Do I need to include calcium chloride in this reaction equation? Yesterday I titrated a solution of calcium hydroxide in aqueous calcium chloride with $16.7~\mathrm{mL}$ of $\ce{HCl}$. More specifically, I took $25\ \mathrm{mL}$ of the calcium hydroxide in aq calcium chloride and mixed it with $25~\mathrm{mL}$ of deionized water and 3 drops of an appropriate indicator. I wrote the reaction equation as

$$\ce{Ca(OH)_2 + 2HCl -> CaCl_2 + 2H_2O}$$

But do I need to somehow take into account the aqueous calcium chloride in this equation? I know that it affects the molar solubility of the calcium hydroxide during titration but I'm not sure if this should also change my reaction equation.

From an instructor's point of view, you have accounted for the reaction that has taken place during titration.  The point you are making surrounds the conditions the titration took place under (as you pointed out, solubility is altered, for instance), but not the chemistry of the reaction, in and of itself, which you are focused on.

In essence, you are doing the first titration in aqueous calcium chloride as well, because as soon as the titration begins, what is formed?

That said, I feel I may be missing something more of the other potential rxn conditions or possible expectations of your evaluator.


Is there anything akin to molecular chirality for crystalline? For example, in hexagonal closed-packed (HCP) structure, the atoms may be arranged in ABAB stacks or ACAC stacks. Can these two be considered enantiomers? Do they possess different solid state properties?

There are some crystal structures where the nature of the crystal structure creates chirality and therefore optical activity.

A trivial example (important because of its history: it was involved in the discovery of optical activity and molecular chirality) is tartaric acid. But the crystal is chiral because the molecule is chiral so that might not be what yyou were looking for.

But there are examples of crystals where the structure as a whole is chiral but the building blocks are not. Quartz, for example, is formed from a network of linked SiO4 tetrahedra (not molecules but a giant interlinked network where each oxygen is shared between two silicons). The overall crystal structure, though, is chiral due to small shifts in the rotation of the linked tetrahedra (see wikipedia for a detailed explanation).

I think this is a crystal analogy to molecular chirality.

Only a few crystal structures can do this and the ones you use as examples can't.


Electrophilic addition to alkenes Consider the addition of HBr (in absence of any radical initiators or environment that promotes radical mechanistic pathway), to 3-bromocyclopentene.

The first step is the attack of $\ce{H^+}$
On the exposed π electron cloud. But my query is whether it forms a classical carbocation like this one to oppose the negative inductive effects of the halogen atom:

Or does it proceed with the formation of bromonium ion like this one to partially complete its octet using lone pairs on Br.

These two would lead to different products but I'm interested in the major one.
The first one would lead to 1,3-dibromocyclohexane, while the second one would lead to 1,2-dibromocyclohexane.

Since it is electrophilic addition, it can go via 2 ways. Either by forming classical carbocation or cyclic carbocation. Since the electrophile will be H+ it will go with classical carbocation. Therefore product will be 1,3 dibromo cyclopentane. Also taken in account the mesomeric effect of bromine atom


"Synthetic wood" from cellulose and lignin? Is there any research into making wood from its constituents?

From what I know, wood is composed of cellulose, hemicellulose, and lignin. We can extract these from raw wood, but is there any way we can combine these to form a wood-like structure that can be made into furniture?

The US published application US20110190402 titled “Synthetic wood composite” appears to be very close to answering your question: “Is there research into making wood from its constituents.”  The full text can be found here: https://patentscope.wipo.int/search/en/detail.jsf?docId=US73316434&tab=PCTDESCRIPTION&_cid=P10-KGFN8M-16074-4
It appears to me that the short answer to your question is, Yes!
Below is part of the Summary of the Invention and Independent Claim 1.
SUMMARY OF THE INVENTION
The present invention is directed to a synthetic wood composite comprising biomimetic macromolecules and methods for the preparation thereof.
In one embodiment, the invention is a composite comprising at least two polymeric macromolecules, wherein at least one macromolecule is lignin and the other macromolecule is selected from the group consisting of cellulose and hemicellulose, or a combination thereof.
In some embodiments, the composite comprises lignin, cellulose and hemicellulose. In other embodiments, the hemicellulose is xylan.
Claim 1 – the main claim

A composite comprising lignin, cellulose and hemicellulose, wherein the cellulose content is about 40 to about 60% (w/w), the hemicellulose content is about 20 to about 40% (w/w), and the lignin content is about 10 to about 30% (w/w), wherein the lignin is selected from a kraft lignin or a lignosulfonate, wherein the hemicellulose is xylan, and wherein the composite is transparent.


Carbene mesomeric structure of vinyl carbocation? Why isn't a mesomeric structure with carbon having only two single bonds and a lone pair used for depiction of vinyl carbocation?

The alternative answer to what @levineds said is that you're asking about a carbene. Carbenes are actually reasonably common in chemistry though we generally have them coordinated to metals for stability. There are a few reactions that proceed through putative naked carbenes.


How does lye appear in ashes? One of the oldest recipes to create lye is this one:

Take well burnt ashes and put them in a barrel
Fill the rest of the barrel with water
Leave the suspension for a week or two
You can now try the lye concentration with egg - if 2/3 remain above the water, the concentration is sufficient for domestic uses
Separate the lye solution from ashes by carefully draining the water. You can also use filtration.

Now I know there are traces of various metals in us, as well in flowers and trees. But how can ashes accumulate such a big amount of sodium? And how is the sodium bound into it. How does it turn into hydroxide?

There is sodium and potassium in wood.  When the wood burns some of the sodium and potassium become bound to oxygen.  The oxides become hydroxides when exposed to water.


Is cycloocta-1,3 diene planar? I know that cyclooctatetraene is non-planar because of

Angle strain and
Antiaromaticity if it was planar.

But what about cycloocta-1,3,5 triene or cycloocta-1,3 diene. It doesn't risk being antiaromatic, but there still is angle strain. Would it lose its planarity and hence inhibit resonance?

In general, when is the angle strain enough, that releasing the angle strain but also inhibiting resonance at the same time lead to a net increase in stability?

Short answer: there's some twisting of the conjugated structure, particularly in the diene

I ran initial calculations for these using B3LYP density functional theory. If anything this method has a slight tendency for extra delocalization.

I'll follow up later with other methods, but I suspect the general answer will be the same.

Cycloocta-1,3,5-triene

Honestly, I expected the delocalization to be retained in the central conjugated section (albeit not perfectly flat) with some twisting on the outer double bonds.

Indeed, the dihedral angle around that central C=C is ~6.9°, but it's not that different around the other CC=CC bonds (~3° and 5°, respectively).

The whole C=CC=CC=C is fairly co-planar, with the remaining two carbons profoundly out of plane, likely relieving ring strain.

It's clear there's some delocalization because the "single" bonds between the double bonds have bond lengths ~1.46Å, smaller than usual.

Thus, in the triene, there's slight non-planarity to the conjugated section, but it's fairly delocalized.

Cycloocta-1,3-diene

Interestingly, the result is a bit different for the diene. While the C=C bonds are planar (i.e., dihedral angle ~2°) the middle C-C bond is strongly non-planar (i.e., dihedral angle ~38.4°). Moreover, it's 1.475Å in length, longer than the comparable bonds in the triene.

Here, I think the delocalization of two double bonds isn't enough to balance the ring strain.


Acid-Base buffer question Calculate the change in pH of a buffer made by combining $\pu{500 ml}$ of $\pu{0.1 M}$ $\ce{CH_3CH_2COOH}$ and $\pu{500 ml}$ of $\ce{CH_3CH_2COONa}$, when $0.04$ moles of $\ce{NaOH}$ is added to the solution. Assume the addition of sodium hydroxide has no effect on volume.
  ($\pu{\mathit{K_a} = 1.3 \times 10^{-4}}$)

My solution:

$$\pu{0.10 M \times ~ 500 ml = 0.05 mol} \ \ce{CH3CH2COOH}$$
$$\pu{0.12 M \times ~ 500 ml = 0.06 mol} \ \ce{CH3CH2COO-}$$

$$\pu{pH = p\mathit{K_a} + \log \left(\frac{0.06}{0.05}\right) = 3.965}$$

Using the equation again where the salt is $\pu{0.1M}$ and the acid is $\pu{0.01M}$ due to the addition of $\ce{NaOH}$ yields a wrong result equal to $1$.

I honestly find the wrong answer to be logical, however I can't think of a way to find the right one.

The reaction is $\ce{NaOH + CH3CH2COOH -> Na+ + H2O + CH3CH2COO-},$ so adding $\pu{0.04}$ moles $\ce{NaOH}$ will result in the solution having $\pu{0.06 +0.04 = 0.1 mol CH_{3}CH_{2}COO^-}$ and $0.05 - 0.04 = 0.01\, \pu{mol} \, \ce{CH3CH2COOH}.$ Then, you can use the Henderson-Hesselbach equation to find the final pH is equal to $-\log(1.3*10^{-4})+\log\left(\frac{0.1}{0.01}\right) \approx 4.886.$


Treatment of allylic alcohol with thionyl chloride: why is the product rearranged? I have read that $\mathrm{S_Ni}$ reactions involving the attack of $\ce{SOCl2}$ on alcohols proceed without the formation of discrete carbocations, and hence there is no rearrangement involved. But that doesn't explain this reaction:

Here, is the allylic carbocation formed, or is there a different explanation for this reaction?

As a disclaimer, it is worth noting that the reaction of an alcohol with thionyl chloride can proceed via multiple different mechanistic pathways (Wikipedia has a pretty good overview), depending on the conditions.
In the $\mathrm{S_Ni}$ mechanism a free carbocation is not formed; instead the key intermediate is an intimate ion pair $\ce{R+...^-OSOCl}$. This is not the same as a typical $\mathrm{S_N1}$ mechanism, where the leaving group is fully dissociated.
Following the formation of the intimate ion pair, the anionic component breaks down to form $\ce{SO2}$ and chloride ion, the latter of which attacks the cationic component $\ce{R+}$ to form $\ce{R-Cl}$. The reaction proceeds with retention of configuration, because the nucleophilic chloride ion originates from the leaving group $\ce{-OSOCl}$, and hence must attack from the same face as the leaving group left from.1

If the carbocation $\ce{R+}$ is allylic, then it is electrophilic at both the α-position (i.e. where the hydroxyl group started) and γ-position (i.e. two carbons away from the hydroxyl group). Therefore, following breakdown of the $\ce{-OSOCl}$ leaving group, the chloride can choose to attack either position. The attack at the γ-position is sometimes called the $\mathrm{S_Ni'}$ mechanism.

While the example above proceeds with extremely high selectivity for the rearranged product, this is not always the case. The above reaction was reported by Young in 1962,2 and carried out in dilute ether solution, which strongly favours the $\mathrm{S_Ni'}$ pathway. According to Gu and Zakarian in Comprehensive Organic Synthesis II (Chapter 6.16),3

Thionyl chloride has been used traditionally as the reagent of choice to ensure a high degree of selectivity in favor of allylic transposition, however, the outcome is often less predictable than desired. [...] The reaction, however, is often more complicated and dependent on reaction conditions and, to a greater extent, the structure of substrate. This is illustrated in the total synthesis of codeine reported in 2011. When allopseudocodeine was treated with thionyl chloride, a nearly equimolar mixture of regioisomeric products was produced.

References

Smith, M. B. March's Advanced Organic Chemistry: Reactions, Mechanisms, and Structure, 7th ed.; Wiley: Hoboken, NJ, 2013, pp 408–409.

