How to find sulfites in dried fruits? I am a student in my last year of high school and currently making a research on how to find sulfites in dried fruits. I've collected data and different methods online and combined few of them into one.
In my method I use hydrogen peroxide and barium chloride to create a white precipitate (result should be barium sulfite).

But the problem is that I never get any precipitate.
During my research it mentioned that I was to leave the dried fruit in distilled water over night (I used dried apricots).
I had 26g, 30g, 40g of dried apricot in 100mL of distilled water.
Once done, I filter out all of the liquids into a different beaker. Then filter it using filter paper. Once done I separate them into different beakers to test. And this is where I find the problem, I've done 6 trials so far, and all 6 have failed. Because I don't know what amount of H2O2 or what amount of the concentration I need to put into the apricot liquid for it to properly react, or for how long I should leave it in for it to react (I've waited 3 and 5 minutes). I'm then supposed to drop a single drop of barium chloride into the solution to produce the white precipitate. But so far I haven't gotten any sign of any precipitate. I've tried different amounts of H2O2 (1mL, 5mL, 10mL, 15mL, 20mL & 25mL with a 35% concentration in the first trials but rest of the trials were with a 3% concentration). Each with a drop of barium chloride.
Seeing that they did not react to a drop, I tried to use 5 drops each on the first few trials. Yet I still got no reaction.

Could I get some help or advice please?

Thank you

The colorimetric method using rosaniline and formaldehyde is more sensitive and it is well documented.

A somewhat safer version of the test (avoiding use of highly dangerous $\ce{HgCl2}$) is documented at MDPI - Open Access Publishing:

  "A concentrated pararosaniline stock solution (3.088 × 10 −3 mol/L) was made by dissolving pararosaniline hydrochloride in a 10% aqueous ethanol solution. The concentrated stock solution was used to make... [a] working solution contain[ing] pararosaniline hydrochloride, hydrochloric acid (HCl) and formaldehyde in the molar ratio of 1:1255:125 respectively. For example, a working solution with PRA concentration of 4.68 × 10 −4 mol/L in 25 mL contained 3,790 μL of concentrated PRA stock (3.088 × 10 −3 mol/L), 1,439 μL of 32% HCl, 189 μL of 36% formaldehyde solution and 25 mL with water."

Pararosaniline is available from Sigma-Aldrich, though it seems a bit pricey at ~$US82/25 g; perhaps your school would obtain some for use by yourself and by other students, as you would need perhaps a gram for your experiments.


How to make acetic acid more volatile so that it can be removed by vigorous aeration of an acetic acid-water mixture? I want to remove acetic acid from a mixture of acetic acid and water by aeration. The concentration of acetic acid is 200 ppm (mg/L). The pH of the water is more than 5. I want to reduce the acetic acid concentration by around 80%.
I understand that acetic acid can be removed by addition of sulphuric acid to bring down the water pH to around 2 and then boiling the mixture to separate the acetic acid.
I however the water quantity I have is very huge, so it is not practically possible to boil the water. The water temperature is more than 60 deg C. Also, I do not want to put lots of sulphuric acid in the water due to various other issues and my practical limit of using sulphuric acid is around 80-100 ppm which lowers the pH to around 3.0. I can also providing aeration to the water at 0.2-0.5 vvm.
I want to know that under these conditions, can acetic acid be removed from the water?
Is there any other other method by which we can remove acetic acid by aeration at 60 deg C, by adding not more than 100 ppm of any additive?
Is there any other chance that acetic acid would be removed by aeration at 50-70 deg C temperature? If yes how to enhance the efficiency by minimum inputs?

Aeration does not seem to be the way to go.
If there are no other ions in the water, electrolysis (Kolbe reaction) would convert the $\ce{HOAc}$ to ethane and carbon dioxide:
$$\ce{2CH3CO2^- -> 2e^- + 2CO2 + H3C-CH3}$$
This process could probably be arranged to be done in a pipe with electrodes on the sides, so that the water was treated as it was used, rather than in a huge tank, where diffusion would slow down the completion of the reaction.


Yeast Fermentation with YPD + Apple Juice I am working on a project for my General Chemistry class.

I am trying to compare the EtOH production (and the rate of EtOH production) between seven types of yeast strains by fermenting the YPD media, apple juice, or both, if necessary. I can describe the difference between the yeast strains if asked, but I don't think it is necessary right now.

Right now, I have only tested plain YPD media, which is 1% yeast extract, 2% peptone and 2% d-glucose mixture. All strains yield around 8-9 mg/mL of EtOH, and the fermentation stops after ~24 hours. My next step is to test YPD + apple juice media (because the project asks for "fun" element), and I was wondering if I should use the same method for analyzing this YPD + apple juice fermentation.

The method I used for YPD media is:

Grow each strain (total of 7 strains) in ~5 mL YPD media at 30 degrees shaker ~250 rpm overnight.
Subculture each strain into new fresh YPD media so that the new media is 50 mL and has 0.5 OD at 600 nm. Use the "special" flask designed for anaerobic growth. This flask is a 250 mL flask with a special cap. The cap is plastic at the outer radius and rubber at the inside. This allows me to grow yeast anaerobically and poke a needled syringe and retrieve small amount of sample without letting the air in.
Flush all flasks with nitrogen gas for four minutes.
After 6 hour, 12 hour, and 24 hour growths, use syringe to retrieve ~1.5 mL of sample.
Measure OD600 of each sample. (This is used to get the number of yeast cells in the media)
Centrifuge samples for 8 minutes in 14,000 rpm.
Pipette 1 mL of the supernatant into a GC vial.
Use HPLC to obtain concentration of ethanol. I am using refractive index detector, ion exclusion column(?) (I forgot the type of column used at this moment), and sulfuric acid as mobile phase.

For YPD + apple juice fermentation, I was thinking of doing the same procedure with 50 mL YPD + 50 mL apple juice in each "special flask". Will this work? Will this be good for the same HPLC setting?

Thank you for an especially clear and detailed question!

  For YPD + apple juice fermentation, I was thinking of doing the same procedure with 50 mL YPD + 50 mL apple juice in each "special flask". Will this work?

I think this would work but it seems like a better comparison to your earlier experiments to keep using 50 mL of total medium, i.e., mix 25 mL of YPD with 25 mL of apple juice.  That way the volume of the fermentation is the same in each experiment.

  Will this be good for the same HPLC setting?

Probably, although it depends on the components of the apple juice you use.  Have you injected just apple juice (unfermented) into your HPLC?  Have you verified that this negative control injection contains no ethanol or other interfering peaks in the region of interest?  If not, those are things you should do to make sure the HPLC method you have developed for YPD medium will still work in apple juice-based media.


How to connect proteins via disulfide bonds computationally? I want to connect proteins together to form a dimer. As seen in the picture, the monomers come close to each other along the edge of the higher-order structure (forget about the sulfate there). How can I find out, where I need to introduce an Xaa->Cys mutation so that the most stable disulfide bond is formed between the proteins? It is easy to get a visual clue where about a mutation might be good, but is there a good computational method to calculate the energies and scan through some possibilities?

(I would also go with hydrophobic or van de Waals interactions. Basically, I just want to glue those proteins together.)

If you have a structural model for the interaction, i.e. atomic coordinates, then you could use our software called Disulfide by Design.  See the following links.

http://cptweb.cpt.wayne.edu/DbD2/

http://www.biomedcentral.com/1471-2105/14/346


Is basic solution an electrode? Question:

  Calculate the potential of hydrogen electrode in contact with a solution of pH 10.

My attempt:

Cell reaction: $\ce{ H^+ + e^- -> \frac{1}{2}H_2 }$

I've attempted to solve this myself by using Nernst equation which yielded the answer -0.591 V. But I believe that Nernst equation is used when there are two electrodes, but my friends are saying that

"how will you make a cell from one electrode?".

So guys, please rectify the problem.

"how will you make a cell from one electrode?".

There is no need to, because there is nothing which prohibits you from writing a Nernst equation for a half cell.* Remember the rule no.1 in electrochemistry, whenever you mention half-cell potential, your next automatic question should be "with respect to what?". If the textbook is quiet and you are required to look up tables, it is understood that it is with reference to the hydrogen electrode under standard condition which is set at 0 V (exactly).

Now at pH 10, your standard conditions have changed and the activity or hydrogen ion concentration differs. It is no longer zero. So proceed with the Nernst equation for the half cell and work it out.

*(Yes, there is no single electrode galvanic cell, but for fun and general knowledge, people have done single electrode reduction using electrostatics (Bard and others). Faraday did it centuries ago. It was published in a very reputable magazine.)


Why does zinc react while its outermost suborbital is full? Why does zinc lose electrons while its outermost configuration is 4$\rm{s}^2$ 3$\rm{d}^{10}$ both outermost suborbital are full.

Why does it react or lose electrons?

You could ask the same question about the alkali earth metals with their $\rm{s}^2$ valence shell. While there is an energy cost to removing electrons, in a spontaneous reaction (e.g. $\ce{Cu^2+_{(aq)} + Zn_{(s)} -> Cu_{(s)} + Zn^2+_{(aq)}}$) there will be a compensatory transfer of energy elsewhere (like the reduction and deposition of $\ce{Cu^2+}$) which leads to a more stable overall system.


What does the word carat really mean? For gold, the word carat defines its purity whereas in diamonds it indicates its weight.  This seems inconsistent -- what does the unit 'carat' actually refer to?

The karat is a unit used specifically for describing the purity of gold alloys:

Various ways of expressing fineness have been used and two remain in common use: millesimal fineness expressed in units of parts per 1,000 and karats used only for gold. Karats measure the parts per 24, so that 18 karat = 18/24 = 75% and 24 karat gold is considered 100% gold.
The karat (not carat as a unit of mass) (symbol: K or kt) (US) or carat (symbol: C) is a unit of purity for gold alloys.

Note that in the U.S., the symbol K or kt is used; in your locale, the symbol C and the word carat might be used, from which confusion might arise when talking about the unit of mass carat.
The carat is a unit of mass used for, well, telling us what the mass of gemstones are:

The carat (ct) is a unit of mass equal to 200 mg (0.2 g; 0.007055 oz) and is used for measuring gemstones and pearls. The current definition, sometimes known as the metric carat, was adopted in 1907 at the Fourth General Conference on Weights and Measures, and soon afterwards in many countries around the world. The carat is divisible into one hundred points of two milligrams each. Other subdivisions, and slightly different mass values, have been used in the past in different locations.


Why Li can't make it's Peroxide when heated with more O2? Every first group metal reacts with $\ce{O2}$ and makes there monoxide at room temperature.

But it says that $\ce{Li}$ does not make peroxide when heated with more $\ce{O2}$. But other metals in that group do.

Why $\ce{Li}$ can't make it's Peroxide when heated with more $\ce{O2}$?

And if we give the heat continuously with more $\ce{O2, K, Rb, Cs}$ give there Super-oxides. In that time Both $\ce{Li}$ and $\ce{Na}$ do not gives there Super-oxides. What is the reason for it?

It has to do with the small size and high charge density of the lithium ion, which effectively polarizes any peroxide ion that might form and thus pulls it apart.  See http://www.chemguide.co.uk/inorganic/group1/reacto2.html.


How are reactivity and dipole moment related? Recently, I came across a question:  which of the two compounds has a greater reactivity; $\ce{NH_{3}}$ or $\ce{NF_{3}}$?

