Ring contraction when cyclohexene oxide is treated with methyl Grignard My book writes that:

  Recall that epoxides rearrange with Lewis acids in a pinacol-like fashion, and that Grignard reagents in THF exist in the following equilibrium:

I approached this question by reasoning that the Grignard reagent would lead to opening of the epoxide:

but the answer given in the book is instead

How is the five-membered ring formed?

Sara Jane: Lest you miss the point of the stimulating discussion between Waylander and Zhe, I would like to sum up their thoughts with a diagram. Your first clue should have been the opening line: "Recall that epoxides rearrange with Lewis acid...". It is common that catalytic Lewis acids rearrange 2,3-dialkyl oxiranes [epoxides] to ketones. In the case of cyclohexane epoxide (1), rearrangement to ring-contracted cyclopentane carboxaldehyde (4) occurs. In the illustration I have used CH3MgBr as the Lewis acid. Alternatively, one might use the other components of the Schlenk equilibrium, i.e., (CH3)2Mg or MgBr2. The rearrangement is shown as two separate steps (2 and 3) although it is likely to be a concerted process.

The aldehyde 4 in the presence of the Grignard reagent will undergo addition. Given sufficient Grignard reagent, addition to the carbonyl group will occur to afford the secondary alcohol 6. The only ways that the aldehyde 4 can be isolated is if there is insufficient methyl magnesium bromide or that the Grignard reagent acts as a base and forms the unreactive enolate of the aldehyde. The aldehyde 4 would be produced upon aqueous workup. While Grignard reagents with β-hydrogens, e.g., isopropyl and ethyl Grignard reagents, are capable of reduction of aldehydes and ketones, methyl Grignard reagents are incapable of this reaction. Therefore, the primary alcohol 9 may be assumed to be a misprint in place of secondary alcohol 6.
The oxidation of alcohol 6 to ketone 7 is straightforward. The final step is the formation of the alkene 8 (most likely of the (E)-configuration) via the McMurry coupling.


Why does alcoholic KOH prefer elimination whereas aqueous KOH prefers substitution? What is the basic difference between aqueous and alcoholic $\ce{KOH}$? Why does alcoholic $\ce{KOH}$ prefer elimination whereas aqueous $\ce{KOH}$ prefers substitution?

In alcoholic solution, the $\ce{KOH}$ is basic enough ($\mathrm{p}K_{\mathrm{a}} =15.74$) to deprotonate a small amount of the alcohol molecules ($\mathrm{p}K_{\mathrm{a}}= 16–17$), thus forming alkoxide salts ($\ce{ROK}$). The alkoxide anions $\ce{RO-}$ are not only more basic than pure $\ce{OH-}$ but they are also bulkier (how much bulkier depends on the alkyl group). The higher bulkiness makes $\ce{RO-}$ a worse nucleophile than $\ce{OH-}$ and the higher basicity makes it better at E2 eliminations.


Understanding how electrons are counted for lone pairs vs. bonds especially when understanding resonance? This is taken from the book Organic Chemistry as a Second Language from page 35.
In the following image:

It is seen that when the electrons from the double bond are pushed into the lone pair electrons, we get 2 electrons as a lone pair.  And when a lone pair electrons are pushed to from a double bond we remove 2 electrons from the oxygen.
The part that confuses me is how the electrons are being counted (especially for octet rule). Let's take the example of oxygen that starts out double bonded (the molecule left of the double headed arrows) and then the double bond electrons get pushed into a lone pair (Oxygen on the right of the double headed arrow with a formal charge of 1).  If I count the first oxygen (double bonded oxygen to the left of the double headed arrow), I get 6 electrons where the double bond counts as 1 electron for each bond, now if I go to the right of the arrow I see that same oxygen I count 7 electrons.  How can a lone pair which is made up of 2 electrons result as being counted as 1 electron when it becomes part of a bond or vice versa?

Count 2 electrons per bond and you'll be fine. The partners of a bond usually share two electrons in their mutual orbitals. While you are correct in thinking 2 divided by 2 is 1, thats not how the octet rule works - count them all.

Left w/ dbl bond: 4 lone pair electrons + 4 bond electrons = 8.

Left w/o dbl bond: 6 lone pairs electrons + 2 bond electrons = 8.

Octet rule satisfied.


Reactivity of thioesters with respect to nucleophilic attack Why are thioesters relatively reactive with regard to nucleophilic attack? Prof says to wait until pchem 3 when we learn about orbital symmetry. He also said that sulfur’s d-orbitals (?!) don't have the correct symmetry to participate in resonance with the carbonyl carbon.