Young, W. G. Unexpected rearrangement and lack of rearrangement in allylic systems. J. Chem. Educ. 1962, 39 (9), 455–460. DOI: 10.1021/ed039p455.

Gu, Z.; Zakarian, A. Functional Group Transformation via Allyl Rearrangement. In Comprehensive Organic Synthesis II; Knochel, P., Molander, G. A. Eds.; Elsevier, 2014; Vol. 6, pp 636–754. DOI: 10.1016/B978-0-08-097742-3.00624-8.


Neutral anions in salts and strong acids are similar - is it a coincidence? Recently I was studying salts and my professor provided a list of anions which are considered neutral in salts (that is, they do not change the $\mathrm{pH}$ of an aqueous solution when added as a salt): $\ce{Cl-}$, $\ce{Br-}$, $\ce{I-}$, $\ce{NO_3^-}$, $\ce{ClO_4^-}$.

Now I'm looking at this and I realized the list seems to be almost identical to a list of strong acids: $\ce{HCl}$, $\ce{HBr}$, $\ce{HI}$, $\ce{HNO_3}$, $\ce{HClO_4}$, $\ce{H_2SO_4}$.

Is this a coincidence? If it's not, what's the reason, and why is $\ce{SO_4^2-}$ not on the list of neutral ions?

Though couched in different terms, the first part this question is essentially the same as this other one:

Does a strong acid's conjugate base act as Brønsted-Lowry base?

No, this is not a coincidence.  The conjugate base of a strong acid will always be a weak base.  The 'neutral salt anions' of your professor's nomenclature are just weak inorganic Brönsted bases.

The reason why $\ce{SO4^2-}$ doesn't appear in this list of 'neutral salt anions' is because it actually can noticeably affect the $\mathbf{pH}$ of a solution when added as a salt.  $\ce{H2SO4}$ is often treated as though it's a diprotic strong acid, but in reality only the first deprotonation is strongly acidic.  This can be seen by examining its acid dissociation constants; from Wikipedia:

  $\mathrm pK_{\mathrm a1} = -3$
  $\mathrm pK_{\mathrm a2} = 1.99$

This $\mathrm pK_{\mathrm a2}$ is about the same as the first $\mathrm pK_{\mathrm a1}$ of $\ce{H3PO4}$ $($$2.148$$)$, and is actually somewhat less acidic than nitroacetic acid $($$1.68$$)$.

So, in some cases sulfate may act more or less as a 'neutral salt anion', when the $\mathrm{pH}$ is sufficiently high that it will all stay doubly deprotonated, but in others it can definitely influence the solution $\mathrm{pH}$. I've actually used sulfate/bisulfate before as an acidic buffer for various projects at work—works great!


How was the bottle crushed by air? In this video, a water bottle is shown to be crushed by the surrounding air.

My understanding:

When the person pours cold water over the bottle which has hot water inside it, the hot water produces water vapor which increases the pressure inside in bottle. When the person places the bottle into the cold water bowl, the temperature of the bottle decreases, which causes the water vapor inside the bottle to get condensed, thus lowers the pressure inside the water bottle. When the pressure inside decreases, the pressure outside is greater which crushes the water bottle.

If my explanation of this is wrong in anyway, can someone correct it?

You do not need the water vapour for this effect, the air that is contained in the bottle is sufficient.

You may use the ideal gas law to estimate the amount of air in the volume of the bottle of about $V=1\ \mathrm l=0.001\ \mathrm{m^3}$ when the air is heated to about $T=60\ \mathrm{^\circ C}=333.15\ \mathrm K$ (using hot water) at normal pressure $p=1\ \mathrm{bar}=100\,000\ \mathrm{Pa}$ (since the bottle is still open):

$$\begin{align}
pV&=nRT\\[3pt]
n&=\frac{pV}{RT}\\[3pt]
&=\frac{100\,000\ \mathrm{Pa}\times0.001\ \mathrm{m^3}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times333.15\ \mathrm K}\\[3pt]
&=0.036\ \mathrm{mol}
\end{align}$$

If you now close the bottle, the amount of air $n$ is trapped.

When this air is cooled to about $T=0\ \mathrm{^\circ C}=273.15\ \mathrm K$ (using ice water) at an approximately constant volume of $V=1\ \mathrm l=0.001\ \mathrm{m^3}$, the pressure in the bottle decreases according to the ideal gas law:

$$\begin{align}
pV&=nRT\\[3pt]
p&=\frac{nRT}{V}\\[3pt]
&=\frac{0.036\ \mathrm{mol}\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times273.15\ \mathrm K}{0.001\ \mathrm{m^3}}\\[3pt]
&=8.2\times10^4\ \mathrm{Pa}=0.82\ \mathrm{bar}
\end{align}$$

Therefore, the ambient air pressure of $p=1\ \mathrm{bar}=100\,000\ \mathrm{Pa}$ compresses the bottle until the pressures are equalized (or the bottle fails). The volume in the bottle decreases according to the ideal gas law:

$$\begin{align}
pV&=nRT\\[3pt]
V&=\frac{nRT}{p}\\[3pt]
&=\frac{0.036\ \mathrm{mol}\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times273.15\ \mathrm K}{100\,000\ \mathrm{Pa}}\\[3pt]
&=8.2\times10^{-4}\ \mathrm{m^3}=0.82\ \mathrm l
\end{align}$$

Nevertheless, any condensation of water vapour would help to achieve this effect. The resulting changes in pressure and volume due to condensation of water can be even more extreme than the contraction of air. For example, if a volume of $V=1\ \mathrm l$ is filled with steam at $T=120\ \mathrm{^\circ C}$ and $p=1\ \mathrm{bar}$ (without any air) and then cooled until the steam is almost completely condensed to liquid water (at about $T=99.6\ \mathrm{^\circ C}$), the resulting liquid volume is only $V=0.00058\ \mathrm l=0.58\ \mathrm{ml}$.


Why does water vapour deviate much more from ideal gas behaviour as it is cooled compared to nitrogen gas? My guess is that it is something to do with the difference in boiling point between the two aforementioned molecules. But I can seem to come up with a logical explanation to why this is the case.

Well, an ideal gas has no attraction force between its molecules and does not have a boiling point, as you cannot make it liquid. Also, the sum of  molecule own volumes is supposed negligible wrt the gas volume.

Water does not form anything close to an ideal gas because of its hydrogen bonds.  You have to heat water to 100 °C to make it boiling to overcome these bonds between molecules of liquid water. Molecules of water vapour interacts with each other as well, especially at  high pressure.

If water had not had these bonds, it would have reportedly boiled at -120 °C. And if water had not been polar, what makes intermolecular attraction as well, its boiling point would  have been comparable with nitrogen.

Nitrogen is a non polar gas, with minimal molecular attraction, compared to water, due weak van Der Waals force, which leads to nitrogen boiling at -196 °C.


How to store chloroform? I'm going to be aliquoting out ethanol-stabilized chloroform into smaller containers. It's presently in a large bottle sealed with a rubber(y) sheet so I can only draw it out with a needle.

My questions are:

How do I seal them?
If I'm aliquoting to minimize exposure of the full bottle to air, can I get rid of the rubber sheet?
Are there any restrictions on the sort of cap I can use on the aliquot bottle?

A septum sealed chloroform bottle is likely to have been packed under nitrogen to increase shelf life. Unless you are doing this under a dry nitrogen atmosphere, I'm not sure that decanting into smaller containers is likely to improve shelf life.

Chloroform needs to be stored in a cool, dry, well-ventilated place, out of direct light. Your choice of storage container is likely to be influenced by how long you intend to store it for, but an amber glass bottle with a ptfe-lined cap is probably best suited for longer storage. Rigid polyethylene and polypropylene do not have the best chemical resistance for long-term storage, but if they are air-tight containers will be probably be ok only for short term storage.

Be sure to read and understand the MSDS for chloroform fully before doing this. You will find storage recommendations and incompatibilities listed. DO NOT use containers or lids made from or containing aluminium, magnesium and other metals.


Flipping a molecule vs. mirror behind compound Why aren't these enantiomers?

If you place a mirror behind the compound and look into the mirror, you get the second drawing.

If you place a mirror behind the compound and look into the mirror,
  you get the second drawing

"If you place a mirror behind compound 1 and look into the mirror" - If by that you mean, generate the mirror image of compound 1, well OK, but that does not produce compound 2. It generates the molecule labeled "mirror image" in the following diagram, and it is a non-superimposable mirror image.


Silver-promoted reaction of alkyl halide with diphenyl sulfide What is the major product formed when 1-chloro-3-iodopropane reacts with diphenyl sulfide in the presence of one equivalent of $\ce{AgBF4}$?

I know that sulfur is nucleophilic and that it can displace a halide in a SN2 reaction, but the resulting molecule will have a positive charge. Is that okay? What possible use does the product have?

In the presence of $\ce{AgBF4}$, diphenylsulfide is alkylated to 3-chloropropyl diphenylsulfonium tetrafluoroborate. The sulfonium salt is stable and can be isolated as a crystalline product.

This sufonium salt can be deprotonated (e.g. with $\ce{NaH}$ in $\ce{THF}$) to the corresponding sulfonium ylide.

In the latter, intramolecular displacement of $\ce{Cl-}$ yields cyclopropyl diphenylsulfonium tetrafluoroborate, which again is a stable solid.

This cyclopropyl derivative, once converted to its sulfonium ylide S may serve as a reagent in the Corey–Chaykovsky reaction.


Are there any significant uses of the compound formed by magnesium and anthracene? In an unusual reaction, magnesium reacts with anthracene in THF to form a grignard-like compound called, prosaically, magnesium anthracene:

I remember using it briefly in the 1980s. At the time there was speculation about its potential use in both synthesis and as a starting material for hydrogen-storage systems.

Are there any significant current uses of the the unusual complex?

It seems like an idea of using magnesium anthracene systems for the $\ce{MgH2}$ production persisted since 1980s [1] till late 2000s, when new more efficient method with better scalability for industrial use was established.

One of the recent reviews in hydrogen-storage applications [2, p. 220] compares the older two-step process of $\ce{MgH2}$ synthesis:

  First, metallic magnesium is dissolved in anthracene to form Mg-anthracene solution. Then, the hydrogenation of the Mg-anthracene solution produces $\ce{MgH2}$ with $\ce{CrCl3}$ or $\ce{TiCl4}$ catalyst.

with the most recent direct synthesis from Mg powders and tablets [3]. Their experimental trial involved the following:

  Powder and tablet Mg were placed on trays inside a 50 kg hydrogenation furnace. After purging with Ar, the furnace was ﬁlled with 99.9999% hydrogen with a pressure of less than 2 MPa. The temperature of the process was controlled so as to remain near the boundary between the regions $\ce{\{MgH2\}}$ and $\ce{\{Mg + H2\}}$ in the thermal equilibrium diagram, and maintained for 30–45 h. The hydrogen consumption behavior was monitored by a pressure gauge. Afterward, the furnace was gradually cooled down to room temperature. The hydrogenation yield (%) was calculated via the weight difference of Mg samples before/after the process.

$\ce{MgH2}$ obtained with this method showed a hydrogen yield of around 96% with great reproducibility and scalability, and was proved as industry-ready.

But still, magnesium-anthracene · 3THF is used at a lab-scale in fine organic, organometallic, and inorganic syntheses [4, pp. 300-307]:

as a source of soluble zerovalent magnesium;
in preparation of finely-divided metal powders and transition metal complexes;
catalytic synthesis of Grignard compounds.