It is known that despite the same structure the dipole moment of $\ce{NH_{3}}$  is greater because of the electronegativity difference. So, I though that the reactivity of $\ce{NF_{3}}$ should be more as it is easy to break the bond due to the high electronegativity of fluorine but the answer turned out to be $\ce{NH_{3}}$ and I can't figure out why.

The reactivity of amines is generally based on how nucleophilic they are.  In turn, their nucleophilicity is dependent upon how much of the nitrogen lone pair of electrons really exists on the nitrogen atom (e.g. the lone pair electron density on nitrogen). In ammonia ($\ce{NH3}$), the lone pair electron density on nitrogen is very high making ammonia quite nucleophilic, quite reactive.  In the case of nitrogen trifluoride ($\ce{NF3}$), the three extremely electronegative fluorine atoms inductively remove much of the electron density from nitrogen; therefore $\ce{NF3}$ is less reactive than ammonia in a nucleophilic sense.


Name of an unsaturated cyclic compound I deduced the name as cyclodeca-1,6-dien-3-yne. Is it correct?

You have the right name unless you need to specify the geometric isomer.

If I may stretch you a little, consider whether to specify the relative positions of the atoms at the double-bonds, e.g. (1Z,6Z)-cyclodeca-1,6-dien-3-yne. It's not possible to tell which geometric isomer it is from your drawing, because you would need a clear depiction of the bond angles around the double-bonds. If this drawing is all you have to go on, you can't specify the isomer and your answer is spot on.

The below image is a model of your molecule showing the (1Z,6E)-cyclodeca-1,6-dien-3-yne form. You can see the difference between E ("opposite" or in this case "trans"), on the left, and Z ("together" or in this case "cis"), on the right, by looking at the shape of the chain of the four carbons nearest each double-bond. Of course, there is more to geometric isomerism than this but I hope this is a useful introduction if you've not encountered it before.


What is the probability of 3 possible products, between two chemical species? I am not a chemist. I hope I will be specific enough.

Suppose there are two chemical species $\ce{A}$, $\ce{B}$ with the following properties:

at temperature $t < T_r$, no reaction occurs between $\ce{A}$ and $\ce{B}$ (in any combination).
at $t\ge T_r$, $\ce{A}$ interacts with itself to create $\ce{A_2}$, $\ce{B}$ reacts with itself to create $\ce{B_2}$, and $\ce{A}$ and $\ce{B}$ are reacting to create $\ce{AB}$.
$\ce{A_2}$, $\ce{B_2}$ and $\ce{AB}$ are never reacting.

In experiment, we first mix $\ce{A}$ and $\ce{B}$ in temperature $t<T_r$. Amounts of species mixed are $a$ for $\ce{A}$, $b$ for $\ce{B}$. Then, we add heat to obtain temperature $t\ge T_r$ and start the reaction.

What amounts of $\ce{A_2}$, $\ce{B_2}$, and $\ce{AB}$ can be expected to be produced?
To obtain the amounts, should probability theory be used? E.g., amount of $\ce{AB}$ equals to probability that species $\ce{A}$, $\ce{B}$ will interact ("collide" or similar interpretation).

Assume the rates of the reactions are equal.

Well, if you assume the rates are known and the reactions' order follows from stoechiometry (e.g. if they are elementary reactions), you can put the chemical kinetics into simple equations:

$$\frac{\mathrm da}{\mathrm dt} = -k_1 a(t)^2 - k_3 a(t)b(t) $$
$$\frac{\mathrm db}{\mathrm dt} = -k_2 b(t)^2 - k_3 a(t)b(t) $$

($t$ here being time, not temperature).

Knowing initial amounts or concentrations $a_0=a(t=0)$ and $b_0=b(t=0)$, you can pretty much integrate  the system to find out what happens.

Edit — solving this system for $k_1=k_2=k_3$ yields the quantities of AA, AB and BB at infinite time to be as follows:

$$aa = \frac{a_0^2}{a_0+b_0}$$
$$ab = \frac{a_0 b_0}{a_0+b_0}$$
$$bb = \frac{b_0^2}{a_0+b_0}$$


Are aromatic amines or amides less basic? When the lone pair of an amino group $\ce{R-NH2}$ is involved in resonance, its basicity decreases. But, between conjugation with carbonyl group $\ce{R}=\ce{R'CO}$ or resonance with benzene $\ce{R} = \ce{Ph}$, which one will reduce the basicity of amino group to a greater extent?

Measures for acid and base strength

First, you need to be clear about 'strong' and 'weak' acids.
In order to measure the strength of an acid relative to water and find out how effective a proton donor it is, you must look at the equilibrium constant for the reaction

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

The position of equilibrium is measured by the equilibrium constant for this reaction $K_{\text{eq}}$.

\begin{equation}
  K_{\text{eq}} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{H2O}][\ce{HA}]}
\end{equation}

The concentration of water remains essentially constant (at $55.56 \, \text{mol} \, \text{dm}^{–3}$) with dilute solutions
of acids wherever the equilibrium may be and a new equilibrium constant, $K_{\text{a}}$, is defined and called the acidity constant.

\begin{equation}
  K_{\text{a}} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}
\end{equation}

Like $\text{pH}$, this is also expressed in a logarithmic form, $\text{p}K_{\text{a}}$.

\begin{equation}
  \text{p}K_{\text{a}} = -\log (K_{\text{a}})
\end{equation}

Because of the minus sign in this definition, the lower the $\text{p}K_{\text{a}}$, the larger the
equilibrium constant, $K_{\text{a}}$, is and hence the stronger the acid. The $\text{p}K_{\text{a}}$ of the acid is the $\text{pH}$ where it is exactly half dissociated. At $\text{pH}$s above the $\text{p}K_{\text{a}}$, the acid $\ce{HA}$ exists as $\ce{A^{–}}$ in water; at $\text{pH}$s below the $\text{p}K_{\text{a}}$, it exists as undissociated $\ce{HA}$.

An acid's $\text{p}K_{\text{a}}$ depends on the stability of its conjugate base.
$\ce{HCl}$ is a much stronger acid than acetic acid: the $\text{p}K_{\text{a}}$ of $\ce{HCl}$ is around $–7$ compared to $4.76$ for acetic acid. This tells you that in solution $K_{\text{a}}$ for hydrogen chloride is $10^{7} \, \text{mol} \, \text{dm}^{–3}$ whilst for acetic acid it is only $10^{–4.76} = 1.74 \cdot 10^{–5} \, \text{mol} \, \text{dm}^{–3}$. But, why are the equilibria so different? Why does hydrogen chloride fully dissociate but acetic acid do so only partially?

\begin{equation}
  \ce{HCl (aq) + H2O (l) <=>> H3O+ (aq) + Cl- (aq)} \quad \qquad \qquad K_{\text{a}} = 10^{7} \, \text{mol} \, \text{dm}^{–3} \\
  \ce{CH3COOH (aq) + H2O (l) <<=> H3O+ (aq) + CH3COO- (aq)} \qquad K_{\text{a}} = 1.74 \cdot 10^{–5} \, \text{mol} \, \text{dm}^{–3} \\
\end{equation}

The answer must have something to do with the conjugate base $\ce{A^{–}}$ of each acid $\ce{HA}$, since this is the only thing that varies from one acid to another. In both the equilibria above, water acts as a base by accepting a proton from the acid. For the hydrochloric acid equilibrium in the reverse direction, the chloride ion is not a strong enough base to deprotonate the hydronium ion. Acetate, on the other hand, is easily protonated by $\ce{H3O+}$ to give neutral acetic acid, which means that acetate must be a stronger base than chloride ion.
A nice thing to remember:

The stronger the acid $\ce{HA}$, the weaker its conjugate base, $\ce{A^{–}}$
The stronger the base $\ce{A^{–}}$, the weaker its conjugate acid $\ce{AH}$

So, the $\text{p}K_{\text{a}}$ is a good measure for the strength of an acid: the lower the $\text{p}K_{\text{a}}$ the stronger the acid.

But what about bases? To say something about the strength of a base it is convenient to take the following viewpoint: to what extent does a protonated base want to keep its proton or in other words: how big is the acid strength of its conjugate acid? For example if you want to know which is the stronger base - say, formate anion or acetylide anion - you look up the $\text{p}K_{\text{a}}$s for their conjugate acids. You find that the $\text{p}K_{\text{a}}$ for formic acid ($\ce{HCO2H}$) is $3.7$, whilst the $\text{p}K_{\text{a}}$ for acetylene is around $25$. This means that acetylene is much more reluctant to part with its proton, that is, acetylide is much more basic than formate. This is all very well for anions - you simply look up the $\text{p}K_{\text{a}}$ value for the neutral conjugate acid, but what if you want to know the basicity of ammonia? If you look up the $\text{p}K_{\text{a}}$ for ammonia you find a value around $33$ but this is the value for deprotonating neutral ammonia to give the amide ion, $\ce{NH2^{–}}$.
If you want to know the basicity of ammonia, you must look up the $\text{p}K_{\text{a}}$ of its conjugate acid, the ammonium cation, $\ce{NH4+}$, protonated ammonia. Its $\text{p}K_{\text{a}}$ is $9.24$ which means that ammonia is a weaker base than hydroxide - the $\text{p}K_{\text{a}}$ for water (the conjugate acid of hydroxide) is $15.74$.
In conclusion: the higher the $\text{p}K_{\text{a}}$ of the conjugate acid the stronger the base.

The answer to your question

The $\text{p}K_{\text{a}}$ of the conjugate acid of aniline is $4.6$. So, since this value is much lower than the $9.24$ you had for ammonia, it is much less basic than ammonia, as is expected since the lone pair of aniline is delocalized into the phenyl ring. But for amides the $\text{p}K_{\text{a}}$ values of the conjugate acid are typically smaller than $0$, so an amide is much less basic than aniline. This is to be expected because the carbonyl group contains an electronegative oxygen atom and thus has a lower lying LUMO than the phenyl ring which can interact more efficiently with the nitrogen's lone pair, thus spreading the negative charge more evenly among the conjugated groups and leading to a higher stabilization of the molecule.


Detection of potassium ions I've bought a white powder. The label on the package states that the powder contains sodium chloride, potassium chloride and potassium citrate. How can I make sure that the powder does indeed contain potassium chloride? I'm mainly interested in the presence of potassium ions.

What do you want?

If you're just looking for a source of potassium ions you could just buy $\ce{KCl}$ or some other compound, they're relatively cheap if not pro analysis. If you actually want to test for $\ce{K+}$:

What else do you know about the product?

Tests are completely different for different amounts of $\ce{K+}$. Does the vendor specifies the amount of each compound on the product's label?

How sure do you need to be?

Do you need to do a QUALITATIVE analysis or a QUANTITATIVE analysis? If you need to be absolutely sure about the concentration then you need to guess it's ridiculously low - you'll probably need at least AAS (Atomic Absorption Spectroscopy) equipment, which can't be accessed by anyone nor find anywhere, but you can order an analysis. Alternatively, you can use an electrode - keep reading .

Some ways of testing that I can think of:

(Of course you'll need to dissolve the powder in at least distilled water to do all the following tests, and do a blank with the same water - the more pure the water the more accurate your results.)

For big amounts of $\ce{K+}$

The flame test suggested by Paul J. Gans on the comments above could work, although I never actually did it.