Wait. I thought that sulfur didn’t utilize its d-orbitals. Was it previously taught that sulfur’s lone pairs were in d-orbitals? Why would sulfur even need d-orbitals when in a thioester? Can't we just explain this based on the poor overlap between carbon's 2p and sulfur's hypothetical 3p orbitals?

According to these lecture notes from Gonzaga University:

  The question arises over why thioesters occupy a prominent place in metabolism while oxygen esters play a relatively minor role. One answer is thermodynamic. The sulfur in the thioester linkage is less able to participate in electron delocalization through the acyl group (the usual explanation given for this is poor overlap between the 2p orbital of carbon and the 3p orbital of sulfur), and this makes the thioester bond less stable than the ester bond. Hence, the thioester bond has a more negative standard free energy of hydrolysis (−7.5 kcal/mole vs. about −5 kcal/mole for most oxygen esters). In many cases, it appears that thioesters are more reactive than oxygen esters, undergoing more facile nucleophilic displacement reactions at the acyl group. The reactivity of a thioester at the alpha carbon compares favorably with that of ketones.

...

  Because resonance structure 2c [see web page - CF] makes a more significant contribution for a thioester than an analogous structure for an ester, the acyl carbon is more positive, hence more susceptible to nucleophilic attack. An attacking nucleophile would be more readily acylated by a thioester than it would be by an ester. Furthermore, the C-S bond is weaker than the C-O bond, and the thiolate (or thiol, if protonated) is a better leaving group than alkoxide (or alcohol). These factors all make a thioester in general a better acylating agent than an ester.

So the answer doesn't seem to involve d orbitals explicitly, but instead the difference between 2p and 3p orbitals (as well as other factors).


Calculate the lattice energy of ammonium carbonate The lattice energy is a measure of the stability of a compound and I need this data for the ammonium carbonate ($\ce{(NH4)2CO3}$) but I don't know how to calculate and I can't find it anywhere. I know that it can be calculated with the Born-Lande equation but I don't know how to apply this equation to solve my problem.

I wonder if anybody knows a database where I can find this value, I need it for an important chemistry assessment.

Thank you for your help

The lattice energy is not directly measurable, and it is difficult to calculate. In a paper about estimating lattice energies, the estimation for $\ce{Na2CO3}$ is 30% off from the reference value (which is from CRC Handbook of Chemistry and Physics). I mention sodium carbonate because ammonium carbonate is not one of their examples. The introduction explains why it is not possible to directly measure the lattice energy:

  Lattice  potential  energy  (UPOT)  is  a  dominant  term  in  the thermodynamic analysis of the existence and stability of ionic solids.  Direct  experimental  determination  is  generally  not possible since, in practice, the crystalline solid dissociates into atoms  and not into  gaseous  ions,  as  is  required  in  the  lattice energy  evaluation.  Therefore,  its  indirect  experimental  determination, computation, or estimation is of considerable interest in modern materials science; indeed, whenever the energetics of condensed-state materials are studied, the chemical processes under consideration may be rationalized if the appropriate lattice energy steps can be incorporated into the thermochemical cycle.

They go one to discuss how you can obtain the lattice energy from experimental or theoretical data:

  A  variety  of  estimation  methods  for  lattice  energies  is vavailable. These include the Born-Haber-Fajans thermochemical cycle (which requires ancillary thermodynamic data) and modern computational methods (which generally require knowledge of the lattice constants and the coordinates of the ions as well as needing an established force field). The computational methods  range  from  direct  energy  calculational  procedures through to programs that produce lattice energies in the course of their modeling of the solid. Quantum mechanical procedures are also available, but are highly computationally expensive and are generally applied only to the simpler systems.

Then they talk about cheaper methods that are less accurate. The paper is about a new quick estimation methods.

  By contrast,  a  few  exceedingly  rapid  estimation  methods based on ionic radii or volume per formula unit (“molecular”volume) have  been  developed.  Originally  only  applicable  to the simpler binary ionic solids (such as MX (with a 1:1 charge ratio)), the volume-based approach has recently been extended by  us to  include  MX2(2:1)  and  M2X  (1:2))  salts.

So check the CRC Handbook of Chemistry and Physics.


Does presence of alkyl groups on alpha carbon increase yield of alkane in Kolbe's Decarboxylation? We know that, during the synthesis of alkanes by Kolbe's Decarboxylation reaction,
the reaction proceeds via a free radical $\ce{R.}$ at the anode.

So, an alkyl group, say methyl, when substituted at the alpha carbon will increase the yield of alkane since the radical is stabilized by hyperconjugation. But my book states that the presence of alkyl group on alpha carbon decreases the yield of alkane. Why is it so?