Bibliography

Bogdanovic, B. Acc. Chem. Res. 1988, 21 (7), 261–267. DOI 10.1021/ar00151a002.
Sasaki, K.; Li, H.-W.; Hayashi, A.; Yamabe, J.; Ogura, T.; Lyth, S. M. Hydrogen Energy Engineering: A Japanese Perspective; Springer Japan: Tokyo, 2016. ISBN 978-4-431-56042-5.
Uesugi, H.; Sugiyama, T.; Nii, H.; Ito, T.; Nakatsugawa, I. Journal of Alloys and Compounds 2011, 509, S650–S653. DOI 10.1016/j.jallcom.2010.11.047.
Fürstner, A. Active metals: preparation, characterization, applications; VCH: Weinheim; New York, 1996. ISBN 978-3-527-61517-9.


Thermochemistry - zero heat energy We were just wondering in our chemistry class: is there any reaction which is neither exothermic nor endothermic?

You could just as well have wondered is there any number which is neither negative nor positive. Yes, there is such number, and it is called zero. It is rarely encountered among the real-life physical values, though, unless enforced by some law (which is not the case here). After all, there are immensely many positive numbers, and equally many negative numbers, and only one zero. You can hardly expect to hit it by choosing numbers at random.

Then again, reaction enthalpies are subject to variation with temperature, so if we pick any reaction with low enthalpy, it is quite probable that there is a certain temperature at which the said enthalpy is exactly 0. That would be the answer to your question. It is for entropy to decide whether the reaction will occur spontaneously at this temperature. Chances are it will.

On a side note, there is a huge number of "void reactions" like $\ce{H2 + H -> H + H2}$ with guaranteed zero enthalpy at any temperature, but many would object against calling these things reactions in the first place.

So it goes.


Do sugar alcohols have the potential to get a person drunk? From what I understand is ingesting high amounts of sugar alcohols can lead to various effects such as abdominal pain and diarrhea, my question is, is there a certain amount of sugar alcohol if ingested cause drunkenness (even if it causes a person to be in a critical state or death) ?

Sugar alcohols are only partially absorbed. It is very unlikely anyone will become intoxicated with sugar alcohols. Also, their effects are less likely to cause intoxications.


Chemical measurement questions I have some basic chemical measurement questions. I have to mix 3% of potassium sulfate with calcium sulfate as an accelerant. How do I calculate the 3%? Base it off 100%? Does this question make sense?

Also, I need to measure out 5cc amounts of powder from a large bucket of powder. Is there a measuring cup for cubic centimeters?

Generally a specification like 3% of potassium sulfate is wt%. Such a solution would have 30 g in a about a liter of water (weighing 1 kilogram). $$30/1030 = 2.91 \%$$ which should be close enough. You could use 970 g of water to be more exact. But just filling a beaker to the 1 liter mark would work fine.

I never though of cooking in Europe, but I'd guess that they do have measuring spoons in some $\mathrm{cm}^3$ (mL) units. Again good enough would almost certainly work. $$ 1 \text{ teaspoon} = 4.92892 \text{ mL}$$


Schrödinger equation for multi-electron atoms Does the Schrödinger equation alter its form when we apply it for an atom having a configuration other than hydrogen-like? I think it should happen because of difference in nuclear attraction and shielding effects.

If there is a change, how significant is that?

That probably depends on what you consider to be the "form" of the Schrödinger equation.

As Todd succinctly said, the Schrödinger equation[s]

$$\mathrm i\hbar \frac{\mathrm d|\psi\rangle}{\mathrm dt} = \hat{H}|\psi\rangle \quad \text{(time-dependent)} \tag{1}$$

or

$$\hat{H}|\psi\rangle = E|\psi\rangle \quad\text{(time-independent)} \tag{2}$$

don't change. The Hamiltonian will, of course, be different, but the question is how different? The general form of the (non-relativistic) Hamiltonian for an atom has three components

$$\hat{H} = \color{red}{\underbrace{-\frac{\hbar^2}{2m_N}\nabla_N^2}_{\text{KE of nucleus}}} \,\,\,\, \color{blue}{\underbrace{- \sum_i \frac{\hbar^2}{2m_e}\nabla_i^2}_{\text{KE of electrons}}} \,\,\,\, \color{green}{\underbrace{-\sum_i\frac{Ze^2}{4\pi\varepsilon_0r_i}}_{\text{nucleus-electron PE}}}\,\,\,\, \color{purple}{\underbrace{+ \sum_i\sum_{j>i} \frac{e^2}{4\pi\varepsilon_0r_{ij}}}_{\text{electron-electron PE}}} \tag{3}$$

where the nucleus has mass $m_N$ and charge $Ze$; and $r_i$ denotes the distance between electron $i$ and the nucleus, and $r_{ij}$ denotes the distance between electrons $i$ and $j$.

Whether you are dealing with a hydrogen atom or a silver atom, this form remains valid. The only special thing about the hydrogen atom (or in a hydrogenic ion) is that there is only one electron, so the first two sums only have one term, and the last sum has no terms. So, the Hamiltonian for the hydrogen atom can be simplified to

$$\hat{H}_\text{H atom} = \color{red}{-\frac{\hbar^2}{2m_p}\nabla_p^2} \color{blue}{- \frac{\hbar^2}{2m_e}\nabla_e^2} \color{green}{- \frac{e^2}{4\pi\varepsilon_0r}} \tag{4}$$

but I'm of the opinion that the form of the Hamiltonian

$$\hat{H} = \color{red}{\hat{T}_n} + \color{blue}{\hat{T}_e} + \color{green}{\hat{V}_{en}} + \color{purple}{\hat{V}_{ee}} \tag{5}$$

doesn't change, it's just that there are fewer terms in each sum. You might disagree, but I don't really care, as long as you write the correct Hamiltonian out.


Multiphasic liquid "flotation" experiment I'm trying to create an experiment where two immiscible liquids are mixed and are then separated by density. Most of the container is occupied by the lighter liquid, and the small amount of heavier liquid sinks.

Then, a third liquid, immiscible with the other two is added, in amounts intermediate between the two. This liquid is lighter than both of the other liquids and floats.

However, this 3rd light liquid can wet the dense liquid, and after mixing it carries it upwards to the top of the container, basically creating a flotation process.

Here's a rough sketch of what I want to achieve:

I don't care if after the last step the dense liquid then sinks again through the light liquid. If the dense liquid is elevated, even temporarily, that's good enough.

This is slightly similar to industrial flotation processes that separate ore minerals from a slurry of crushed rock. Air bubbles "wet" surfaces of heavy minerals such as copper sulfides and lead sulfides, and float them to the top of a container. In my case I'm talking about three liquids and not a liquid, solid, and a gas.

The reason I'm asking is that I created something similar at high pressure and temperature (~1 GPa, 1000 C) in a sealed pressure vessel and I'm looking for a good way of demonstrating this phenomenon in a more accessible way.

Is there such a thing? I have access to common (and some uncommon) chemical reagents. The liquids have to be transparent, obviously, to see the entire thing. Preferably coloured, or possible to colour them.

https://www.youtube.com/watch?v=GRPufqPgTeI

I played around a bit with this in the lab and this is what I came up with.
Basically, I made a three phase system with hexanes on top, water with fluorescein and green food colouring in the middle, and a perfluoroalkylether HT-110 on the bottom.
Just finding three mutually immiscible liquids was pretty challenging and unfortunately, like most highly fluorinated liquids, HT-110 is almost twice as dense as the other two liquids, but if you look at the video carefully, you can see that some bubbles of HT-110 are getting trapped in the hexanes as its surface energy isn't quite as high in hexanes as it is in water.
When I tilt the vial on its side, you can see these bubbles snapping back across in the interface.

It's not perfect, uses a goofy solvent, and I probably could have played with the liquid ratios/container shape to trap more HT-110 at the top, but it seems that something at least similar to what you're after can be made fairly simply.


Fluorine and fossils They are now assisted by chemists who are able to fix dates through the analysis of pollen, radioactive carbon, and fluorine found in connection with fossils..

Here the text is talking about basic paleontology.

Could anyone teach me in a simple way how fluorine is connected with fossils findings?

(This is originally about English test (TOEFL), but the subject in question does not in my opinion belong to ELL)

I checked the literature, it has been reported that fluorine ions replace hydroxyl ions in osseous tissue (bone) so by making a measurement of the flourine content of bone it is possible to estimate the age of the bones.

One early paper on the subject is Cook, S. F.; Ezra-cohn, H. C. An Evaluation of the Fluorine Dating Method. Southwestern Journal of Anthropology 1959, 15 (3), 276–290. DOI: 10.1086/soutjanth.15.3.3628977.

One method would be to use neutron activation anaylsis on bones to measure the flourine-19, using the n,gamma reaction. This method does have a problem as sodium-23 can form flourine-20 by a n,alpha reaction. As a result a correction will have to be made for the sodium content of the bones.

This is discussed in Cheng, T. P.; Anderson, H. D.; Mills, D. S.; Spate, V. L.; Baskett, C. K.; Morris, J. S. Determination of the fluoride distribution in rabbit bone using instrumental neutron activation analysis. J. Radioanal. Nucl. Chem. 1997, 217 (2), 171–174. DOI: 10.1007/BF02034437.


On what principle do electric alcohol detectors work? I know that normal alcohol detectors work on the principle that potassium dichromate oxidizes alcohol . On what principle does an electric alcohol detector work?

I believe you are referring to breathalyzers. When you breathe into a breathalyzer, any ethanol present in your breath is oxidized to acetic acid (the main component of vinegar apart from water) at the anode (there is a diagram below).  At the cathode atmospheric oxygen is reduced. The overall reaction is the oxidation of ethanol to acetic acid and water. The electric current produced by this reaction is measured by a microprocessor and displayed as an approximation of overall blood alcohol content (BAC).

This website can give you more information.

Hope this helps!


Are there two compound resins that tolerate over 1000°C for glass/carbon fiber? I'm a hobby enthusiast that loves to play with electronics so my major is not in chemistry, there for I'm hoping that this forum is the right place to find aid in this question.

I've searched day in and day out, and when someone refers to "epoxy" with "extreme high temperature tolerance" they're generally speaking about 200-300°C, which to me isn't that much.

I can understand that depending on the context that is a lot, but there has to be something that is actually "extreme" in the sense that it can tolerate up to 1000 °C since formula 1 cars typically hit around 1000°C at the end of a break line and typically the thickness and heat dispersion of the setup just can't take more because of the limits within FIA.

So my question is, what is the technical term for high temperature epoxy's and what other options is there if not?

In an oxygen-containing environment, no carbon-based polymer (which includes all epoxies) that I know of can tolerate 1000 C, at least not without significant degradation in mechanical strength.

There are two reasons for this:

One is that the carbon-carbon bonds will start to break down in large numbers at higher temperatures, and since the carbon-carbon to carbon-oxygen reaction is exothermic, if oxygen is around there will be no going back upon cooling.

The second is that even if the polymer doesn't chemically degrade, it will undergo a physical phase transition - it will melt, or at least soften. This can be compensated for by reinforcing with fiber or particles to make a composite. However, if it softens enough, a fiber composite will "delaminate" - the physical bonds holding the resin onto the reinforcment will come apart, and the composite will no longer be a composite.

Here is an interesting study that talks about both mechanisms of degradation, and lists several systems with their correpsonding melt and thermal decomposition temperatures. Polytetrafluoroethylene has the highest thermal decomposition temperature in this study, at 775 K (502 C).