Another way is doing the following titration:

$$\ce{KCl(aq) + AgNO3(aq) -> AgCl(s) + KNO3(aq)}$$

$\ce{AgNO3}$ can be found at any store that sells chemicals. We used 0.1 M for small amounts of $\ce{Cl-}$.

What you are actually testing for is $\ce{Cl-}$ ions, but if the amount of K = the amount of $\ce{Cl}$ the job is done. Now, since the powder contains potassium citrate this is just a qualitative test, although it can be very accurate for small concentrations if properly executed (see Mohr method). Silver chloride is a white solid that is barely visible in small amounts, and a similar  titration's end point is just a few drops (Mohr method), and there are a lot of possible interference so you should not expect precision unless you do some pro work.

For small amounts of $\ce{K+}$

I think the best way of doing this is using a potassium-selective electrode. I never used one of those, but other ion-selective electrodes work great on water analysis, detection levels were a few mg/L on our old ones!

http://en.wikipedia.org/wiki/Potassium_selective_electrode

And of course, you can use AAS.


Does burning graphite give off carbon monoxide? Put $\pu{3.7V}$ through a graphite rod, I know graphite doesn't burn but it did smoke. So I automatically assumed it to have gave off $\ce{CO}$, so I've aired out the house $\ce{CO}$ Detector, but it did not go off. I know they only detect above $\pu{30PPM}$, but does burning heating graphite rods give off $\ce{CO}$?

It could do - but I doubt you had enough carbon in the rod for any significant risk . Hence no detector signal.
( carbon monoxide is normally formed where oxygen is restricted in the combustion process )


Among HClO4 and HClO2 which is the stronger acid if we do not know Ka values? I have two acids: $\ce{HClO4}$ and $\ce{HClO2}$ how do I determine which is the strongest acid without knowing $K_a$? The two hydrogens are bound both to an oxygen so there isn't anything to differentiate them, how can I  proceed?

$\ce{HClO4}$  is stronger acid due to the formation of more stable conjugate base. Also there are more number electronegative oxygen atoms in case of $\ce{HClO4}$ , so acidity should be more.


What's the chemistry behind only charging a Lithium-ion battery to 80% capacity at most, to increase its lifespan? What's the chemistry behind only charging a Lithium-ion battery to 80% capacity at most, to increase its lifespan?

Over on the Skeptics StackExchange, a question has come up about charging Lithium-ion cells to just 80% of their maximum, in order to extend their lifetime.

Is there a sound basis in chemistry for such a recommendation, and if so, what is it?

Most commercially-available lithium-ion batteries use graphite as the anode (negative electrode, where the lithium ions go when the battery is charged) and $Li Co O_2$ as the cathode (positive electrode, where the lithium ions go when the battery is discharged). Both graphite and lithium cobalt oxide have a structure composed of layers stacked on top of one another; the lithium ions fit in between these layers in a process called intercalation.

The lithium sitting between the layers props the layers apart. If too much of the lithium is removed from either the cathode (over-charging) or the anode (over-discharging) then the layers can collapse back together and it becomes hard to put the lithium back in. Normally batteries have control circuitry to prevent the cells from over-charging or over-discharging but this circuitry is not always perfect, particularly for smaller, simpler battery packs.

That said, AFAIK the circuitry is already set to limit charging and discharging to well below the theoretical maximum capacity, so I would not recommend limiting the charge any further. Perhaps unscrupulous vendors will cut the amount of active material inside their batteries while keeping the rated capacity the same and therefore run the cells closer to their theoretical maximum, in which case keeping charge and discharge within their rated limits would be a good idea, but it should not be necessary for high-quality batteries.


Chlorination of Water Does passing chlorine gas through water form hydrochloric acid or bleach? Also, will a mixture of hydrogen and chlorine gas from electrolysis of brine do the same?

Passing chlorine gas through water yields a solution of $\ce{Cl2}$, hydrochloric acid ($\ce{HCl}$) and hypochlorous acid ($\ce{HOCl}$). The latter two are formed by reaction of chlorine with water under disproportionation.

$$\ce{Cl2 + H2O \rightleftharpoons HCl + HOCl}$$

Bleaches are chemicals which remove color by oxidation of dyes. $\ce{Cl2}$ and $\ce{HOCl}$ themselves are already strong oxidizers, but adding alkali metal hydroxides to an aqueous chlorine solution increases the solubility of $\ce{Cl2}$, and chlorine bleach is obtained, which contains alkali metal hypochlorite as active ingredient.

$$\ce{Cl2 + OH- \rightleftharpoons ClO- + Cl- + H2O}$$

Sodium hypochlorite is produced by this reaction and common household bleach is an aqueous 3-8 mass-% solution of the salt. The necessary chlorine gas is actually produced by electrolysis of $\ce{NaCl}$.

While gaseous $\ce{Cl2}$ itself can act as a bleaching agent by oxidation, the use of hypochlorite solutions has the benefit of avoiding handling of the poisonous gas. $\ce{H2}$ is a much better reducing agent, and therefore unlikely to be useful for bleaching.

A mixture of $\ce{H2}$ and $\ce{Cl2}$ will react to form gaseous hydrogen chloride when exposed to UV light. Hydrochloric acid is obtained when gaseous $\ce{HCl}$ is dissolved in water.

$$\ce{HCl + H2O \rightleftharpoons H3O+ + Cl-}$$


Can I react a Grignard reagent (ethylmagnesium bromide) with water and then ammonia? I came across this question where I had to convert $\ce{CH3-CH2-Mg-Br}$ to $\ce{CH3-CH2-NH2}$.
If I need to do so can I just react the given compound with water to get ethane and then react it with $\ce{NH3}$.

If the answer is no then why can't I?
Is there any set of principles in organic chemistry which govern as to which substances can be replaced with what?
Kind of like the reactivity series of metals?
For e.g.: If I react an alkane with a halogen to get a haloalkane and then react this back with hydrogen to get the alkane then I don't think so it would be correct to do so. And I don't think that organic chemistry works that way.
But how do I understand while solving a question whether I am going in the right direction.
Please help.

NB: This is not a homework question its just a question I stumbled across while my preparation for a college entrance test. (edit the title maybe)

I came across this question where I had to convert
  $\ce{CH3-CH2-Mg-Br}$ to $\ce{CH3-CH2-NH2}$. If I need to do so can I
  just react the given compound with water [...]

Always be cautious about unwanted, side reactions that may occur. Proton exchanges are one huge class of possible side reactions. Protons transfers occur readily in many circumstances because the protons is small and can be shuffled around without much to get in its way, physically.

So in the case of your hypothetical reaction between a Grignard reagent and water - realize that Grignards are strongly basic and therefore will deprotonate water. Unfortunately protonated Grignards are no good in terms of reactivity. They just fall back apart into hydrocarbons and a basic halide salt.

Further reading regarding Grignards - It is tempting to use Grignards to create carbon-carbon bonds, but anything that is acidic as water or more acidic will be deprotonated by the Grignard before any carbon-carbon bonds are created. When preparing Grignards, glassware must be perfectly dry. Not just wiped dry with a paper towel but heated dry. Grignards also live in ether solvents since ethers are only very weakly acidic.

Grignards may be used to attack aldehydes and ketones in creating carbon-carbon bonds, but realize that if the aldehyde or ketone contains a functional group that is about as acidic as water or more so - then the first equivalent of Grignard you add isn't going to the aldehyde or ketone carbon - it's going to perform a proton-transfer.

  Is there any set of principles in organic chemistry which govern as to
  which substances can be replaced with what?

In general, it is good to keep in mind the motif nucleophile attacks electrophile in introductory organic chemistry.

Exceptions to motif: nucleophile attacks electrophile

Other heuristics you can use in predicting what reactions might commence ...

Bond energies. This is mainly useful for studying radical reactions - to which the motif nucleophile attacks electrophile doesn't apply as well. Are we going from weaker to stronger bonds? That's a favorable move, enthalpically.
Polarities. This falls under nucleophile attacks electrophile.
Kinetics. It's good to learn the difference between thermodynamics and kinetics early in organic chemistry. Sometimes chemical entities may be thermodynamically unstable but kinetically stable. Maybe a carbon is particularly electrophilic, but it's too well "hidden" by neighboring groups (steric hindrance). So will a nucleophile attack this carbon? Probably not to a great extent.


Alloy-component weight based on desired molecule I'm a gold smith and I want to start experimenting with non-standard metal-alloys. Now high school is a long time ago for me so while I think I've got this right, I'm just not as sure as I should be when working with precious metals costing 40 dollar a gram. So could anyone confirm that what I'm doing is right:

I have two metals
Metal X, with an atomic-weight of 20
Metal Y, with an atomic-weight of 3

If I want to create an intermetallic alloy of $\ce{XY2}$, where one atom of X is combined with two atoms of Y and I'd want 10 grams of that alloy. I would alloy (20*10)/26=7.692g of metal X and (6*10)/26=2.308g of metal Y, correct?

Yes, you are correct.

7.692 g / (20 g/mol) = 0.3846 mol X (or 2.316 * 10^23 atoms X)

2.308 g / (3 g/mol) = 0.7693 mol Y (or 4.633 * 10^23 atoms Y)

Moles are just a count of atoms, so if you have two moles of Y per mole of X, you have two Y atoms per X atom.


Mechanism for Friedel-Crafts Acylation in Synthesis of Sertraline In the first step of the synthesis of Sertraline, aluminium chloride catalyst was used in a Friedel-Craft's reaction to give the following product. My question is how does the exact mechanism work? 1-napthol does not seem look like an electrophile to me, and I do not understand how AlCl3 is able to turn 1-napthol into an electrophile. The best guess I have is that the phenol tautomerise to give a ketone, and AlCl3 forms a complex with it. Yet again, I do not understand how the γ carbon is made electrophilic using AlCl3. Any help would be appreciated.

The alleged original RSC source for this reaction was not accessible online.1. However, Olah et al. have reported2 that α-naphthol 1 in superacid medium forms a dication species 4. The use of excess AlCl3 provides a mechanism for the formation of superacid 3. Benzylic dication 4 undergoes substitution of o-dichlorobenzene 5 which ultimately forms the sertraline precursor, tetralone 8. For reaction conditions, see ref. 3. No reducing agent is required.

1) Kavitha, C. V.; Mantelingu, K.; Sarala, G.; Naveen, S.;
Anandalwar, S. M.; Prasad, J. S.; Rangappa, K. S. J. Chem. Res. 2006,
11, 730.
2) Koltunov, K. Y.; Chernov, A. N.; Prakash, G. K. S.; Olah, G. A. Chem. Pharm. Bull. 2012, 60, 722.
3) Lee, S. H.; Kim, I. S.; Qing, R. L.; Dong, G. R.; Jeong, L. K.; Jung, Y. H. J. Org. Chem., 2011, 76, 10011.


Why do atoms on the periodic table have the number of valence electrons that they do? I'm studying year 11 and am learning about valence electrons and how these electrons drive intermolecular bonds.
What I have not been able to find an explanation to is why all the elements on the periodic table have the number of valence electrons that they do.
To me, it seems intuitive that if the halides are so eager to gain an electron, why would their natural state not be to have the 8th electron in their shell?

