from touching each other and seal the top with wax or epoxy or some other method to your choosing.  what this will do is it will give you the ability to pull an AC and DC voltage.  with this placed in the water with the leads coming out even partially in the water you will be able to measure the frequency of that water at that volume and then determine the properties necessary for your electrolysis to make it the most efficient.


What are the differences between monotropic and enantiotropic polymorphic transformations? I'm studying polymorphism and came across the terms "enantiotropic" and "monotropic" transformations. The first is described as reversible, whereas the latter, irreversible due to metastability. But what does "reversible" and "irreversible" means practically? What exactly is metastability? What makes these two types of polymorphism distinguishable between each other?

For instance, it's used as an example of monotropic polymorphism the transformation of graphite to diamond. However, the same source claims the reverse operation is possible. As for enantiotropic polymorphism, it used the example of silica polymorphs (cristobalite and alpha-quartz) which also can be transformed from one another. If both are reversible, why are they different?

In enantiotropic there exist an transition temperature between two crystalline state in equilibrium but reverese is true for monotropic


Why does the Birch reaction not yield fully saturated products? Why isn't benzene reduced all the way down to cyclohexane when performing a Birch reaction? (Why is it not reduced further?)

Similarly for the reaction with sodium dissolved in liquid ammonia, why isn't an alkyne reduced fully to an alkane?

The mechanism of the Birch reaction includes single-electron transfers from the sodium dissolved in ammonia to the aromatic compound. These transferred electrons will attempt to occupy the lowest possible orbitals.

In the case of benzene, we have a delocalised $\unicode[Times]{x3C0}$-system. Delocalisation can be understood here to mean higher occupied orbitals and lower unoccupied orbitals. Therefore, the electron will find an orbital at a lower energy and it will gladly hop in to create the radical-anion intermediat.

Once this is protonated, the remaining radical system can still be seen as a delocalised pentadienyl radical. Thus, the energy level a new electron will occupy is still rather low and another anion can be created. Further protonation yields the product of the Birch reaction, a cyclohexa-1,4-diene — most importantly, a non-conjugated system. This can be regarded as a pair of isolated double bonds.

The energy of the unoccupied $\unicode[Times]{x3C0}^*$ orbitals in an alkene seems to be just too unfavourable for a free electron to occupy them. I don’t have the facilities to do calculations (and neither am I firm enough to be able to do them) so I can’t provide you with values to compare, but it has to be the case otherwise the reaction would continue. If it were not so, there would be no reason for the reaction to stop.

The radical anion after the first step of the Birch reduction needs to be conjugated with (=stabilised by) another pi-system, otherwise it simply will not be formed from a pair of sp2 carbon atoms.


How adding one more electron increases the ionization energy? Elements of group 6A, compared to 5A, require less ionization energy due to the paired electrons of 6A.

But a question arises: why does group 7A, compared to group 6A, require more ionization energy in spite of the paired electrons in group 7a too.

Regardless of the general trend that the ionization energy increases as we go from left to right in periods, why didn't we apply the same procedure as the first comparison?

It's not just adding one more electron. Another proton is also added to the nucleus. The overall trend across a row is driven by shielding effects.

In general, as you go across a row, the shielding effect is fairly small compared to the effect of increasing nuclear charge.

As you note, there is a slight deviation in this trend between 5A (i.e., $\ce{N}$) and 6A (i.e., $\ce{O}$) because the extra electron in oxygen is paired. This increases the electron-electron repulsion, so the ionization energy decreases slightly.

As you note, $\ce{F}$ (7A) resumes the overall trend of increasing ionization energy driven by the attractive effects of the higher effective nuclear charge $Z_{eff}$.

Elements of group 6A (also called group 16) require less ionization energy than group 5A (group 15) elements because group 5A elements have a half filled p-orbital.

Group 7A (group 17) DOES NOT have paired electrons. Ionization energy of group 7A elements is more than that of group 6A elements due to higher effective nuclear charge on group 7A elements.


Bonding of Lithium and Argon I saw a meme that was joking around about Lithium and Argon bonding (see pic below). It got me wondering: Can Lithium and Argon bond in any circumstance?

Argon is a noble gas. Meaning, it does not simply "bond" with anything. These special gasses have a VERY strong resistance to joining with other atoms. So, to answer your question, no, you can not bond Argon and Lithium under any circumstance. Anyone who says they have done so is a LiAr ;)

Also, since Lithium is highly reactive and flammable and Argon is used in neon, I would not suggest trying to combine these two without a proper location and gear.


Separating iron from blood How can I separate iron from blood, without doing any damage to the blood/haemoglobin?

I have thought about using a magnet, but apparently iron in blood is not magnetic, and I have also looked at filtration techniques but not sure if that will affect the haemoglobin.

Updated Dec 2017: Let me add some context, I was/am looking for an alternative for chelation therapy such as Desferaloxamine infusions to remove iron in patients with iron overload as a result of repeated transfusions due to anemia

It is not possible to separate iron from blood without damaging the haemoglobin. Iron is a part of haemoglobin, and if iron is removed, the haemoglobin becomes something else. The blood then becomes incapable of transferring oxygen.

Using the reactivity series, just displace the iron with a more reactive metal.  The Haber process uses carbon to do this but as the heme molecule is mostly made of carbon with a central iron atom, this wouldn't do the trick.  Add heat, pressure and the more reactive metal and you'd get the iron.
http://www.bbc.co.uk/education/guides/zqjsgk7/revision