Non-carbon polymers might be an option, although the tradeoff is that they are weaker and composite options are a lot more limited. Silicon-based polymers generally withstand higher temperatures than carbon analogues, but I wasn't able to find any that can approach 1000 C. Fluoro-polymers are another possibility - as you can see from the reference, they tend to have much higher stability in terms of thermal decomposition. There might be a fluoro-epoxy resin out there. One thing to keep in mind, though, is that the products of fluoro-polymer decomposition are particularly toxic and reactive, and so the applications where they could safely be used at high temperatures are probably very, very limited.

Edit in response to questions

  One question regarding the "oxygen around", would this also include if layers of epoxy were applied as a extra layer ontop of the carbon or would it need to be a complete vacuum in order to prohibit oxidization?

If you had a layered carbon/polymer composite, the chemical reaction with the oxygen wouldn't start at the polymer-carbon interface, it would start at the surface exposed to oxygen. However, since the reaction is exothermic, once it starts burning, it would go quickly. The carbon would burn too at these temperatures. An inert atmosphere would prevent that reaction, but it would still thermally decompose - the difference would be that hopefully the decomposition wouldn't be as exothermic, and therefore would go more slowly. Now for epoxies, they have oxygen in the polymer structure. At high temperatures, that oxygen will react and the polymer will break apart whether there is an inert atmosphere or not.


How do non-sublimable solids exist in equilibrium with vapour? How do non-sublimable solids exist in equilibrium with vapour?

All substances exist in equilibrium with their vapour, something I learned from Jan's comment. But my question is how does that happen with solids that are non-sublimable, like ice, piece of iron, or, a piece of cotton?

Thermodynamically, even cotton exists in equilibrium with its vapour, even though this is completely silly in any practical sense. The reason is that thermodynamics does not involve time; it may take $10^9$ years, but so what! The equilibrium may only consists of one molecule / km$^3$ but $technically$ its still equilibrium! This is because you can't arbitrarily pick and choose the cases you want thermodynamics to apply to, but you can apply reasonable physical insight based on observation rather than applying theory blindly.


Are electron orbitals in a helium atom degenerate? In a hydrogen atom, the presence of only one electron allows various orbitals' energy states to be dependent only on the principal quantum number and not on angular momentum. Orbital degeneracy and all that fun stuff. (This has been answered many times on SE)

In helium, would this still be true? How come? I can't seem to find a satisfactory explanation.

The energy of orbitals for hydrogen atom is dependent only on the principal quantum number: $$E_{n}(\rm{eV})=-\frac{13.6}{n^2}$$
For atoms with more than one electron (all the atoms except hydrogen atom and hydrogenoid ions), the energy of orbitals  is dependent  on the principal quantum number and the azimuthal quantum number according to the equation: $$E_{n,l}(\rm{eV})=-\frac{13.6Z^{*2}}{n^2}$$
Where $Z^{*}$ is the  effective nuclear charge: $Z^{*}=Z-\sigma$.

where$Z$ is the atomic number and $\sigma (n,l)$ is the screen (shielding) constant and it depends on the principal quantum number and the azimuthal quantum number.

According to this, the atomic orbitals $2s$ and $2p$, that are at the same energy in the case of hydrogen atom, are no longer at the same energy for the atom of helium.


Production of hexavalent chromium during electrolysis using a stainless steel as the cathode I want to do rust removal by electrolysis (sodium carbonate, water solution) Can I use a stainless steel basket as a cathode or will it produce hexavalent chromium ?

I know if I use stainless steel as the anode it does produce hexavalent chromium, but what if it is a stainless cathode ?

I have a solution of sodium carbonate, water and hexavalent chromium. What can I do to eliminate the chromium or make it safe, preferably safe enough to dump in the drain if that is possible.

In anode, things get oxidized. That is why the chromium in anode would become hexavalent chromium (hexavalent chromium is the oxidized form of chromium).

In cathode, things get reduced (opposite of oxidized). That is why you could remove rust, because rust is the oxidized form of iron, so if you reduce rust, you would get iron.

You will not obtain hexavalent chromium, because that is not possible through reduction of chromium; that is only possible through oxidation of chromium, which is impossible at a cathode.

You can reduce hexavalent chromium to trivalent chromium with sulphite in acidic environment:

$$\ce{Cr2O7^2- + 8H+ + 3SO3^2- -> 2Cr^3+ + 3SO4^2- + 4H2O}$$

Then you can precipitate out the trivalent chromium with sodium hydroxide:

$$\ce{Cr^3+ + 3NaOH -> Cr(OH)3(s) + 3Na+}$$


A substance that is immisiscible with water and that can be made cloudy I need to identify a substance that is immiscible with water that is or can be made cloudy.

At the moment I'm using glycerol, but I don't know how to make glycerol cloudy nor is it perfectly immiscible with water.

Clarity:

I need to make a bottle containing two liquids. One part being water mixed with rhodamine and the other part another liquid that is immiscible with the rhodamine water mix that is cloudy. The other chemical can either be cloudy already or be mixed with another substance so that it can become cloudy.

It smells but I think chloroxylenol 48 mg in 250 ml ethanol mixed with water would be what your looking for, I'm not sure what it's called where you are but it's a common antiseptic sold everywhere, in Australia the most common brand is Dettol.

Also , if something is completely immiscible with water it will form two phases , both relatively clear, eg water/xylene.

If you mix the solution it will go cloudy but then seperate, the longer it takes to do this depends on how much it was mixed, an emulsion can be formed but that's not what your looking for.

You want something that's only very slightly or sparingly soluble but doesn't seperate into a different phases once it's mixed.

Dettol came straight into my mind but I'm sure there's many more. Arrowroot or cornflour mixed with water would produce a milky/cloudy solution.


How does the Hammett acidity function work and how to calculate it for [H2SO4] = 1,830? Any references? I'm modelling the concentrations in the Belousov Zhabotinsky reaction. For this model, I need to convert [BrO3-] to [HBrO3]. I discovered in several papers that the following correlation applies:
$$ [\ce{HBrO3}] = \frac{h0}{h_0+0.2} * [\ce{BrO3^-}]_0 $$
See for example this dissertation on page 28 or this paper on page 2.
I have a solution which has $[\ce{H2SO4}]$ = 0.915M (I assume $[\ce{H+}]$ = 1,830 M).
My $[\ce{BrO3^-}]$ was 0,2411 M.
I researched the Hammett Acidity function, but I still don't completely understand the concept. What I do know is that $h_0$ substitutes for $[H+]$ and that $h_0$ has the unit of M.
Can anyone explain the concept of this function in a bit more detail and explain in what way it exactly relates to pH (and therefore to H+)?
On the internet I found this paper, which says that $h_0$ = 3 when $[\ce{H2SO4}]$ = 1.5M. Are there any other tables or references with $h_0$ related to the concentration of $\ce{H+}$ or $\ce{H2SO4}$? There is one more question on SE about Hammett acidity. This question links to tables with $h_0$ for certain % of H2SO4. Can I perhaps use that? How does $[\ce{H2SO4}]$ relate to a % of H2SO4?
Is there a way to calculate the $h_0$ when $[\ce{H2SO4}]=1.830 M$?

I have a solution which has $[\ce{H2SO4}]$ = 0.915M (I assume $[\ce{H+}]$ = 1,830 M).

While $\ce{H2SO4}$ is a strong acid, $\ce{HSO4-}$ is not ($\mathrm{p}K_\mathrm{a}$ around 2). So your pH should be about zero, and you might not need to use the Hammett acidity at all.


How to extract elemental bismuth from pepto bismol? I was trying to isolate elemental bismuth from pepto bismol (bismuth subsalicylate) using the process seen here but after I added hydrochloric acid, instead of getting a pink foamy substance, I got a white liquid with some white foam. The only reason I can think of, for why this would happen, is that the pepto bismol was expired by a month.

After I realized the reaction didn't work, I added a homemade universal indicator (red cabbage - phenolphthalein - preserved with a little isopropyl alcohol) and started adding sodium hydroxide to neutralise any remaining acid. When I started adding the sodium hydroxide, the solution started heating up and then changed color fairly rapidly from pink to a dull grey green which didn't change when I tried adding more acid.

Shortly after the color change, something started to precipitate out. I filtered off the precipitate and got a clear orange liquid. I then added a little hydrochloric acid to the orange liquid and something immediately started precipitating out. The remaining liquid then turned clear and smells slightly like mint.

My only guess for what I have made is bismuth hydroxide (the first precipitate) but I have no idea what the second precipitate is. Here are pictures of the final precipitates.

I'm not sure you did do the experiment wrong. I performed this experiment and it closely matched this video which had a light yellow liquid and foam.

Here are my results

I filtered it a couple more times to remove more foam. Then added the aluminum foil.

This is what I got at the end before I filtered the liquid out at the end.

In the end I was able to extract 6.2 grams of bismuth powder from 67 tablets of pepto (a bit less than the theoretical ~10 grams I was hoping to get).


Why is Cubic closed packing called "cubic"? How is Cubic Closed Packing "cubic" and also how is hexagonal closed packing "hexagonal"?
I have been thinking over this for a few hours but I am unable to visualise.

I understand that when placing the third layer on top of the second layer ,$2$ situations arise:
If the third layer is placed on the tetrahedral voids it forms an A-B-A-B... type structure which is called hcp (why?)
and If the third layer is placed on the octahedral voids it forms an A-B-C-A-B-C... type structure which is called ccp (why?)

Start with the hexagonal A-B-A-B structure.  It is hexagonal because, choosing one atom in a layer, it is surrounded by 6 equivalent atoms. So, you have a 6-fold symmetry axis, hence the 'hex-'. The fact that it is close-packed can be gotten at mathematically (or at least shown that it is closer-packed than other crystal structures).

Cubic close packing is, at least in most materials science and metallurgy circles, much more commonly called face centered cubic. The 'standard' unit cell is a cube of atoms with additional atoms at the center of each face.  It is somewhat harder to visualize how this is related to close-packed planes, although there are many pictures available on-line (one, chosen somewhat randomly, is from Florida State). The point is that the stacking direction of the close-packed planes are along the 111 body diagonal of the cube.

Rotate the cube so that it is standing on one point with the opposite point on the body diagonal pointing up. Your close packed planes are:
(1) the bottom point, (2) the top point, (3) the three corner atoms closest to the bottom point as well as the face atoms closest to the bottom atom, (4) the three corner atoms closest to the top atom as well as the face atoms closest to the top atom. It may take some squinting...


Systematic name of OCS (COS) What is the systematic name of carbonyl sulfide? My professor commented that the systematic name was "very odd" so he didn't bother to mention it. Is it really that odd? And what is it, actually?

I found the name "Thioxomethanone". Is that really odd? From my limited knowledge, I've seen the root "thiol" before in describing something with a sulfur in it. So at least the root doesn't seem weird. I also see "oxo." There's an O in OCS.

I'm not sure if there's definitive answer to not-yet-asked question, whether this compound can be considered organic or inorganic compound.

While organic nomenclature books do not mention it (acording to the current rules it might be called sulfanylidenemethanone, which is somewhat absurd), official IUPAC inorganic chemistry nomenclature (Red Book 2005, table IX, p. 292) lists two names for this compound:

carbonyl sulfide
oxidosulfidocarbon

Your mentioned (organic) systematic name thioxomethanone is just a variant of already mentioned sulfanylidenemethanone, as S= group's preferred name ("preselected prefix") is now sulfanylidene-* instead of thioxo- (rule P-15.5.3.3)

*) see analogy H2C=/H3C− methylidene-/methyl-, and
S=/HS− sulfanylidene-/sulfanyl- (older name: mercapto-)


Isolation of cis-3-hexanol from grass If I wanted to isolate cis-3-hexanol (cis-3-hexen-1-ol;  IUPAC name: (Z)-Hex-3-en-1-ol ) from grass I could start the extraction by dissolving the grass in a solution of ethanol. EtOH is a common solvent for extractions, and dissolving grass in it would leave some remaining solid, as well as a crude solution containing the cis-3-hexanol. Cis-3-hexanol is a liquid at room temperature and is soluble in ethanol but not (only to a small degree) soluble in water.