Atoms have protons (and neutrons) in the nucleus that define the element. For example, Fluorine always nine protons. The number of electrons must balance the number of protons in the atomic state. If fluorine had ten electrons, i.e. a full inner and outer shell, it would be a negative ion with net charge -1, so the electrostatic charge would repel the last electron.
It is only with the help of other atoms nearby that charges can be (almost) balanced. In $\ce{LiF}$, the fairly positively charged $\ce{Li+}$ ion attracts the comparatively negatively charged $\ce{F-}$ ion, so there is no net unbalance in a crystal of $\ce{LiF}$, though in each pair, the valence electron spends most time in the outer shell of fluorine.


What is the oxidation state of copper in Gilman's Reagent? What is the oxidation state of copper in Gilman's reagent? In $\ce{R-Cu}$ bond the oxidation state is +1. Thus in $\ce{R-Cu-R}$ bond the oxidation state has to be +2. But if that so why positive $\ce{Li}$ act as counter ion in Gilman's reagent? And why is it named as dialkyl cuprate (-ate for anionic complexes)?

Don't count bonds.  Count electrons.

Copper has $11$ valence electrons as a neutral atom, as the $d$ orbitals are able to participate in reactions.  In the oxidation state formalism the alkyl groups, whose atoms are more electronegative than copper, are each considered to take an electron away from the copper to complete their carbon octets.  But then the lithium, bonded to copper (bonding lithium to the alkyl groups would give a separate organolithium molecule instead), adds an electron.  So the copper goes from $11$ valence electrons down to $10$ in the oxidation state formalism, losing one overall; the oxidation state of copper is thereby $+1$.


How to calculate Ka for hydronium and Kb for hydroxide? I am wondering how to calculate the $K_a$ for hydronium, $H_3O^+$, and the $K_b$ for hydroxide. Because hydronium undergoes the equilibrium

$\ce{H_3O^+(aq) + H_2O (l) <=> H_2O (l) + H_3O^+ (aq)}$ ,

isn't the constant  exactly 1? (And the same thing for hydroxide?)
However, Wikipedia gives the $pK_a$ for hydronium to be -1.7. Does it have something to do with the "concentration" of water, which is 55.6 M, and upon taking the negative natural log, is -1.74?

To an extent, you are right; the calculation of the $K_\mathrm{a}(\ce{H2O})$ does include $[\ce{H2O}]$.

Consider the following equation:
$$\ce{H2O(l) +H2O(l) ->H3O+(aq) + OH- (aq)}$$

The equilibrium for this equation is:

$$K_\mathrm{a}(\ce{H2O})=\frac{[\ce{H3O+}]\cdot [\ce{OH-}]}{[\ce{H2O}]\cdot [\ce{H2O}]}$$

Since $K_\mathrm{w}=[\ce{H3O+}]\cdot [\ce{OH-}]=1.01 \times 10^{-14}$ at $25^\circ \mathrm{C}^{[1]}$, we can make such a substitution in our equation:

$$K_\mathrm{a}(\ce{H2O})=\frac{1.01 \times 10^{-14}}{[\ce{H2O}]\cdot [\ce{H2O}]}$$

Now here you will have to bear with me. In a usual $K_\mathrm{a}$ calculation, $[\ce{H2O}]$ is disregarded, as it remains mostly constant ($55.5084\ \mathrm{M}$), except when working with extremely concentrated solutions. In this equation, there are two $[\ce{H2O}]$ terms, and only one of them is disregarded (as a solvent). The other one can be considered concentration of the acid. Therefore:

$$K_\mathrm{a}(\ce{H2O})=\frac{1.01 \times 10^{-14}}{[\ce{H2O}]}=\frac{1.01 \times 10^{-14}}{55.5084}=1.819544 \times 10^{-16}$$

Thus we have calculated the $K_\mathrm{a}$ for water, or the $K_\mathrm{b}$ for $\ce{H3O+}$. The relation between these is:

$$K_\mathrm{w}=K_\mathrm{a}\cdot K_\mathrm{b}$$

So:

$$K_\mathrm{a}(\ce{H3O+})=\frac{1.01 \times 10^{-14}}{1.819544 \times 10^{-16}}=55.5084$$

and:
$$-\log(K_\mathrm{a}(\ce{H3O+}))=\mathrm{p}K_\mathrm{a}(\ce{H3O+})=-\log(55.5084)=-1.7404$$

The calculation for the $K_\mathrm{b}(\ce{OH-})$ is exactly the same for the calculation of $K_\mathrm{a}(\ce{H3O+})$ (pretty much just replace all the $\ce{H3O+}$ with $\ce{OH-}$) and yields the same result.

[1] Starkey, R.; Norman, J.; Hintze, M. Who Knows the Ka Values of Water and the Hydronium Ion? J. Chem. Educ. Journal of Chemical Education. 1986, 63, 473.


What does the phrase "reducing atmosphere" mean in quantitative terms? There is a debate about "how reducing" the atmosphere of the early Earth was, and this question is my attempt to grasp toward an understanding of what that that phrase really means.  As a bit of context, the Miller–Urey experiment produces a wide variety of complex organic molecules, but people say it requires a very reducing atmosphere to work, and it's disputed whether the early Earth's atmosphere was reducing enough for analogous processes to occur. Ultimately I would like to understand what it is about reducing atmospheres that allows complex organic molecules to be created so easily, but the first step is to understand what it really means to say one mixture of gases is "more reducing" than another.

If anything can be more reducing than something else, it seems like $\ce{CH4}$ gas should be more reducing than $\ce{O2}$, so I'm trying to understand the extent to which this can be quantified. However, I'm currently self-learning chemistry, and I'm having difficulty seeing connections between some of the concepts. (So apologies in advance for the rambling nature of this question.)

It seemed like one place to start might be to draw the net reaction of methane oxidation:

Obviously, if you add up the total oxidation state of every atom, you get zero - the carbon gets oxidised, the oxygen gets reduced, and the hydrogen stays the same. This must always be the case for any reaction (right?) since electrons are conserved. At first this made me think that maybe you can't really say that one thing is objectively "more reducing" than another. Though of course you can still talk about its tendency to oxidise or reduce carbon, and maybe that's what people are talking about when they say an atmosphere is "reducing" or "oxidising".

But then I came across the concept of redox potential, which does seem to be about things being objectively more reducing or oxidising than others. The problem is, the discussions of the concept that I can find online seem to be in terms of electrons "being transferred" from one molecule to another. But in the reaction above, as far as I can see, electrons aren't being transferred from the $\ce{O2}$ to the $\ce{CH4}$ but instead both molecules are being destroyed and replaced with completely different ones. So I'm having trouble seeing whether, or how, the concept of redox potential can be applied in this case.

Given this confusion, I guess my specific questions are:

When people talk about an atmosphere being "more reducing" than another, are they referring to redox potential or something else, such as the degree to which the atmosphere reduces or oxidises carbon? If it's the latter, is it something that can be quantified, or is it more of a subjective judgement?; and
What is the relationship between the concepts of the redox potential of a substance on the one hand, and the oxidation states of atoms on the other? Can the concept of redox potential be applied to gases like $\ce{CH4}$ and $\ce{O2}$?

I’m an environmental chemist and we frequently refer to ‘oxidizing’ or ‘reducing’ conditions in the environment. I’ll use soils as an example. The ‘oxic zone’ is the area of soil that oxygen can reach by diffusion and is seen as being under ‘oxidizing’ conditions because the reduction of $\ce{O2}$ is the dominant redox reaction that occurs in that zone. Elements that have multiple oxidation states tend be in the higher oxidation states. For example, sulfur will exist primarily as sulfate ($\ce{SO4^2-}$). As you get further away from the Earth’s surface, the oxygen is depleted (by microbes) and the redox conditions change: the dominant reduction reaction changes. Usually in this order: $\ce{O2}$, nitrate, manganese(III,IV), iron(III), sulfate and finally $\ce{H2}$. When the conditions become so reducing that the microbes are using $\ce{H2}$ as an electron acceptor, they making methane via a process called methanogenesis. This order arises because of reduction potentials. Microbes get the most energy for the least effort from reducing $\ce{O2}$, so they do that until it runs out, which is when they switch to the next best thing available.

Ultimately what it means to be a ‘reducing environment’ is that if you added a compound with a higher reduction potential then the dominant redox pair, it would become reduced. For example, if you could put $\ce{SO4^2-}$ in a methanogenic environment, it would be reduced to $\ce{H2S}$ very quickly.

I’ve read that the composition of Earth’s atmosphere in pre-life earth was very different from today. If I recall correctly, there was no $\ce{O2}$ as it was formed by microbes later on. Without $\ce{O2}$ in Earth’s atmosphere, it makes sense that conditions would be very reducing based on an analogy to the soil redox zones.


Use of concentrated sulfuric acid in Fischer esterification Concentrated sulfuric acid is often used as a catalyst in Fischer esterification reactions. To my knowledge, the role it plays in the reaction is not as a reducing agent or dehydrating agent but only serves the role of creating an acidic environment, as seen in the mechanism.

Thus, dilute sulfuric acid would be sufficient. In fact, it would be better since it ionises to give an even more acidic solution. However, chemists choose to use the concentrated one instead, suggesting that the role of sulfuric acid is more of a reducing agent. Is there a reason for this?

In fact, sulfuric acid is not a must. Many other H-acids, as well as Lewis acids and their combinations would be sufficient.

The trick here is not to get concentrated acid, but to reduce amount of water in the system. Consider the simpliest transformation, which is typically performed with a large excess of methanol and trace amounts of acid:

$$\ce{R-C(O)OH + CH3OH_{(excess)} <=>>[HCl] R-C(O)OCH3 + H2O}$$

This is an equilibrium reaction. The catalyst allows us significantly faster achieve the equilibrium state, but it does not shift it. Catalyst is equally promoting both direct and reverse reactions. A large excess of alcohol, in accordance with the law of mass action, shifts the equilibrium toward the formation of the ether and thus increases the degree of conversion of the acid into ether.

But if water is not eliminated from the system, or is initially present, hydrolysis occurs:

$$\ce{R-C(O)OCH3 + H2O_{(excess)} <=>>[HCl] R-C(O)OH + CH3OH }$$

Hydrolysis of ether will happen even faster, if a strong base is present, as in this case reaction equilibrium shifts further to the right due to the salt formation.

The bottom line is, in order for the reaction to go to the end, you either add an excess of acid (or alcohol), or remove water as it is produced. Starting with concentrated or water-free acid you are not only simplifying your life, but you are also getting better yield.


remove inhibitor from monomer I am working with a commercial monomer (monomethacryloxypropyl terminated polydimethylsiloxane) with an inhibitor of BHT @1%.

I need to remove it for making an ATRP reaction. What is the good way to do this?

The BHT is only present to prevent small amounts of potential radicals polymerising the monomer. One option is to simply consume it i. e.  Put in one extra per cent of your initiator.
Otherwise,  similarly to basic alumina, use a basic resin. Once you consume the BHT,  your monomer is prone to polymerisation, so you may need to run it into your reactor under your reaction conditions e.g. flushed with nitrogen, pre-cooled, etc


Proton NMR Spectrum for Nicotinic Acid I have been able to slightly unpack the NMR spectrum for the nicotinic acid to understand the order of signals and what ones correspond to what protons but I am having a problem with why some of the signals produced a different number of peaks I can not explain. I understand how simple and complex spin-spin splitting works to an extent when producing a number of different types of groups such as, s, d, t, m, dd, dt and many others.

The spectrum below is one for nicotinic acid, and the peaks are in order of most downfield to least, signals for H on carbons 2, 6, 4 then 5.

Looking at the environment proton two is in it has no equivalent neighboring pairs so why is it producing a doublet?

Same goes for the proton at carbon six.  Why is it producing a double of doublets when it only has one neighboring pair?

As for the signal for proton five that is simple to understand and it should produce a double of doublets because the neighboring protons are in-equivalent, but yet again this is still not the case for the splitting of the signal. The same for proton 4 which produces a double triplets.

Would some one be able to explain why this is occurring?

First, a great source of information on this kind of problem is the website of Hans Reich at UW Madison, see e.g. here. Couplings can occur with protons at ring positions ortho, meta and para, with typical values of $\pu{7-9}$, $\pu{2-3}$ and $\pu{<1 Hz}$, respectively.

This  is (roughly) the information I extracted from your spectrum:

$\pu{  9.15 ppm, \space d,  \space 1.5 Hz}$
$\pu{  8.83 ppm,\space dd, \space 5 Hz, 1.5 Hz}$
$\pu{  8.3 ppm, \space dt (apparent), \space 2 Hz (t), 8 Hz }$
$\pu{  7.6 ppm, \space m (ddd),   \space 5 Hz, 8 Hz, < 1 Hz}$

You should report the field at which the spectrum was acquired since otherwise the magnitude of the J couplings is ambiguous. Assuming that the largest J is $\pu{8 Hz}$ leads to the guess that it is a 400 MHz ($\ce{^1H}$) magnet. You should also report the solvent.

Second, your spectrum is very nice, showing great resolution. Compare for instance to this spectrum (also at 400 MHz but presumably much higher concentration -  $\pu{100 mM}$). Either the concentration of the solvent and temperature may be reasons you obtain higher resolution, resolve very small couplings. The BMRB spectrum was acquired in D2O, pH 7.4, 298 K. Your spectrum was acquired perhaps in $\ce{H2O}$, and perhaps at low pH, resulting in protonation of the pyridine N, but probably not.

Based on the structure you expect 2 to have the weakest couplings to other H. It displays a small coupling, not easily quantifiable because you clipped the spectrum in the figure. 5 Is expected to be a doublet of doublets, with both couplings similar and large. It also displays a smaller third coupling. Therefore lets start by tentatively assigning 2 to the resonance at 9.15 ppm, and 5 to 7.6 ppm. The resonance at 8.3 ppm is an apparent doublet of triplets, the triplet splitting being small, 2 Hz, the larger 8 Hz matching up with that from 5, so this H is vicinal to 5. One of the smaller couplings is presumably to 2, the other to 6, explaining the apparent triplet multiplicity. The last resonance, at 8.8 ppm, has a coupling matching up to 5, and another smaller one to 4, so we assign this to 6. So, to summarize, here are the assignments:

$\pu{  9.15 ppm, \space d,  \space 1.5 Hz, \space }$  2
$\pu{  8.83 ppm,\space dd, \space 5 Hz, 1.5 Hz, \space }$ 6
$\pu{  8.3 ppm, \space dt (apparent), \space 2 Hz (t), 8 Hz , \space }$ 4
$\pu{  7.6 ppm, \space m (ddd),   \space 5 Hz, 8 Hz, < 1 Hz, \space}$ 5

This matches your assignments and those in the BMRB. So this analysis explains the small coupling for 2. The simplest explanation: that's a meta coupling to 4. It does not explain the small coupling for 5, though. That might be a trans coupling to 2.

The magnetization transfer pathways between the assigned H can be verified by inspecting the $\ce{^1H-^1H}$ TOCSY available at the BMRB. In particular, there is no pathway between signals at 9.1 and 8.8 ppm.


Why does this reaction not proceed by sn1/sn2? this is an assignment question but I'm stumped.

we have a polar protic solvent (also a weak base) interacting with the secondary carbon -- it looks like a good example of an sn1/sn2 to me. So why wouldn't the answer be "I and III"?

The major reasons I would give for going with the SN1 products are:

The amount of steric hindrance on tertiary alkyl halides in this context highly limits the amount of second-order reactions that can occur (which require participation from two molecules).
EtOH is not a strong nucleophile in this context. In contrast, EtO- would be, but it's not present. Had EtOH been a strong nucleophile, SN2 would be a tad more likely, but a primary alkyl halide situation, for instance, is much more likely to favor SN2. Therefore, an SN2 product cannot form as a major product. However, considering both I AND III, the product is not formed through SN2 (unless you pick only III, in which case you assume I is not formed).
III is one product in the racemic mixture that would result from forming the tertiary carbocation (which is nearly planar around carbon 3, allowing for both retention and inversion of the stereochemistry). Also, the 20 degrees C does not convince me that elimination is a major product since lower temperature makes elimination more difficult. It is harder for the molecules to orient themselves quickly enough before all reactant is used up if the environment is cold. Therefore, the answer is probably I AND III, assuming that somehow the answer key is incorrect.

I also agree with orthocresol that product II is strange. I would expect by Zaitsev's rule that the new pi bond would form across carbons 3 and 4 in the main chain since his rule says that typically, the elimination occurs by acquiring a proton from the carbon with fewer protons, forming typically the more substituted alkene.


Why six C atoms are usually seen in cyclic compounds? When it gets to Carbon-based molecules, one very possible structure when there are more than six C atoms is the hexagon; though not mostly perfect, it emphasizes that six Carbon atoms tend to bond with each other.* Why?
I assume this is because of the "stability" that is attributed to the hexagon. Explainers have discussed mainly about this issue.
Neither me nor my teacher were convinced by the description; (Actually, all of this is because of my teacher!) since he asked that for example, why not octagon? No response.

I wondered that if I asked this in here what ideas I am going to face.... Sorry, but I'm accustomed to doing lots of "whys".
*For example; look at the pic below:

In the image above, the number of hexagonal rings is twice the number of pentagonal rings and eight times more than octagonal rings. This macro-molecule is the micro of what's happening in the nature. I don't have citations; but it takes a minute or two to see how much the number of hexagonal rings is surpassing the number of other rings. Perhaps it was so obvious, nobody mentioned it.
Update: According to @ron, and with a little payment of thought, I reached the fact that these rings aren't necessarily geometric; the important thing is that six C atoms, instead of five or eight or whatever else number of C atoms are the majority.
Image credit: Pseudopotential model for Dirac electrons in Graphene...

It is all about minimizing the energy of a molecule.
In the case of carbon, the only molecule that adopts a perfect hexagonal geometry in its ground state is benzene (and its derivatives that possess a 6-fold rotational axis).  In this case the hexagonal geometry is adopted because all of the carbons are $\ce{sp^2}$ hybridized.  The ideal geometry (lowest energy) for an $\ce{sp^2}$ hybridized carbon involves three 120 degree bond angles around the carbon atom.  This can be accomplished in the case of benzene by placing 6 carbon atoms in a plane and connecting them as shown to produce the benzene structure.

Other carbon-based molecules may project as hexagons, but they are not hexagons when viewed in 3 dimensions.  For example, cyclohexane is not flat like benzene, but rather typically exists in what is called a "chair conformation" in its ground state where the C-C-C angle is around 109.5 degrees.

why not octagon?

Some other regular polygon shapes can be found in molecules.  For example, the cyclopentadiene anion exists as a regular pentagon, and the cyclooctatetraene dianion exists as a regular octagon.  Both of these molecules contain only $\ce{sp^2}$ hybridized carbons.

As mentioned above, these atoms prefer to exist in a geometry with 3 equivalent 120 degree bond angles.  However, the internal angle in a regular pentagon and a regular octagon are 108 and 135 degrees respectively.  Since these angles deviate from the 120 degree ideal, these molecules will be strained, unlike benzene.

What other ring shapes are preferred, low-energy minima?

We've already discussed the $\ce{sp^2}$ case where the preferred 120 degree angle is found in benzene which exists as a hexagonal structure.  $\ce{sp}$ hybridized carbon is most stable when the X-C($\ce{sp}$)-X bond angle is 180 degrees.  Clearly small or medium size rings comprised solely of $\ce{sp}$ hybridized carbon cannot exist as stable entities.  In the case of $\ce{sp^3}$ hybridized carbon, the closer we get to the ideal tetrahedral angle (an angle whose cosine = -1/3, ca. 109.47 degrees) the lower the energy of the molecule.  We've already mentioned cyclohexane and while it's C-C-C angle is close to the ideal tetrahedral angle, it is not exactly the tetrahedral angle (because each of its carbon atoms does not exist in a perfect tetrahedral environment).  If we fuse a few more cyclohexane rings onto cyclohexane, we can produce the molecule adamantane with 4 cyclohexane faces all in the chair geometry.

There is less strain energy in adamantane (per carbon) than in cyclohexane.  The geometry at each carbon atom is closer to the ideal tetrahedral geometry, but still not quite there.  How can we further elaborate adamantane into a ring system where all carbons have the ideal tetrahedral angle? Does adamantane look vaguely familiar?  By continuing to add more and more cyclohexane rings we build the following ring structure...

... Diamond, where all of the $\ce{sp^3}$ carbon atoms are in the ideal tetrahedral environment, an extremely stable arrangement.
EDIT: response to OPs comment

So, the lesser energy state must be the reason that after six number
of atoms, five is the common one, right?

Kinda.  Here's a table showing estimated ring strains for saturated hydrocarbon rings of various sizes.

[Image source  Also, see this link for a discussion about why the ring strain for cycloheptadecane (-3.4 Kcal/mol) is suspect and probably should be higher.]
5- and 7-membered saturated hydrocarbon rings have similar strain energies, yet 5-membered rings are more common.  We must consider the free energy of the process, enthalpy and entropy changes. In addition to strain (an enthalpic factor) we must also consider entropy.  In ring forming reactions it is entropically more difficult for a 7-membered chain to bring both ends together for closure to a ring, then it would be for a smaller 5-membered chain.


What are the different types of van der Waals forces? As per different websites on internet van der Waals force are of

Three types (Keesom force, Debye force and London dispersion force) @ Wikipedia
Two types (Dipole-dipole force/Keesome force and London dispersion force) @chemguide and chemwiki

While an expert website  also explained H-bonds to be a van der Waals force.

I'm confused. What is originally correct?

The definition of van der Waals forces according to IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8., is as follows:

  The attractive or repulsive forces between molecular entities (or between groups within the same molecular entity) other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules. The term includes: dipole–dipole, dipole-induced dipole and London (instantaneous induced dipole-induced dipole) forces. The term is sometimes used loosely for the totality of nonspecific attractive or repulsive intermolecular forces.

The phrase "other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules" would seem to imply that according to the IUPAC Gold Book, if a force is not an ionic or covalent chemical bond or electrostatic interaction, then it is a van der Waals force.


Strong activation of phenoxide ion compared to others I was told that in phenoxide ion, benzene ring is very strongly activated. The order my teacher told me was phenol < amine < $2^\circ$-amine < $3^\circ$-amine < phenoxide (for activation of benzene).
Example: Reimer–Tiemann reaction happens only in sodium phenoxide
So, what is so special in phenoxide that makes it so much activating? Also, why does oxygen contribute more in resonance than nitrogen, when it is more electronegative than the latter?

I assume by your notes that your teacher was mentioning electrophilic aromatic substitution reactions. These are two-step reactions.  The product-determining step (whether it is ortho-, meta- or para-) in the substitution mechanism is the first step, which is also the slowest (high activation energy), and hence the rate determining step (see [energy Diagram] below).  The second step is relatively faster, because positively charged intermediate ion loose a proton to get back the resonance stabilized aromatic ring again. Accordingly, the rate of the reaction is roughly depend on the influence of each of the given substituents to stabilize the benzenonium intermediate (positively charged adduct). You can best see it by looking at the capability of each substituent to delocalize the positive charge within the benzenonium intermediate.