After filtering out the remaining undissolved solids, I think the next best step would be to boil off the remaining ethanol and water with both of their boils points being lower than cis-3-hexanol's boiling point at around 156.5 degrees C. This would presumably leave a crude solution containing cis-3-hexanol and other organic components extracted from the grass.

What would be the best method to further purify the remaining solution for a higher purity? Please correct me if my thought process is wrong.

Cis-3-hexanol is used in perfumes and the demand for it has increased in industry.

First, how much compound do you want to make?
Second, how much cis-3-hexanol is there in grass? It is probably around 0.01%, so be ready to process a lot of grass.
Third, it might be reactive (that is it can decompose).

Your best bet is to:
(1) Figure out theoretical yield
(2) Do steam distillation rather than extraction (bp is 156.5 °C - good for distillation).
(3) Extract collected fraction (mostly water+ some organics including hexenol) with some light organic solvent (hexane, ether). You can try ethyl acetate (aka MEK substitute, bp = 77C), but that is the highest bp you want to go for.
(4) Split organic solution in several fractions and try washing it with water solutions of 5% baking soda and then brine (saturated solution of NaCl). Hexenol should stay in organic solvent unless it decomposes.
(5) Concentrate it under reduced pressure. If you don't have access to rotavap - let it sit in open air. You can direct fan at it if you will, but you might catch fire if you are not accurate.

You will get a drop of something that is a mix of 20 compounds. How you need to do column chromatography, which is pain even with equipment.

It doesn't sound like you have access to chemistry lab. In this case for practice purposes try extracting eugenol from cloves first. Grind cloves, place in distillation apparatus, add water, distill. Oil on top is eugenol.

Steam distillation: Shred grass, add to a gallon of water and distill it off using a lab distillary apparatus.


How is the octane percentage defined? Normally octane percentage is mentioned on petrol (95% octane). How is this defined? Is it percentage of chemical or some other parameters?

See the Wikipedia article on octane rating. 95% octane is the same as an octane rating of 95, and it means the petrol has the same knocking characteristics as a mixture of 95% iso-octane and 5% heptane.


Does a bulky nuclephile react faster by Michael or direct addition? Between Michael or direct addition on a $\alpha,\beta$-unsaturated ketone, I think that Michael addition is faster than the direct addition as the nucleophile can attack orthrogonally to the $\ce{C=C}$ which is a less hindered approach than the Bürgi-Dunitz angle required to attack the $\ce{C=O}$.

First, you got wrong thinking Michael addition is faster than the direct addition. Actually, Michael addition (thermodinamic control) is comparatively slower than the direct addition (kinetic control) here.

Second, to my understanding, Bürgi-Dunitz angle (Ref.1) has nothing to do here (may be it affect both additions, you never knows). However, there are several factors control the outcome.  The way that nucleophiles react depends on three factors:

The conditions of the reaction;
The nature of the $\alpha,\beta$-unsaturated carbonyl compound; and
The type of the nucleophile.

An example for the conditions of the reaction: The reaction of but-3-en-2-one and $\ce{KCN}$  in the presence of an acid catalyst gives 1,1-cyanohydrin (2-cyanobut-3-en-2-ol) at $\pu{10-20 ^\circ C}$ by 1,2-addition (kinetic product), but would give 4-cyanobuta-2-one at $\pu{80 ^\circ C}$ by 1,4-addition (thermodynamic product). Thus, it is clear that the Direct addition to the carbonyl group (which forms at low temperature) is faster than conjugate addition (forms at higher temperature), giving cyanohydrin. However, that reaction is reversible and, as a result, converted to more stable cyanobutanone at high temperatute, which is essentially an irreversible reaction.

Why is direct addition (1,2-addition) faster than conjugate addition (1,4-addition)?: Simple molecular orbital calculation shows both, the carbon atom of the $\ce{C=O}$ group and the carbon atom $\beta$ to it, carry some positive charges. However, the carbon atom of the $\ce{C=O}$ group carries comparatively more charge, so electrostatic attraction for the charged nucleophiles will encourage it to attack the carbonyl group directly rather than undergo conjugate addition.

On the other hand, with direct 1,2-addition, we gain a new $\ce{C–C}$ $\sigma$-bond, but lose the $\ce{C=O}$ $\pi$-bond and keep the $\ce{C=C}$ $\pi$-bond. Meantime, with conjugate 1,4-addition, we still gain a new $\ce{C–C}$ $\sigma$-bond, but lose the $\ce{C=C}$ $\pi$-bond instead of $\ce{C=O}$ $\pi$-bond, and keep the $\ce{C=O}$ $\pi$-bond instead of $\ce{C=C}$ $\pi$-bond. Overall, the $\ce{C=O}$ $\pi$-bonds ($\pu{369 kJmol-1}$) are stronger than $\ce{C=C}$ $\pi$-bonds ($\pu{280 kJmol-1}$), hence the conjugate addition gives comparatively more stable product.

An example for the nature of the $\alpha,\beta$-unsaturated carbonyl compound: The reaction of 2-butenal (an $\alpha,\beta$-unsaturated aldehyde) and butyllithium in THF at $\pu{-70 ^\circ C}$ gives oct-2-en-4-ol by 1,2-addition exclusively. Yet, under the identical conditions, N,N-dimethyl-2-butenamide (an $\alpha,\beta$-unsaturated amide) would give N,N-dimethyl-2-butylbutanamide by 1,4-addition exclusively.

An example for the type of the nucleophile (Hard or Soft): The reaction of 4-methylpent-3-en-2-one (an $\alpha,\beta$-unsaturated ketone) and thiophenol at $\pu{25 ^\circ C}$ gives 4-methyl-4-thiophenoxypent-2-one exclusively by 1,4-addition (Remarkably, no acid or base catalyst is needed for this reaction,which would be needed for the alcohol additions, instead).

  Here, it has been explained that the attraction between nucleophiles and electrophiles is governed by two related interactions: (1) Electrostatic attraction between positive and negative charges, and (2) Orbital overlap between the HOMO of the nucleophile and the LUMO of the electrophile. Successful reactions usually result from a combination of both, but sometimes reactivity can be dominated by one or the other. The dominant factor, either it be electrostatic or orbital control, depends on the nucleophile and electrophile involved. Nucleophiles containing small, electonegative atoms (such as $\ce{O}$ or $\ce{Cl}$) tend to react under predominantly electrostatic control, while nuclophiles containing larger atoms (including the sulfur of thiols) are predominantly subject to control by orbital overlap. The terms hard and soft have been coined to describe these two types of reagents. Hard nucleophiles are typically from the early rows of the periodic table and have higher charge density, while soft nucleophiles are from the later rows of the periodic table—they are either uncharged or have larger atoms with higher-energy, more diffuse orbitals (Ref.2).

Few examples:

$$
\begin{array}{c|c}
\hline
\text{Hard nucleophiles} & \text{Borderline nucleophiles} & \text{Soft nucleophiles} \\
\hline
 \ce{F-, \mathbf{HO^-, RO^-}, SO4^2-, Cl-} & \ce{N3-, \mathbf{CN^-}} & \ce{\mathbf{I^-, RS^-}, RSe-, S^2-}\\
\ce{\mathbf{H_2O, ROH}, ROR', RCOR'} & \ce{\mathbf{RNH_2, RR'NH}} & \ce{\mathbf{RSH, RSR'}, R3P}\\
\ce{\mathbf{NH_3, RMgBr, RLi}} & \ce{\mathbf{Br^-}} & \text{Alkenes, aromatic rings}\\
\hline
\end{array}
$$

Electrophiles can also classified in this manner. For example, $\ce{H+}$ is a very hard electrophile because its small size and charged. Bromine $\ce{Br2}$ is, on the other hand, a soft electrophile due to its large size and having no charge.

In general, hard nucleophiles prefer to react with hard electrophiles, and soft nucleophiles with soft electrophiles.

Overall, Hard/Soft reactivity:

Reactions of hard species are dominated by charges and electrostatic effects
Reactions of soft species are dominated by orbital effects
Hard nucleophiles tend to react well with hard electrophiles
Soft nucleophiles tend to react well with soft electrophiles

References:

H. Bürgi, J. Dunitz, J. Lehn, G. Wipff, "Stereochemistry of reaction paths at carbonyl centres", Tetrahedron 1974, 30(12), 1563-1572 (http://dx.doi.org/10.1016/S0040-4020(01)90678-7).
Jonathan Clayden, Nick Greeves, Stuart Warren, Organic Chemistry; 2nd Edn; Oxford University Press: Oxford, United Kingdom, 2012.


Is astatine a non-metal or a metalloid? Is astatine a non-metal or a metalloid?

A metalloid is an element which exhibits some of the properties of a metal as well as those of a non metal.

Is Astatine a non metal or a metalloid?

Unknown

According to the Wikipedia article on astatine, the longest-lived isotope has a half-life of 8.1 hours. The element is so radioactive that large samples would melt and vaporize under the heat generated by its own radioactivity. Such behavior makes it difficult to assess whether astatine has the appropriate traits to be a metalloid.

Metalloids tend to be lustrous, brittle, semiconductors with other properties intermediate between metals and nonmetals. Nonmetals are dull/matte, brittle, and poor conductors of electricity and heat. All of these properties require a visible solid (or liquid) sample to assess, and thus astatine is impossible to assess and classify.

Astatine is usually placed with the metalloids or the nonmetals by different authors based on its position in periodic table, but in truth, astatine is so unstable in its elemental form, we have no idea which one it is.


Best reagent to convert 4-chloro-4H-pyran into an aromatic compound My question:

My attempt: (I am not really good at mechanisms, so a little help would be good. Thank-you)

I do not know what kind of reactions these will give.

In order to choose the best reagent to convert the compound to an aromatic one, let's ask the question of what is preventing this compound from being aromatic? At carbon-4, the pi-system is disrupted by an sp3-hybridized atom. If carbon-4 could be converted to sp2 by some reaction, then the compound could become aromatic. At C-4, we see a chloride. We know that chlorides are good leaving groups, so let's look at the product of that step.

After chloride leaves, there's a cyclic, fully conjugated pi-system containing six electrons (two for each double bond, two from the oxygen, and zero from the carbocation). Drawing a resonance structure shows that this is an oxygen analog of benzene (known as pyrylium ion).

What kind of reagent would promote this process, i.e., a halide leaving? A Lewis acid, such as $\ce{AlCl3}$ could promote this in two ways. First, it could interact with the chloride in the starting material, building up positive charge on the chloride, making it a better leaving group. Secondly, the Lewis acid can coordinate the chloride on the product side, preventing it from attacking the pyrylium ion.


Strain Energy in di-axial vs di-equatorial I am dealing with two conformations of cis-1,4-dimethylcyclohexane. One conformation is di-axial, and the other one is di-equatorial. Why does di-axial have higher strain energy. I would make sense if the molecule was 1,3-dimethylcyclohexane because then it would the steric hinderance that would cause higher energy. Is it similar with 1,4-.. conformation? If its not steric repulsion, what factors cause di-axial to have higher strain energy?