Each given substituent ($\ce{-OH, -NH2, -NHR, -NR2}$) except for the phenolate ion has a lone pair. Thus, each can share it with the benzenonium nucleus to reduce the positive charge, reducing the energy of intermediate. However, phenolate ion ($\ce{ph-O^-}$) has full negative charge, which can neutralize benzenonium intermediate completely, to give least energy among all substituents considered here.


Questions in MO theory Τhe electrons in the antibonding MO are the unpaired electrons of the molecule? E.g. $\ce{O2}$, by writing down the diagram for the MO we notice that there are only 2 electrons in the top antibonding orbitals thus 2 unpaired electrons, yet by writing down the Lewis structure, the molecule turns out to have 0 unpaired electrons..
So, what do the electrons in the anti bonding orbitals do if they don't represent the unpaired electrons of the molecule?

The antibonding orbital doesn't necessarily tell you the number of unpaired electrons - it's coincidental that for O$_2$, the HOMO (highest occupied molecular orbital) is a (1$\pi_g)^2 $ orbital - or, if you prefer to write it like you have for N$_2$, ($\pi ^1 _{2px} = \pi ^1 _{2py}$).
$\pi$ orbitals are doubly degenerate, so the lowest energy configuration is for the 2 electrons to occupy different orbitals (one in each of the 2 $\pi$ orbitals). This means they're both unpaired, as shown in this MO diagram of O$_2$ from Wikipedia. For an example of an MO diagram where there are unpaired electrons in an orbital which is not antibonding, try drawing the B$_2$ MO diagram (remember that s-p mixing occurs).

In terms of what they do, antibonding electrons destabilise the bond, "cancelling out" the energetically favourable effect of bonding electrons - try drawing an MO diagram for He$_2$ or Ne$_2$. It should be apparent why they don't form stable molecules.

As far as N$_2$ goes, I think you may have neglected to take into account the effect of the $\sigma ^* _{2s} $ electrons. Remember that antibonding electrons cancel out bonding electrons.


Why are many silver salts insoluble? For example, silver chloride, silver sulfide, and silver oxide are three exceedingly stable silver salts in water solution.

Why is it that silver forms such stable salts?

Does this have to do with charge density of the component ions? This doesn't seem to hold water because we know that silver oxide is much more soluble than silver sulfide, and the oxide anion is obviously a lot smaller than the sulfide ion, so the oxide ion should have a much higher negative charge density.

So simple charge density rationalizations go out the window. I've Googled, I've looked in books, etc., but I can't find an explanation.

On the other hand, can a Lewis acid/base model be used to rationalize the stability of the silver compounds? For example, the sulfur, by nature of being less electronegative than oxygen, is a better Lewis base/electron pair donor/nucleophile; the sulfide ion is more willing to "donate" its electrons. Also the sulfide ion should be bigger and has empty 3d valence orbitals which may or may not be accessible to a great extent. Nonetheless, metals, by nature of possessing a large number of valence d-electrons, may be able to increase the stability of the sulfide ion by populating its d-valence.

This Lewis acid/base model seems to get a lot of mileage. I also know that silver is a good Lewis acid, as are many metal cations, because of the poor shielding afforded by the d electrons. Also, the silver cation has an empty 5s orbital. Thus, we can form a sulfide-to-cation sigma coordinate covalent bond.

We may also form a pi-type coordinate covalent bond as the far-away d-electron density "drifts" to empty 3d valence orbitals on the sulfide anion, making the sulfide anion now more nucleophilic than ever.

Let me know if my rationalizations are incorrect or if there are better explanations of silver salt stability.

This is just an unproven guess on my part.  Whether compounds are soluble or not depends on 1) how strong the bonding (including lattice) forces are that hold the compound together (e.g. how much energy do we have to put in to liberate the ions, and 2) how much energy is recovered through solvation of the resulting ions.  I don't see anything odd about the ionic radius of the silver ion so I'm guessing that the energy of solvation of the resultant ions would be fairly normal; so it is probably not factor "2" that makes most silver salts insoluble.  Here is a link to an article that suggests that bonding in silver compounds may have a covalent component, it says, for silver salts "the structure has some contribution from covalent bonding. This appears to be bonding the structure together more tightly, as in a giant covalent structure."  The words "more tightly" catch my eye.  This seems to suggest that factor "1" may be in play in that silver salts may have increased bond\lattice energies.  If true, this would help explain the insolubility of many silver salts.  Finally I'll add that some of the soluble silver salts ($\ce{AgF, AgNO3, Ag(CH3CO2)}$) seem to involve anions that would be expected to be strongly hydrated.  The extra energy generated through solvation would further assist in solubilizing these silver salts.


What are the limits of size differences in a host–guest complex? We learn in my chemistry course that there must be a snug fit for a host-guest complex to form. So, something like benzene (0.6 nm in diameter) can form a host-guest complex with a β-cyclodextrin host (0.7 nm cavity).
Other than this vaguely-worded ‘snug-fit’ rule, we are not given a lot of information of what molecules would fit.
Would something like methane (0.38 nm diameter, according to Wikipedia) form a stable host–guest complex with β-cyclodextrin, or would it be too small?
I know that methyl red fits in a β-cyclodextrin cavity, but what about something slightly larger like anthracene?
These questions are a small part of something that contributes a small amount towards my overall course grade. Please, if you are able to, give me some kind of guidance so that I can work out answers on my own. I've tried to look independently online but I come up with nothing.

The size of the cavity is one of the many factors collaborating to host-guest chemistry with cyclodextrin (CD). For example, depending on the size and the properties of the guest molecule, CD may form complexes with either the entire molecule or only a part of the molecule (The diameter and the volume of the cavity of β-CD are around $\pu{7.8 Å}$ ($\pu{0.78 nm}$) and $\pu{262 Å^3}$, respectively, and β-CD complexes well with aromatic rings; Ref.1). It is accepted in general (Ref.2) that the binding forces involved in complex (host-guest) formation are:

Van der Waals type interactions (or hydrophobic interactions) between the hydrophobic units of the guest molecules and the CD cavity.
Hydrogen bond between the polar functional groups of the guest molecules and the hydroxyl groups of the CD rims.
Release of high energy water molecules from the cavity during the complex formation process.
Release of strain energy into the ring structure system of the CD.

Note that the role of hydrogen bonding is not universal since stable complexes are formed with hosts and hydrophobic guests such as benzene derivatives (Ref.3), which do not form hydrogen bonds.

Also, regardless of which types of stabilizing forces are involved, the most important factors in determining the stability of the inclusion of the complexes are: (i) The geometric capability (which is the OP's concern); (ii) the polarity of the guest molecule; (iii) medium; and (iv) the temperature. The reference 2 states that:

Geometric, rather than the chemical factors, are critical in determining the type of guest molecules that can penetrate into the cavity. If the guest is too small, it passes easily through the cavity and the bond will be weak or will not occur. The formation of complexes with the molecules significantly larger than the cavity is also possible, but only some limited groups or the side chains penetrate into the cavity.

It is also important to note that cavity size of CD seems to depend on the method used by the researcher (Ref.4). For instance, Ref.2 reported that inner diameters of $\pu{5.7 Å}$ for $\alpha$-cyclodextrin, $\pu{7.8 Å}$ for $\beta$-cyclodextrin, and $\pu{9.5 Å}$ for $\gamma$-cyclodextrin, but did not verify each was the upper or the lower or middle diameter. Meantime, Ref.3 reported that top and bottom inner-diameters of $5.3$ and $\pu{4.7 Å}$ for $\alpha$-cyclodextrin, $6.5$ and $\pu{6.0 Å}$ for $\beta$-cyclodextrin, and $8.3$ and $\pu{7.5 Å}$ for $\gamma$-cyclodextrin. Therefore, it is hard to predict which molecule would bind to appropriate CD without knowing exact parameters of their cavity. Regardless of the reported cavity sizes, Ref.3 shows that the grafting of $\beta$-cyclodextrin to silica gel conveniently removed water soluble organic molecules with different sizes such as p-nitrophenol, p-nitroaniline, m-nitrophenol, p-chlorophenol, and phenol.
It is also evident that cyclodextrin-catalyzed organic synthesis can be achieved using $\beta$-cyclodextrin as the catalyst. In these reactions, you may find part of starting materials with different sizes making stable host-guest complexes with $\beta$-CD (Ref.5).
On the other hand, an identical molecule can be bound to all three CDs ($\alpha$-, $\beta$-, and $\gamma$-cyclodextrin) grafted to a biomolecule with different impact (Ref.5). For example, $\mathrm{^1H \ NMR}$ studies suggested that methyl-$\beta$-CD forms inclusion complexes with the lipid guest molecules such as cholesterol, linoleic acid, and $\beta$-sitosterol within the CD cavity regardless of their sizes (Ref.6):

Note: Although it is not relevant to the question in hand, it is worth noting that  the chiral nature of the cavity in these molecules makes them useful for enantiomeric separation applications (Ref.7).

References:

Vijaykumar Parmar, Gayatri Patel, Nedal Y. Abu-Thabit, “Chapter 20:  Responsive cyclodextrins as polymeric carriers for drug delivery applications,” In Stimuli Responsive Polymeric Nanocarriers for Drug Delivery Applications, Volume 1: Types and Triggers (Woodhead Publishing Series in Biometerials); Abdel Salam Hamdy Makhlouf, Nedal Y. Abu-Thabit, Eds.; Woodhead Publishing, an imprint of Elsevier: Duxford, United Kingdom, 2018, pp. 555-580 (ISBN: 978-0-08-101977-9).
Julia Martin, Enrique Jacobo Díaz-Montaña, Agustín G. Asuero, “Chapter 1:  Cyclodextrins: Past and Present,” In Cyclodextrin: A Versatile Ingredient; Poonam Arora, Neelima Dhingra, Eds.; IntechOpen: London, United Kingdom, 2018, pp. 3-44 (ISBN: 978-1-78923-068-0).
A. Bibby, L. Mercier, "Adsorption and separation of water-soluble aromatic molecules by cyclodextrin-functionalized mesoporous silica," Green Chem. 2003, 5, 15-19 (DOI: 10.1039/B209251B).
Avilasha A. Sandilya, Upendra Natarajan, M. Hamsa Priya, "Molecular View into the Cyclodextrin Cavity: Structure and Hydration," ACS Omega 2020, 5(40), 25655–25667 (DOI: https://doi.org/10.1021/acsomega.0c02760).
Chang Cai Bai, Bing Ren Tian, Tian Zhao, Qing Huang, Zhi Zhong Wang, "Cyclodextrin-Catalyzed Organic Synthesis: Reactions, Mechanisms, and Applications," Molecules 2017, 22(9), 1475 (17 pages) (DOI: https://doi.org/10.3390/molecules22091475).
Lajos Szente, Ashutosh Singhal, Andras Domokos, Byeongwoon Song, "Cyclodextrins: Assessing the Impact of Cavity Size, Occupancy, and Substitutions on Cytotoxicity and Cholesterol Homeostasis," Molecules 2018, 23(5), 1228 (15 pages) (DOI:  https://doi.org/10.3390/molecules23051228).
Mikhail V. Rekharsky, Yoshihisa Inoue, "Complexation Thermodynamics of Cyclodextrins," Chem. Rev. 1998, 98(5), 1875-1918 (DOI: https://doi.org/10.1021/cr970015o).