You can't have both in a chair conformation!

If the methyl groups in 1,4-dimethylcyclohexane adopt a bis-axial orientation, they are trans and they stay trans after ring flip:

If your 1,4-dimethylcyclohexane is cis-substituted the orientation of the methyl groups is axial and equatorial, respectively:


Why aren't these two resonance structure same? Draw the resonance structures of Fluorobenzene.

In the solutions the structures below were given, I wonder why these structures are not just same. Both of these structures have a negative formal charge on the ortho position.

Are the above structure of Fluorobenzene same ?

There are called equivalent resonance structures since the number of pi bonds, lone pairs and sigma bonds are the same. On the other hand, resonance is the delocalization of pi electrons. These structures show how pi electrons are delocalised from ortho position to para position and then back to ortho(on the other side of the molecule). So both the structures you mentioned are different in this context and do have a purpose in explaining the resonance phenomenon.0

Note: This rule only applies to hydrocarbons. In other cases, the two resonance structures need to have the same structural formula to be equivalent.


Naming of a molecule $\ce{H2C=C=O}$ what is the group of the  molecule? How to characterize it and how  is this dissimilar to ketone?

If we add $\ce{O2}$ with it then will we get $\ce{CO2}$?

As, your question isn't much clear and it needs to be focused little more.
From current state of this question, I can provide you the answers for specific questions that I get to understand from its current state.

$\ce{H_2 \ \ C=C=O}$
What is the group of the  molecule?

Well this is known as ketene.

how to characterize it and how this is dissimilar to ketone?

Ketenes are generally very reactive, and participate in various cycloadditions. They will also undergo [2+2] cycloaddition reactions with electron-rich alkynes to form cyclobutenones, or carbonyl groups to form beta-lactones.

Ketones engage in many organic reactions. The most important reactions follow from the susceptibility of the carbonyl carbon toward nucleophilic addition and the tendency for the enolates to add to electrophiles.

See this and this

If we add $\ce{O2}$ with it then will we get $\ce{CO2}$?

No! I hope structure of ketene and wikipedia articles make it clear to you why. :)
Also see this.


What is the momentum of a particle described by the following equation? $$\psi(x) = Ae^{ikx}  \ + \ Be^{-ikx}$$

The above is not an eigenfunction and to obtain its momentum my professor applied the momentum operator separately to the first and second parts. The final result was that he obtained two different values of momentum,
$$\hbar k \text{ and } -\hbar k\text{.}$$

My question: is this method correct? If yes, how to explain the separation of values?

The idea of superposition of eigenstates is quantum 101, so you should probably go and read a proper book on QM. I personally recommend Griffiths.

In quantum mechanics, observables such as $p$ are represented by Hermitian operators such as $\hat{p} = -\mathrm{i}\hbar(\mathrm d/\mathrm dx)$.

The momentum of a particle is not well-defined unless it is in an eigenstate of the momentum operator $\hat{p}$. If it is, then a measurement of the momentum will always return the corresponding eigenvalue. For example, if the wavefunction of the particle is simply $\exp(ikx)$ , then a measurement of the momentum will always return the value $\hbar k$, and the particle is said to exist in a state of well-defined momentum.

Now, if the wavefunction of the particle is $A\exp(ikx) + B\exp(-ikx)$, then a measurement of the momentum will sometimes yield $\hbar k$ and sometimes yield $-\hbar k$. So, the actual momentum of the particle is not well-defined, as it could be either of the two values (before you observe it).

The expectation value of the momentum, though, is well-defined. It is simply

$$\langle p \rangle = P_+(\hbar k) + P_-(-\hbar k)$$

where $P_+$ and $P_-$ are the probabilities of measuring the momentum to be $\hbar k$ and $-\hbar k$ respectively. Assuming that your wavefunction is normalised, i.e. $|A|^2 + |B|^2 = 1$, basic quantum mechanics tells us

$$P_+ = |A|^2;\quad  P_- = |B|^2$$

and so the expectation value in this case is found to be

$$\langle p \rangle = \hbar k (|A|^2 - |B|^2)$$

However, that simply means that if you prepare many particles in the state above, measure their momenta, and average their momenta, the average will tend toward this value. Each individual measurement is guaranteed to give either the value $\hbar k$ or $-\hbar k$.

And yes, each individual particle does not have one single well-defined momentum. That's quantum mechanics for you.


Basicity of hypoboric acid Why is the basicity of hypoboric acid $\ce{H4B2O4}$? Is it $2$ even though it has $4$ hydroxyl groups?

My teacher told us that the basicity of this acid is $2$ but I am not remembering the exact reason for it.

This is a Lewis acid, not a Brønsted acid. Like boric acid $\ce{B(OH)3}$, which form an adduct $\ce{B(OH)3(H2O)}$ which then lose a proton from the water molecule of the adduct: $$\ce{B(OH)3(H2O) -> B(OH)4- + H+ }$$

Above molecule will do this at each B atom, and so lose 2 $\ce{H+}$ in total, not 4. IE will have:
$$\ce{B2(OH)4 + 2H2O -> B2(OH)6^{2-} + 2H+}$$
as the overall reaction.
Note $\ce{B(OH)3}$ acts as a Lewis acid due to vacant B p orbital, which can accept lone pair of $\ce{H2O}$ and so form an adduct. From this adduct lose proton, and hence acidic.


Niobium has a density of 8.57 g/cm3 and crystallizes with the body-centered cubic unit cell. Calculate the radius of a niobium atom Here is my question

  Niobium has a density of $8.57 \pu{g/cm^3}$ and crystallizes with the body-centered cubic unit cell. Calculate the radius of a niobium atom.

Here is my solution

This is how I would do the calculation...
Given data:
$$\text {Weight of crystal} = \pu{92.90638 g/mole} \ce{->[Round] \pu{92.906} g/mole}$$
$$\text {Density} = \pu{8.57 g/cm^3}$$
Ok, the density is good to only 3 significant figures so the answer shouldn't have any more than that. But doing the whole problem, I'll carry 5 significant figures throughout all the intermediate calculations to try avoid rounding errors within  the multiple calculations. I'll round the final result to 3 significant figures.
$$\text{Volume of mole} =\text{V}_{mole} = \frac {92.906}{8.57} = \pu{10.841 cm^3/mol}$$
In BCC, there are 2 atoms in one unit cell. Also note that the accepted figure for Avagadro's constant to five significant figures is now $6.0221\times10^{23}$
$$\text{Volume of unit cell} = \text{V}_{cell} = 10.841 \times \frac {2}{6.0221 \times 10^{23}} = 3.6004 \times 10^{-23}\text{ cm}^3$$
A unit cell is a cube with each side being $a$
$$\text{V}_{cell} = a^3$$
$$\therefore a = \sqrt[3]{3.6004 \times 10^{-23}} = 3.3021\times 10^{-8}\text{ cm}$$
But the BCC crystal, spheres of radius, $r$, packed inside a cube with side, $a$, the geometric relationship between $a$ and $r$ is: $$a = \dfrac{4}{\sqrt{3}}r$$ so
$$r = \dfrac{\sqrt{3}\times(3.3021\times 10^{-8})}{4} = 1.4299\times 10^{-8}\text{ cm}$$
Now rounding $r$ to 3 significant figures gives
$$r = 1.43\times 10^{-8}\text{ cm} = 143\text{ pm}$$
You shouldn't chase values through websites and literature. You can only calculate given whatever data you're suppose to take as given. For instance no temperature was given for the density which certainly changes with temperature. Also, to make the point again, with a calculator you should carry extra significant figures in the intermediate calculations to try to avoid round off errors in the final result(s).


Mechanism of substitution reaction with no change in stereochemistry When (2S,3R)-3-iodobutan-2-ol undergoes a substitution reaction with sodium azide the only organic product from the reaction is (2S,3R)-3-azidobutan-2-ol. Give a mechanism for the reaction.

My Attempt

Now I know this is not a normal $\mathrm{S_N1}$ or $\mathrm{S_N2}$ reaction since the stereochemistry remains the same and only one of the enantiomers are formed. I am guessing that at first there is some intramolecular reaction where the iodine atom attacks the other carbon bonded with the hydroxyl group. Then the azide substitutes the iodine atom.

I can explain the reason why the configuration does not change by using neighboring group participation of the hydroxyl group. This is the mechanism that I propose:

The hydroxyl group attacks C-3 and $\ce{I-}$ leaves.
Then $\ce{N3-}$, the nucleophile, attacks C-3.
Since two successive inversions occur at C-3, the net result is retention.

As you can see, the configuration has not changed.

@bon brought up a point: The formation of a mixture of products when an epoxide opening takes place. If the nucleophile were to attack C-2 instead,

C-3 will still have the S configuration.

As noted by @Jan in the comments, the epoxide intermediate is symmetrical and both epoxide carbons are equivalent. This can be shown by a $C_2$ rotation:


How to determine the minimum pH at which iron(III) hydroxide dissolves completely? Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $\pu{10 mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.

My idea was to calculate the amount of $\ce{Fe^3+}$ and $\ce{OH-}$ ions, and then add enough acid containing $\ce{H+}$ ions in order to neutralise the $\ce{OH-}$ ions corresponding to the amount of $\ce{Fe^3+}$. To do this I used the equilibrium constant leaving the hydroxide ions as the unknown. Then I calculated the difference between the initial hydroxide concentration (corresponding to $\pu{10 mg}$ of salt) and the needed concentration to reach $K_\mathrm{s}$. This corresponds to the concentration of hydronium ions in the acidic solution, and I can calculate pH.

Doing this I get $\mathrm{pH} = 3.03$. Is this how it should be done?

The OP first stated the problem as:

  Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $\pu{10 mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.

I took that to mean that acid was added to a solution containing 10 mg of iron (iii) hydroxide and the pH of the solution when the iron (iii) hydroxide was totally dissolved was desired.*

Knowing that $K_\mathrm{sp} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ then
\begin{align}
  K_\mathrm{sp} &= \ce{[Fe^{3+}][OH^-]^3} = 6\cdot10^{-38},&
\text{ so }&&
\ce{[OH^-]} &=\sqrt[3]{\frac{6\cdot10^{-38}}{\ce{[Fe^{3+}]}}}
\end{align}

You need to solve for $\ce{[Fe^{3+}]}$ which you can calculating knowing that there is $\pu{10 mg}$ of $\ce{Fe(OH)_3}$ in 0.1 liters of solution. It has a molecular weight of $\pu{107 g/mol}$, so there is
$$0.010/107 = \pu{9.35* 10^-5 mol//L}\text{ of }\ce{Fe(OH)_3}.$$

Going back to the equation for the $K_\mathrm{sp}$:
$$\ce{[OH^-]} =\sqrt[3]{\frac{6\cdot10^{-38}}{\ce{[Fe^{3+}]}}} = \sqrt[3]{\frac{6\cdot10^{-38}}{9.35\cdot10^{-5}}} = 8.63\cdot10^{-12}$$

Knowing $\ce{[OH^-]}$ you can calculate $\ce{[H^+]}$ via

$$\ce{[H^+]} = \frac{1\cdot10^{-14}}{\ce{[OH^-]}} = \frac{1\cdot10^{-14}}{8.63\cdot10^{-12}} = 1.16\cdot10^{-3}$$

and then pH:

$$\mathrm{pH} = -\log\left(1.16\cdot10^{-3}\right) = 2.94.$$

Now the OP has modified the problem statement to be:

  Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $10\ \mathrm{mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.

We again end up with the final pH being just acidic enough to dissolve $\pu{10 mg}$ of iron (iii) hydroxide in $\pu{100 ml}$ of water, but obviously the solution must start out more acidic. The final solution will again have a pH of $2.94$ or a $\ce{[H^+]} = 1.16\cdot10^{-3}$.

In $\pu{100 mL}$ of a solution of pH $2.94$ there are $\pu{0.116 mmol}$ of acid.

We calculated before that there were $\pu{9.35*10^{-5} mol}$ of $\ce{Fe^{3+}}$, but there are three $\ce{OH^-}$ anions for every cation of $\ce{Fe^{3+}}$, so there are $3 \times 9.35\cdot10^{-5} = \pu{2.81*10^{-4} mol}$ of $\ce{OH^-}$.

The total amount of substance (in millimoles) of acid needed is
$$0.116 + 0.281 = 0.397.$$

For $\pu{100 ml}$ to contain $\pu{0.397 mmol}$ the solution would need to be 3.97 millimolar in $\ce{[H^+]}$, or the pH = $2.40$.

If you start with $\pu{100 ml}$ of solution with pH $2.40$ and dump in $\pu{10 mg}$ of $\ce{Fe(OH)3}$, then the final pH will be $2.94$ and all of the $\ce{Fe(OH)3}$ will be just dissolved.


How is energy released when a chemical bond is formed? Chemical reactions are classed as endothermic or exothermic.

Defn: exothermic

  (of a reaction or process) accompanied by the release of heat.

The heat is released when the chemical bonds in the product are formed.

Defn: heat

  heat is energy in transfer to or from a thermodynamic system, ... The
  mechanisms include conduction, ...; or radiation between separated
  bodies; or ....

So, I have three questions:

1 - if the heat is released via conduction, is this equivalent to an increase in the kinetic energy of the product(s) compared to the kinetic energy of the reactant(s)?  Also, how does the formation of a chemical bond produce kinetic energy?

2 - if the heat is released via radiation, how is radiation created in the formation of a chemical bond?

3 - are conduction and radiation the only forms of heat transfer when a bond is created?

Edit - to clarify the question - the issue is addresed in the OPENSTAX AP chemistry text - which covers one example of 2 H atoms coming together to form H2 in detail as follows:

  When the atoms are far apart ..... by convention the sum of their
  energies is 0. As the atoms move together their orbitals begin to
  overlap.  Each electron begins to feel the attraction of the nucleus in
  the other atom.  In addition, the electrons begin to repel each other,
  as do the nuclei.  While the atoms are still separated, the
  attractions are slightly stronger than the repulsions, and the energy
  of the system decreases. (A bond begins to form.) ... At some specific
  distance between the atoms, which varies with the atoms involved, the
  energy reaches its lowest  (most stable) value.  This optimum distance
  between the two bonded nuclei is the bond distance between the two
  atoms.

The text doesn't specify what energy is decreasing, and why.  But I can answer these questions I think.  The energy that is decreasing is the potential energies of the attractive fields of the atoms, and that energy is being transformed to kinetic energy as the atoms accelerate toward each other.

So we have the nuclei accelerating towards each other.  What happens next?

According to the text ... continuing where we left off

  The bond is stable because at this point the attractive and repulsive
  forces combine to create the lowest possible energy configuration.

So there is no mention of the release of energy, and the atoms just arrive at the stable combined state.

Here is what I would expect, given that the nuclei are accelerating towards each other - at some point the repulsive forces become stronger than the attractive forces, and the atoms begin to decelerate.  If there is no damping force on the motion the atoms will oscillate about the point where the attractive and repulsive forces are equal.  So, there must be some damping force on the atoms, that causes them to come rest at the bond distance of H2.

Now comes the mystery, which is, how does the formation of a chemical bond release energy?

What happens to the kinetic energy of the two nuclei accelerating towards each other?  If the two hydrogen atoms are coming together head on there will be a collision that would appear to cancel out the motion, i.e. kinetic energy, of both atoms.  If there is a damping mechanism that generates the heat, what is it?

Since the energy is released as heat, I think it must be in the form of kinetic or electromagnetic energy, but what is the specific mechanism?

Or, am I trying to get too much out of the kinetic model?

It is easier first to think of what happens when a bond is broken in an exothermic reaction. The energy is initially released into vibrations,  rotations and translational motion of the two fragments. This energy is then lost as these fragments collide with other gas molecules or with solvent molecules and so eventually ends up as heat.

In exothermic bond formation the extra energy vibrationally and rotationally excites the new molecule and this is then lost via collisions with surrounding gas or solvent molecules and again ends up as heat.

(If the new molecule were isolated in space (so no collisions) it would remain 'hot' until the energy is radiated away from vibrational levels and until it (radiatively) comes into equilibrium with the surroundings. This is a  slow process, by many orders of magnitude,  compared to collisions in solution or in a gas at normal atmospheric pressure )

In a very few types of reaction an electronically excited state is produced in the product and then photons radiate (as visible light) some of the energy away. This is called chemiluminescence.


Where does the energy required for excitation of electrons in hybridization come from? We're commonly taught that when atoms want to hybridize in for instance methane, an electron is excited from $2s$ to $2p$ to allow for four unpaired electrons, which is necessary for bonding.

Where does the energy for this excitation come from?

Ab initio hybridisation is a concept taught in schools to help understand how carbon with its one s- and three p-orbitals can achieve a tetrahedral environment. As part of this process, it is required to formally excite an electron.

Formally is the key word here. The entire process is a formal process that has practically nothing to do with real-life atoms (especially because these real-life atoms rather rarely end up in a vacuum without anything around them as the theory assumes). Thus, in real life this formal energy is not needed.


Specific Case of the Rate law My textbook says that for a chemical reaction with two reactants, $\ce{A}$ and $\ce{B}$ with equation $\ce{aA + bB -> Products}$, where $a$ and $b$ are coefficients, the general rate law is:

$\text{Rate} = k[\ce{A}]^m[\ce{B}]^n$

where $m$ and $n$ are the reaction orders for $\ce{A}$ and $\ce{B}$ respectively. Then it says that $m=a$ and $n=b$ if  the reaction between $\ce{A}$ and $\ce{B}$ occurs in a single step and with a single activated complex. Why is this true?

According to collision theory, for a chemical reaction to occur, the reacting molecules must collide with each other.

For a bimolecular reaction (as given by you), collision occurs in the single step and thus becomes the rate determining step, as there are no other steps involved.

If you consider reactions like bromide ions reacting with bromate ions in the presence of acid, it involves 12 reacting particles. If all the 12 reacting particles collide simultaneously you can write rate law expression considering it to be elementary, but those events are extremely rare. In reality such reactions are found to be complex i.e reactions which take place through a sequence of two or more consecutive steps. Here, there are many steps involved, and one should consider slowest step to write rate law expression. But, why? That will become another SE question. In case of reactions taking place in single step, as there are no other steps left to compare with, whether it is slower or not, so you need to consider  the theoritical equation.

One more thing to be noted is that elementary reactions are associated with single transition state and not single activated complex. For more info read this page: Activated complex


Galvanic Cells, theoritical voltage This reaction proceeds in the galvanic cell:

$$\ce{Zn + Pb^{2+} -> Zn^{2+} + Pb}$$

I need the equation for the half-reactions of this cell and the theoretical voltage produced by this cell if concentration of $\ce{Zn^{2+}}$ is $\mathrm{0.1\ mol/L}$ and concentration of $\ce{Zn^{2+}}$ is $\mathrm{10^{-4}\ mol/L}$.

$$Pb^{2+} + 2e^{-}->Pb  E_{0}(Pb^{2+}) = -0.1263$$
$$Zn->Zn^{2+} +2e^{-}   E_{0}(Zn^{2+}) = -0.76 $$

$$[Zn^{2+}] = 0.1 mol/L [Pb^{2+}] = 10^{-4} mol/L$$

$$Qc= \frac{[Zn^{2+}]}{[Pb^{2+}]}$$
$$E_{0}(cell) = E_{0}(Pb^{2+})- E_{0}(Zn^{2+})= -0.1263 - (-0.76) $$
$$ E(cell)= E_{0}(cell) - \frac {RT ln Q_{c}}{nF}$$

Small piece of advice : Tag homework and exercises on these type of questions , read through "Help Center" specifically "Asking" . Try to attempt the question by yourself and show your own attempt of an answer. This site isnt "Solve my homework site " good luck cheers


What am I doing incorrectly with this IR/ Mass spec data? Hello, I am trying to determine the chemical formula for the following compound.  From the IR spec, I see there is a carbonyl present.  Moving on to the mass spec, I see an M+ peak of 142, with what appears to be no Cl/Br present.

This brings me to a chemical formula of C10 H6 O (12 x 10 + 6 + 16 = 142), with a degree of unsaturation of 8, which seems off.

I feel I am doing something wrong and any help would be greatly appreciated, thank you!

What I was trying to say was that firstly the molecular formula has to be determined correctly. From the mass spectrum of the given molecule, the molecular ion peak ($\ce{M+}$) was determined as $\frac{m}{z}=142$, which corresponds to the molecular formula $\ce{C9H18O}$. Note that, it can also correspond to $\ce{C10H6O}$, but in your $^1\mathrm H$ NMR spectra, a total integration of at least $9$ is there, thus the latter consideration is anyway not possible to be the correct formula.

After determining the chemical formula, it's good to look for index of hydrogen deficiency(i.e. degree of unsaturation) of the molecule. If the formula is $\ce{C9H18O}$, the degree of unsaturation is $1$. From the IR spectra also, we can see a sharp distinct $\ce{C=O}$ stretch near $1700\ \mathrm{cm^{-1}}$, and also no aldehydic proton peak near $9.5{-}10\ \mathrm{ppm}$, therefore we can conclude there is a keto group present in the molecule.

Now, we come to analyse the $^1\mathrm H$ NMR spectrum, which shows only total relative integral of $9$, but remember the integrals in proton NMR are always relative, they are never absolute. So, you can scale them up or down relative to your choice. Here to match the total protons in our system, we have to scale it up by a factor of $2$. Therefore the integrals now correspond to $4,8$ and $6$ protons respectively, and also by this scaling we can also say that the molecule is symmetric in nature.

Now peak at $\delta_\mathrm H=2.5\ \mathrm{ppm}$ is splitted into a triplet and these correspond to the protons attached to your carbonyl, resulting in the most downfield shift, near $\delta_\mathrm H =1.5\ \mathrm{ppm}$, two peaks merge and therefore they are actually the methylene protons, and lastly near $\delta_\mathrm H\approx0.9\ \mathrm{ppm}$, the triplets correspond to methyl proton peaks and thus we arrive at the final most likely possible structure as,

Analysis of the splitting pattern to find out the $^3J$ values, I leave it to you...


Collisonally induced intersystem crossing What makes a molecule a good initiator of collisionally induced ISC? $\ce{N2}$, $\ce{O2}$ and $\ce{CH4}$ have been mentioned as examples.

I imagine the most important factor is the ability to quench vibrational excitation so that the Franck-Condon factors for non-radiative decay are optimized.

There are two factors that I recall that can enhance intersystem crossing (ISC, the conversion of an excited singlet state to an excited triplet state, or vice-versa; $\ce{S_1 <=> T_1}$).  One is the presence of a "heavy atom" such as bromine or iodine in the colliding molecule (or the excited molecule itself).  The other is the presence of a paramagnetic species.