How can I find the Van der Waals constants (a and b) without any critical values? I'm given a container of gas that is kept at constant temperature, and whose volume is decreased by applying pressure, from A to E, as the graph shows. I am given the temperature, the total moles in the container, and the following data:
At B the molar volume is 1.153, and the pressure is 17.76.
At C the molar volume is .8167, and the pressure is 17.76.
At D the molar volume is .032, and the pressure is 17.76.

Given this data, and I emphasize, only this data (no critical P, T, or Vmolar), find the van der Waals constants, a and b, for the gas.

I've been working on it for 3 hours and so far all I can come up with is a single equation in terms of a and b at point B: 17.76=RT/(Vmolar-b) - a/Vmolar^2
At point C, the equation would be the same as what I came up with for B, since the temperature and pressure remains the same as B, and we're in the region of co-existence, so Vmolar is at its maximum for the gaseous phase and therefore also is the same as in B (any further increase in Vmolar causes precipitation).

Any and all help is greatly appreciated.

I would approach this a little differently.  I would integrate the pressure between $V_D$ and $V_B$, such that the average pressure is 17.76 bars:  $$\frac{\int_{V_D}^{V_B}{PdV}}{(V_B-V_D)}=17.76$$or,
$$17.76(V_B-V_D)=RT\ln{\frac{(V_B-b)}{(V_D-b)}}-a\left[\frac{(V_B-V_D)}{(V_B-b)(V_D-b)}\right]$$
This would provide an equation for expressing a in terms of b.  So the pressure could then be expressed exclusively in terms of b.  Then, to determine the value of the constant b, I would set $P(V_B)$ equal to $P(V_D)$ and solve the resulting equation for b.


What is the shiny structure under a metal coating? I performed a hot dip aluminizing on steel for 15 seconds. Then, after I bent the steel to remove the coating easier, the steel under was rough, shiny, and metallic, and was different from the original steel.The layer seems to be random and has bumpy structures. I can't mechanically removed the layer. I couldn't have analyze it chemically. Could I conclude that it is an intermetallic layer of Al and steel?

Likely an intermetallic , I think there are a couple different ones. I am more familiar with "spray and diffuse"  or vapor deposition ( Alon). Alonizing is very long ( > 24 hr) but gives a high integrity coating , can't remember the thickness.  As I remember the "diffuse" cycle takes a couple hours. I expect your process produced a thinner coating but still with some intermetallics.


How much CO₂ should one expect in an IR spectrum? I ran an IR spectrum ($\ce{KBr}$ pellet) and found a sharp absorbance at $2360 \text{cm}^{-1}$, which was unexpected. As far as I can tell, this indicates either a nitrile or $\ce{CO2}$.

I know that there is no nitrile in my product. Would you expect atmospheric $\ce{CO2}$ to show up to a noticeable degree in a typical IR spectrum?

From what I've read (and personal experience with IRs I've run), a relatively intense $\ce{CO2}$ asymmetric bond stretching signal at approximately $\ce{2350cm^{-1}}$ is an extremely common background artifact in IR spectra (along with the broad, weak signals downfield of $\ce{3000cm^{-1}}$ attributable to $\ce{O-H}$ bonds from ambient moisture). Depending on your software, it's often possible to run a "dry" IR reading with no sample present in order to establish a baseline spectrum of the surrounding air, which can then be checked against by your software (so that overlapping signals from the baseline air spectrum are deleted from the spectrum of your sample). Alternatively, you can take measures to dry the optics (I've once used $\ce{N2}$ for that purpose; alternatively, air directed through $\ce{CaCl2}$ drying tubes could possibly be used, but I can't attest to the efficacy of that). That manual drying procedure takes some effort, and might be futile in any case depending on the conditions of the lab. My one experience with that approach was disappointing.


Why should AAS use element lamps? AAS (Atomic Absorption Spectroscopy) is a quantitative analytical technique used to measure very small concentrations of ions in substances. The main idea is that the sample is atomised in a flame and then light is shot through the atomised sample and the absorbance is measured.

However I don't get why they use an atomic emission lamp (a lamp made of only 1 element). What's wrong with using a normal incandescent lamp or some lamp that produces all the wavelengths at once (it would save you having to change the lamp for every test)? I've heard that multi-element lamps make the machine less sensitive due to noise but I don't understand how. If we tune the monochromator to the particular wavelength we want, it should be the same as with using an atomic emission lamp?

Absorption bands are very small: smaller than the band which a typical monochromator is capable of isolating.

If a continuum light source was employed, the absorbed band would be much smaller than the band which would be isolated by the monochromator and which would be read by the detector. This implies a low SNR (signal-to-noise ratio), and a very poor resolution.

Using a single element lamp (or lamps with combined elements but with distinct emission frequencies) overrides the problem: a monochromator is well capable of isolating the different frequencies, and absorption, which occurs in a narrow band of frequencies, occurs in the same narrow band of light from the source: this implies a much higher SNR, and a much better resolution.

To make an easy to conceive example, let's say that the effect of absorption by the sample is the effect of a star shining in the sky: using a continuum light source implies watching it through the midday sky. A HCL lamp would be a midnight sky, in the metaphor.

Note that, aside from what textbooks say, continuum souce AAS (HR-CS AAS) does exist, but it requires, as expected, a sophisticated monochromator (high resolution monochromator, with a resolution of a few pm!)


Are any commercial plastics made from biomass/plant matter? Are any commercial plastics made from plant matter currently. I expect it can be done in a lab but I am wondering about cost competitive tonnage quantities.

The original answer only gave a link to a commercial bioplastic producer.  The answer has been edited to provide more information.

Yes there are.  Here is an example: Biome Bioplastics

European Bioplastics (an industry group) estimates that the global production of bioplastics was about 2 million metric tonnes in 2017.  They forecast that this will increase to almost 2.5 million metric tonnes by 2022.Source

One estimate puts the market value of bioplastics at around 6 billion USD in 2018 and growing to almost 15 billion USD by 2023.  Source

Common sources of bioplastics are starches (like corn), cellulose and aliphatic esters.

Commonly, bioplastics are used to make compostable/biodegradable packaging and films, as well as disposable cups and utensils.


In Bohr's model what is angular momentum in the lowest orbital? Angular momentum is given by: $n\frac{h}{2\pi}$ where $n=1,2,3...$

And angular momentum is also given by: $\sqrt{l(l+1)}\frac{h}{2\pi}$

By the first equation angular momentum should be $\frac{h}{2\pi}$ but by the second it should be $0$.

And if you say that in first equation $n$ is also $0$  then first radius of hydrogen should be $0\pu{A^\circ}$ instead of $0.53\pu{A^\circ}$.

You are trying to mix the Bohr model with quantum mechanics. The Bohr model is a semi-classical treatment of the hydrogen atom in which an electron is circling around a proton. Classically, this electron has angular momentum, even in the lowest orbital because it has a well defined position vector $\vec{r}_n$ and velocity $\vec{v}_n$. The classical angular momentum is thus given by

$$
l_n=m_e|\vec{r}_n\times\vec{v}_n|,
$$
where $m_e$ the electron mass. Bohr postulated that the angular momentum must be quantized (this is the semi-classical part) resulting in a set of orbitals.

Though Bohr's model works remarkably well it is not perfect because electrons are not really particles in the classical sense, but also behave like waves. A better treatment (but not the best) of the hydrogen atom is therefore to use quantum mechanics. In quantum mechanics the electron is described by a wavefunction that is spread out over all space and has a certain shape. For $s$ orbitals, such as the lowest orbital in the hydrogen atom, the wavefunction is spherical symmetric and has no angular momentum. The second expression you mention is the quantum mechanical operator to obtain the angular momentum.

To conclude, you cannot compare the angular momentum of the Bohr atom to the Schrodinger atom because they are based on different assumptions. Note further that for principal quantum number $n$, $\ell=0, 1, \ldots, n-1$.


Why there is no need for reference standards in High Resolution Mass Spectrometry (HRMS) I have a question about HRMS (High  Resolution Mass spectrometry). HRMS is a powerful technique that can discriminate between possible target compounds without the need of reference standards, unlike QqQ (triple quad systems). My question is simple why do HRMS don't need reference standards for analysis?

A short explanation would really help I cannot find the answer online or anything related to it.

Thank you, hopefully I can be helped with this part. I put a section of a book as an attachment maybe that will help.

Greets,
Student

The statement is not 100% correct. Ordinary mass spectrometry will not give you exact masses to 5 or 6 decimal places, say, m/z 415.14 for a singly charged molecular ion. Now there will be hundreds of compounds with this mass. On the other hand HRMS gives you a accurate and precise mass say 415.14509. This detailed number drastically reduces the number of matches in a spectral library. Not only that, the isotopic distribution pattern and their masses reduces the choice to one compound. Hence the claim that HRMS will not need a reference.

Like any claim, there are always exceptions. Structural isomers can be a problem. If we have isotopologues (isotopic isomers), even HRMS will fail.  Say you have two analytes but both are acetonitriles. Say one is

$^{13}$CH$_3$CN, and the other analyte is CH$_3$$^{13}$CN, HRMS will give the identical molecular mass to five decimal places. One might look up their fragmentation pattern but MS is not that great at low masses (high noise). Here you would need another spectroscopic technique such as microwave rotational spectroscopy. These techniques are used an in emerging field called position specific isotopic analysis.


Does anyone have information about Piquet and Calitet (chemists)? I was given a set of names of chemists to research and one of them was actually a duo that I understand to have been French chemists (at least their last names suggest it). I have queried and queried not simply Google, but also academic databases (Wiley, ProQuest, Elsevier, etc.), yet I've found nothing.

I believe I am spelling the names correctly, but for clarity, phonetic expression of their surnames is:

  "PEA-ket and CAL-ee-TET"

Hath anyone information on these names within chemistry (or similar natural science discipline)? It would allow me to end a nearly two month search – one way or another.

Most appreciative for any corrections if I have erred in spelling, nation of origin, etc.

Can you exclude spelling errors? If not, there is

Amé Pictet, known from the Pictet-Spengler cyclization.
Claude Piguet, an inorganic chemist at the University of Geneva.

Both were/are is Swiss, not French.

However, the second name does not ring a bell for me.


How to dispose of pure potassium I have about 2g of 99.5% Potassium.

I want to get rid of it, right now it's in a glass jar sealed.

What is the proper protocol? I was going to combust it in water and wash it down the drain but the by-product potassium hydroxide is caustic and hazardous so I don't think that's best.

Could I just throw it away and it goes to a garbage facility where the vial probably breaks and it reacts with rain water and the potassium hydroxide is so small and buried that it never really has an impact?

In a suggested lab, the RSC recommends using propan-2-ol:

  If there is any unused potassium remaining at the end of the experiment, remove it with the tweezers and place it in a beaker containing about 100 cm3 of propan-2-ol to dissolve away – it will fizz, giving off bubbles of hydrogen. When the fizzing has stopped, dispose of the resulting alkaline solution down the sink flushing it away with plenty of water. The acidic solution remaining from the experiment may also be disposed of down the sink flushing it away with plenty of water.

The same is recommended for sodium.