Both types of molecules enhance ISC by relaxing the quantum rules which forbid  singlet-triplet conversions. When a heavy atom or paramagnetic molecule collides with a molecule in its excited state it can facilitate the electron spin flip (e.g. converting paired electrons to unpaired electrons, or vice versa) through spin-orbit coupling.  It is this spin-orbit coupling that relaxes the rules forbidding direct conversion between $\ce{S1}$ and $\ce{T1}$ states.  In essence spin-orbit coupling leads to a mixing of the singlet and triplet wavefunctions, thereby allowing singlet-triplet conversion.


Percolation network analysis with 1H MAS solid state NMR Assume you have a composite rubber that was produced in an extrusion process and it consists of a basic polymer with some additives.

One of the additives does not get
mixed with the basic resin but forms a percolation network in the resulting resin. Is it possible to get some quantitative information
on this percolation network using 1H MAS solid state NMR spectroscopy?

What if the percolating additive is very similar to the basic resin?
If 1H MAS solid state NMR spectroscopy is in no way suitable for such an analysis, what other methods can you think of to get some
information like this?

Solid-state (MAS) NMR can most likely help you understand more about the interaction between the percolating additive and the resin. If you have not yet carried out solid-state NMR analysis, it can be a useful complement to information that you have determined by other methods. However, it is hard to be 100% sure that it would answer your exact question in terms of guiding you on how to modify the percolation threshold.

Should you do $\ce{^1H}$ MAS solid-state NMR?

Unlike in solution-state NMR, $\ce{^1H}$ MAS solid-state NMR is not yet a routine procedure and has much fewer applications (compared to the widespread use of $\ce{^1H}$ in solution-state NMR) until fairly recently (within the last 5-10 years).

$\ce{^1H}$ has a high natural abundance and high sensitivity. However, unlike in solution-state NMR where the free tumbling leads to averaging of such anisotropic interactions, in solid-state, anisotropic interactions are retained. Thus, the high abundance and sensitivity of $\ce{^1H}$ lead to a dense network of anisotropic dipolar interactions in the solid-state, meaning that a "straight-forward" $\ce{^1H}$ spectrum would yield a broad unresolved hump rather than resolved, informative (sharp) signals.

Thus, the approach taken more commonly is to detect heavy atoms with better inherent chemical shift resolution, such as $\ce{^13C}$. Techniques such as cross polarisation (CP) to tackle the problems of lower sensitivity with heavy atoms have been around for decades and can now be considered routine in solid-state NMR. So, if you may want to consider $\ce{^13C}$ CP-MAS solid-state NMR instead of $\ce{^1H}$ MAS solid-state NMR.

What access do you have to $\ce{^1H}$ MAS solid-state NMR facility and expertise?

$\ce{^1H}$ MAS solid-state NMR has undergone some dramatic technological improvements in the past few years. While it is still not routine, it is now possible to acquire (more) resolved $\ce{^1H}$ solid-state NMR spectra. However, highly specialised hardware is required.

The hardware required is a high magnetic field (700 MHz to 1 GHz) and very high MAS rates (at least 40 kHz but ideally 60-100+ kHz). Both of these types of hardware have been available commercially since about 2009. However, there are not many labs that I know of that have both of these pieces of hardware, and nearly all are national or pan-EU facilities (and unfortunately the ones I can think of nearly are nearly all focussed on structural biology). It is also not straightforward to operate these hardware, so you will need to interest these labs in a collaboration.

There are some labs that are trying to do $\ce{^1H}$ MAS solid-state NMR with lower magnetic fields and/or MAS rates by developing specialist pulse sequences to carry out the $\ce{^1H}$ homonuclear decoupling required. It is not an area of ssNMR that I am very familiar with though I believe that there are people doing this in Israel and India.

Generally, the groups doing $\ce{^1H}$ MAS ssNMR are not going to be doing it like a "service". You cannot just give someone a sample and expect them to give you a spectrum. If you can find a group that specialises in $\ce{^1H}$ MAS ssNMR to collaborate with, then it is worth doing $\ce{^1H}$. Otherwise, if you find a ssNMR group with other interests, you may be able to ask them to carry out a (relatively routine) $\ce{^13C}$ CP-MAS spectrum for you.

Sample amounts and types

I know you did not directly ask about this, but I feel it may be helpful to give an idea of how much and what types of sample you will likely need.

For very fast MAS spinning and $\ce{^1H}$ detection, you need only a small amount of sample. Depending on the density of your polymer it is likely that no more than 10 mg is required. (My estimates are based on biopolymers - mostly carbons and protons, so adjust as appropriate for your polymer.)

For heavy atom detection, more sample is usually advised because that will help with the lower sensitivity. Around 50-100 mg would be a good amount.

It will also be helpful to you (and your collaborator) if you can provide the monomer of the resin, and the pure additive (as monomer if necessary). It can be helpful if you have several rubbers that are more simple "models" of what you are trying to investigate (known to have one particular type of polymer links, etc).

...and what information can you expect to obtain?

In the best case scenario, you will have a fairly resolved spectrum of your polymer sample, and by comparing the spectrum of your monomers and your polymer you can assign nearly all the signals. By looking at which signals have shifted/broadened the most, you can determine the atoms which are interacting (or have become bonded). Knowing which atoms are interacting, you can then have some guidance on what functional groups may allow you on to adjust the percolation point as you wish.

But! If you are used to looking at solution-state NMR spectra, solid-state NMR spectra can look incredibly broad. For polymers, the resolution is often poor and signals are fairly broad (even by solid-state NMR standards). In my experience (detecting $\ce{^13C}$) by CP on a 400 MHz spectrometer with a MAS rate of 10 kHz), the linewidths of polymers I have seen are around 150-300 Hz.

The resolution is a very sample-dependent parameter, and is one of the main reasons why I cannot answer definitely whether ssNMR will help you or not. If, as you said, the additive and the polymer are very similar, their signals may well overlap or be insufficiently resolved for you to determine if there has been any changes.

Moreover, polymers are often heterogeneous. This means that one signal in the monomer may broaden and split into multiple signals in the polymer. If this happens at too many atoms, it will make it difficult to assign your polymer spectrum.

There are ways, again, to overcome these problems. It is possible to employ multidimensional ssNMR spectroscopy detecting $\ce{^1H}$ and/or heavy atoms, leading to heteronuclear correlations or homonuclear double-quantum colrreations. The additional dimension often improves resolution and therefore helps with assignment. However, sensitivity will most likely suffer. Alternatively, the molecular dynamics within the polymer can give insight on where the interaction may be occurring in cases where you have sufficient resolution to assign but it is unclear where the interaction has occurred from observing the chemical shift alone. These approaches are again going into non-routine territory where you will need to find a collaborator rather than having it as a "service".

Here is a review if you are interested in some examples of recent multidimensional $\ce{^1H}$ MAS ssNMR work on polymers. Here is a more dated paper which shows how dynamics can help in identifying the nature of interactions at a rubber-filler interface.


Why do Ca-Ca distances get shorter in ionic structures, as compared to the metal? I saw in the book Chemistry of the Elements by Greenwood & Earnshaw, 2nd ed., p 66, that the distance between calcium atoms decrease when it's forming an ionic structure.

  The closest $\ce{M-M}$ approach in these compounds [metal hydrides] is often less than for the metal itself: this should offer no surprise since this is a common feature of many compounds in which there is substantial separation of charge. For example, the shortest $\ce{Ca-Ca}$ interatomic distance is $\pu{393 pm}$ in calcium metal, $\pu{360 pm}$ in $\ce{CaH2}$, $\pu{380 pm}$ in $\ce{CaF2}$, and only $\pu{340 pm}$ in $\ce{CaO}$ (why?). [sic]

Why does this happen?

Simply put, the distance is measure from the center of the atoms. In the metal, the uncharged Ca atoms are kept at the van der waals contact distance.

In the ionic compounds the atomic radii of the Ca2+ ions have shrunk due to the loss of the outer shell electrons. Furthermore, negative ions cancel out the repelling charges and allow the Ca2+ ions to pack even closer together.


When is a dipole-dipole interaction strongest? Are there any conditions that'll improve the interaction? Just like, I assume, hydrogen bonds are strongest when the difference in electronegativity is biggest.

An electric dipole is formed in a molecule as the vectorial sum of all the partial charges on the atoms multiplied by their separation. Thus if, these charges are large and well separated and not symmetrically situated about the molecule the dipole will be large. For example chloro-benzene has a far larger dipole than toluene. However, 1,4-dichlorbenzene has zero dipole as the charges cancel out. (Molecular point groups can be used to decide if a molecule has a dipole, but not its magnitude. Only molecules in groups $C_1, C_s,C_n,C_{nv}, n>1$ have dipoles). Hydrogen bonds have no special significance with respect to dipoles.

Dipoles interact through space, and the strength of their interaction depends on the distance between the dipoles,  relative orientation and relative permittivity (dielectric constant) of the intervening medium. Each of these have a dramatic effect for any given pair of dipoles.

The interaction energy E for dipole vector $\mu_1, \mu_2$ at separation R and where $\bar R$ is the vector joining the centres of the dipoles is

$$E = \frac{1}{4\pi\epsilon_0\epsilon} \left( \frac{\mu_1\cdot\mu_2}{R^3} - 3\frac{(\mu_1\cdot \bar R)(\mu_2\cdot \bar R)}{R^5}    \right)  $$

The dot product between two vectors e.g. $\mu_1\cdot \bar R = |\mu_1||\bar R|\cos(\theta_1)$ where $\theta_1$ is the angle between the vectors, see figure.

If the vectors are now represented as unit vectors e.g. $\bar \mu_1=\bar u_1 d_1$ where d is the magnitude of the dipole, and $\bar R = \bar u_R R$ then the energy becomes
$$E = \frac{K}{4\pi\epsilon_0\epsilon}\frac{d_1d_2}{R^3}   $$
where
$$K= \bar u_1\cdot \bar u_2 -3(\bar u_1 \cdot \bar u_R)(\bar u_2 \cdot \bar u_R) $$

The figure shows the geometry.

If the dot product $\bar \mu_1 \cdot \bar \mu_2 =0 $ and one  of the two dot products with $\bar u_R $ are $0$ then the energy is zero, and this can occur when dipoles are at 90 degrees and one points to the centre of the other. Physically this makes sense as each end of a dipole sees equal positive and negative charges on the other dipole.


Dilution question, "70 wt. % in H2O"? If I have a solid and want to make an aqueous solution of 70%, how much water do I need? Is it as simple as weighing out 70 $\mathrm{g}$ of solid and adding water until I have 100 $\mathrm{mL}$? Will this solution be 70 wt. % $\ce{H2O}$?

To make a solution that is 70% solute by weight with $\ce{H2O}$ as the solvent, take 70 $\mathrm{g}$ of solute and add 30 $\mathrm{mL}$ of water. The resulting solution will have a mass of 100 $\mathrm{g}$, 70 $\mathrm{g}$ of which is the solute; hence, 70 wt. % in $\ce{H2O}$.


When the intermolecular forces in a liquid increase, does the equilibrium vapor pressure of the liquid at a given temperature increase or decrease? I assumed it would increase, but I think I am wrong. I know that when the forces are strong, the vapor pressure would decrease. Does this same principle apply to the IM forces increasing?

Yeah your hunch is right. Higher the intermolecular forces between the liquid particles, harder it is for it to escape into the vapor phase, ie., you need more energy to convert it from liquid to the vapor
phase, in other words, higher its boiling point.

If it is harder for a liquid to escape into vapor, because it is held back into the liquid by the neighboring particles attraction, you have less vapor and hence low vapor pressure.

So in nutshell, if the intermolecular forces are increased, the equilibrium vapor pressure of the liquid will decrease.