What is the mechanism for the hydrolysis of the boron-alkoxide complex in NaBH4 reduction? According to this question, the stoichiometry of the reduction of a ketone is such that sodium borohydride can reduce up to 4 ketones by forming a boron-alkoxide complex.

The next step would involve the hydrolysis of the complex, and I've been searching through textbooks to no avail. What would the mechanism for this hydrolysis look like?

I don’t know whether the mechanism of hydrolysis of the boronic acid esters has been studied. However, the dissociative mechanism immediately suggests itself if only by comparison to similar structures.

Since the workup is typically written $\ce{H+/H2O}$, you may consider some buffered source of external protons. These can protonate any of the alcoholates which can then dissociate. (The $\ce{B=O}$ double bond I drew in should be considered no more than additional stabilisation from oxygen lone pairs; boron is electron-deficient by definition.) Another water molecule can associate: the first bond is cleaved. Repeat three times and end up with four molecules of alcohol and tetrahydroxidoborate $\ce{[B(OH)4]-}$.


How is a counterpoise corrected geometry optimization done? I understand the problem of basis set superposition error (BSSE) and I know how the counterpoise correction for single point energies is calculated.

Today I found out that many software packages allow for counterpoise correction during optimization calculations, but how does this actually work, especially for methods where analyical gradients are used for optimization?

During geometry optimization we calculate the first derivative of the energy to get energy gradients which we follow down to our minima. I understand that I could use counterpoise correction if I calculate those gradients numerically, which is quite easy to understand but very expensive to do, but it seems that counterpoise can also be used in combination with analytical gradients. How is the counterpoise implemented to get counterpoise corrected gradients?

Background

For a system consisting of two molecules (monomers or fragments are also used) X and Y, the binding energy is

$$
\Delta E_{\text{bind}} = E^{\ce{XY}}(\ce{XY}) - [E^{\ce{X}}(\ce{X}) + E^{\ce{Y}}(\ce{Y})]
\label{eq:sherrill-1} \tag{Sherrill 1}
$$

where the letters in the parentheses refer to the atoms present in the calculation and the letters in the superscript refer to the (atomic orbital, AO) basis present in the calculation. The first term is the energy calculated for the combined X + Y complex (the dimer) with basis functions, and the next two terms are energy calculations for each isolated monomer with only their respective basis functions. The remainder of this discussion will make more sense if the complex geometry is used for each monomer, rather than the isolated fragment geometry.

The counterpoise-corrected (CP-corrected) binding energy [1] to correct for basis set superposition error (BSSE) [2] is defined as

$$
\Delta E_{\text{bind}}^{\text{CP}} = E^{\ce{XY}}(\ce{XY}) - [E^{\ce{XY}}(\ce{X}) + E^{\ce{XY}}(\ce{Y})]
\label{eq:sherrill-3} \tag{Sherrill 3}
$$

where the monomer calculations are now performed in the dimer/complex basis. Let's explicitly state how this works for the $E_{\ce{XY}}(\ce{X})$ term. The first molecule X contributes nuclei with charges, basis functions (AOs) centered on those nuclei, and electrons that will count to the final occupied molecular orbital (MO) index into the MO coefficient array. There is no reason why additional AOs that are not centered on atoms can't be added to a calculation. Depending on their spatial location, if they're close enough to have significant overlap, they may combine with atom-centered MOs, increasing the variational flexibility of the calculation and lowering the overall energy. Put another way, place the AOs that would correspond to molecule Y at their correct positions, but don't put the nuclei there, and don't consider the number of electrons they would contribute to the total number of occupied orbitals. This means that for the full electronic Hamiltonian

$$
\hat{H}_{\text{elec}} = \hat{T}_{e} + \hat{V}_{eN} + \hat{V}_{ee}
$$

calculating the electron-nuclear attraction $\hat{V}_{eN}$ term is now different. Considered explicitly in matrix form in the AO basis,

$$
\begin{align*}
V_{\mu\nu} &= \int \mathop{d\mathbf{r}_{i}} \chi_{\mu}(\mathbf{r}_{i}) \left( \sum_{A}^{N_{\text{atoms}}} \frac{Z_{A}}{|\mathbf{r}_{i} - \mathbf{R}_{A}|} \right) \chi_{\nu}(\mathbf{r}_{i}) \\
&=\sum_{A}^{N_{\text{atoms}}} Z_{A} \left< \chi_{\mu} \middle| \frac{1}{r_{A}} \middle| \chi_{\nu} \right>
\end{align*}
$$

there are now fewer terms in the summation, since the nuclear charges from molecule Y are zero (the atoms just aren't there), but the number of $\mu\nu$ are the same as for the XY complex. This and the $\hat{T}_{e}, \hat{V}_{ee}$ terms aren't really mathematically or functionally different then; this is more to show where the additional basis functions enter, or to show where nuclei appear in the equations [3].

These atoms that don't have nuclei or electrons, only basis functions, are called ghost atoms. Sometimes you also see the term ghost functions, ghost basis, or ghost {something} calculation. Adding the basis of monomer Y to make the full "dimer basis" means taking the monomer X and including basis functions at the nuclear positions for Y.

Geometry optimization

Now to calculate the molecular gradient, that is, the derivative of the energy with respect to the $3N$ nuclear coordinates. This is the central quantity in any geometry optimization. For the sake of simplicity, consider a steepest descent-type update of the nuclear coordinates
$$
R_{A,x}^{(n+1)} = R_{A,x}^{(n)} - \alpha \frac{\partial E_{\text{total}}^{(n)}}{\partial R_{A,x}}
\label{eq:steepest-descent} \tag{Steepest Descent}
$$
where $n$ is the optimization iteration number, $\alpha$ is some small step size with units [length2][energy], and the last term is the derivative of the total (not just electronic) energy with respect to a change in atom $A$'s $x$-coordinate. Even Newton-Raphson-type updates with approximate Hessians (2nd derivative of the energy with respect to nuclear coordinates, rather than the 1st) need the gradient, so we must formulate it.

Formulation of the energy

We're in a bit of trouble, because we want to replace $E_{\text{total}}$ in the gradient with $E_{\text{total}}^{\text{CP}}$, but all we have is $\Delta E_{\text{bind}}^{\text{CP}}$. The concept of CP correction can still be applied to a total energy, but the BSSE must be removed from each monomer. The BSSE correction itself for each monomer is
$$
\begin{split}
E_{\text{BSSE}}(\ce{X}) &= E^{\ce{XY}}(\ce{X}) - E^{\ce{X}}(\ce{X}), \\
E_{\text{BSSE}}(\ce{Y}) &= E^{\ce{XY}}(\ce{Y}) - E^{\ce{Y}}(\ce{Y}),
\end{split}
\label{eq:2}
$$
which, when subtracted from $\eqref{eq:sherrill-1}$, gives $\eqref{eq:sherrill-3}$. More correctly, considering that the geometry for each step is at the final cluster geometry and not the isolated geometry, the above is [4]
$$
\begin{split}
E_{\text{BSSE}}(\ce{X}) &= E_{\ce{XY}}^{\ce{XY}}(\ce{X}) - E_{\ce{XY}}^{\ce{X}}(\ce{X}), \\
E_{\text{BSSE}}(\ce{Y}) &= E_{\ce{XY}}^{\ce{XY}}(\ce{Y}) - E_{\ce{XY}}^{\ce{Y}}(\ce{Y}).
\end{split}
\label{eq:sherrill-10} \tag{Sherrill 10}
$$

The CP-corrected total energy is the full dimer energy with BSSE removed from each monomer is then
$$
\begin{split}
E_{\text{tot}, \ce{\widetilde{XY}}}^{\text{CP}} &= E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY}) - E_{\text{BSSE}}(\ce{X}) - E_{\text{BSSE}}(\ce{Y}), \\
&= E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY}) - \left[ E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{X}) - E_{\ce{\widetilde{XY}}}^{\ce{X}}(\ce{X}) \right] - \left[ E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{Y}) - E_{\ce{\widetilde{XY}}}^{\ce{Y}}(\ce{Y}) \right].
\end{split}
\label{eq:sherrill-15} \tag{Sherrill 15}
$$
Note that I have modified which geometry is used for each monomer in $\eqref{eq:sherrill-15}$. All monomers are calculated at the supermolecule geometry. This is convenient for two reasons: 1. We are only interested in removing the BSSE, not the effect of monomer deformation, and 2. a isolated monomer geometry without deformation doesn't make sense in the context of a geometry optimization. I also added the tilde to signify that the supermolecular/dimer geometry used may not be the final or minimum-energy geometry, as would be the case during a geometry optimization. We simply extract all structures consistently from a given geometry iteration. Perhaps $\ce{XY}(n)$ would be better notation.

Formulation of the gradient

As Pedro correctly states, the differentiation operator is a linear operator. Because there are no products in $\eqref{eq:sherrill-15}$, the total gradient needed for $\eqref{eq:steepest-descent}$ will be a sum of gradients [5]:
$$
\frac{\partial E_{\text{tot}, \ce{\widetilde{XY}}}^{\text{CP}}}{\partial R_{A,x}} = \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY})}{\partial R_{A,x}} - \left[ \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{X})}{\partial R_{A,x}} - \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{X}}(\ce{X})}{\partial R_{A,x}} \right] - \left[ \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{Y})}{\partial R_{A,x}} - \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{Y}}(\ce{Y})}{\partial R_{A,x}} \right],
$$
so each step of a CP-corrected geometry optimization will require 5 gradient calculations rather than 1. Note that the nuclear gradient should be included for each term as well, which is a trivial calculation.

Extension to other molecular properties

Although not commonly done, counterpoise correction can be applied to any molecular property, not just energies or gradients. Simply replace $E$ or $\partial E/\partial R$ with the property of interest. For example, the CP-corrected polarizability $\alpha$ of two fragments is
$$
\alpha_{\text{tot}, \ce{\widetilde{XY}}}^{\text{CP}} = \alpha_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY}) - \left[ \alpha_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{X}) - \alpha_{\ce{\widetilde{XY}}}^{\ce{X}}(\ce{X}) \right] - \left[ \alpha_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{Y}) - \alpha_{\ce{\widetilde{XY}}}^{\ce{Y}}(\ce{Y}) \right]
$$
where I believe it now makes even less sense to have each individual fragment calculation not be at the cluster geometry. In papers that calculate CP-corrected properties, no mention is usually made of which geometry the individual calculations are performed at for this reason.

References

Boys, S. Francis; Bernardi, F. The calculation of small molecular interactions by the differences of separate total energies. Some procedures with reduced errors. Mol. Phys. 1970, 19, 553-566.
Sherrill, C. David. Counterpoise Correction and Basis Set Superposition Error. 2010, 1-6.
One implementation note: Most common quantum chemistry packages should allow for the usage of ghost atoms in energy and gradient calculations. However, as Sherrill states, they do not properly allow for composing the full gradient expression to perform CP-corrected geometry optimizations. Gaussian can, and Psi4 may. For programs that can calculate gradients with ghost atoms, Cuby can be used to drive the calculation of CP-corrected geometries and frequencies.
There is a typo in the Sherrill paper; the subscripts for all 4 energy terms should be $AB$, which here are $\ce{XY}$.
Simon, S.; Bertran, J.; Sodupe, M. Effect of Counterpoise Correction on the Geometries and Vibrational Frequencies of Hydrogen Bonded Systems. J. Chem. Phys. A 2001, 105,, 4359-4364.