I have done some electrosynthesis work using the Kolbe reaction, what I found was with straight chain carboxylic acids that my platinum anode was quickly fouled.

When I had some branching at the beta site, I had much less electrode fouling and I got a good yield of my alkane.


Confusion about the choice of axis for Molecular Orbital Theory This video from MIT OpenCourseWare says that pi bonds and anti-bonds are for 2px and 2py, i.e., $\pi_{2px}$, $\pi_{2py}$ and $\pi_{2px}^*$, $\pi_{2py}^*$ and sigma bonds and anti-bonds are for 2pz i.e., $\sigma_{2pz}$ and $\sigma_{2pz}^*$.

But in my reference books it says pi bonds and anti-bonds are formed for 2py and 2pz i.e.,  $\pi_{2py}$, $\pi_{2pz}$ and $\pi_{2py}^*$, $\pi_{2pz}^*$ and sigma bonds and anti-bonds are for 2pz i.e., $\sigma_{2px}$ and $\sigma_{2px}^*$.

Which is actually the right one? Please explain.

Although we could not see what is written in OP's "reference books", the mentioned difference most likely is the result of different choices for the bond axis.

Conventionally, and as it is (implicitly?) done in the MIT video lecture, it is chosen to be the $z$-axis, in which case $\pi$ bonds are indeed formed from $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals while $\mathrm{p}_z$ can take part in $\sigma$-bonding. If, however, the bond axis is chosen to be the $x$-axis, then $\pi$ bonds are formed from $\mathrm{p}_y$ and $\mathrm{p}_z$ orbitals while $\mathrm{p}_x$ takes part in $\sigma$-bonding.


Electroplating metal inside polymer foam with acetone In my research we have a method of making polymer open-cell foams with high tortuosity and pore sizes on the order of 50–200 nanometers. I can make thicknesses of 100 µm to 2 mm thick (and up to 5 inch diameter disk). I am looking for some cool things to do with this material, and one idea was to try to fill it with metal and then dissolve the polymer away to leave a nanoporous metal structure.

The problem is that so far I cannot get water to penetrate inside the pores, I tried up to 4 MPa pressure and it just won’t go through (small pores and hydrophobic) but acetone will get absorbed in very quickly. My question is, could there be a way to use acetone as an electrolyte and deposit metals (copper, nickel, whatever) to fill these open-cell pores?

I haven’t taken chemistry since intro chem back in undergrad so I really don’t know much of anything about electroplating, so I will be very grateful for any help on this experiment. Thank you very much, I hope I was clear in explaining what I am trying to do.

Electroplating might be possible, but getting good coverage of internal pores is tricky, even with aqueous electrolytes, as deposition will likely occur preferentially at the outer edge of the workpiece, leaving the internal passages lightly- or un-coated.

A more effective approach is likely electroless nickel deposition.  As long as you can find a suitable acetone-soluble reducing agent, it might work out.

I would also recommend testing other solvents for their ability to permeate the polymer structure -- if you could find a protic solvent that wets it (even something like 50/50 ethanol/water!), you'd have a much better chance of success at electrodeposition proper, and also at making electroless work for you, since (I think) most electroless recipes are aqueous-based.


What happens to the NADH produced by lactate oxidation? I first want to note that I'm a layman attempting to gain an understanding of metabolism by reading resources on the internet with only a little understanding of chemistry. I've hit a snag:

Oxidation of lactate into pyruvate reduces NAD+ to NADH. Subsequent glyconeogenesis can then turn two pyruvate into a new glucose molecule at the cost of 6 ATP. However, the NADH is presumably used to generate ATP via oxidative phosphorylation producing ~2.5ATP, making an overall reaction of 2 lactate -> 1 glucose cost only ~1ATP. This clashes with other things I have read - that glyconeogenesis costs 3x the energy glycolysis produces - as this would seem to discount the reduction of NAD+ to NADH in glycolysis. I feel I am clearly misunderstanding something, but I'm not sure where.

What you have read about the direct energy costs/benefits of glycolysis and gluconeogenesis  is not wrong. Glycolysis produces 2 ATP and gluconeogenesis consumes 6 ATP.

The net reaction for coversion of glucose $\ce{C6H12O6}$ into pyruvate $\ce{C3H3O3-}$ by glycolysis is (note that I am keeping track of the charges and the magnesium counterion/cofactor on ADP/ATP):

$$\ce{C6H12O6 + 2NAD+ + 2[MgADP]- + 2HPO4^{2-} ->}$$
$$\ce{ 2C3H3O3- + 2NADH + 2[MgATP]^{2-} + 2H+ + 2H2O}$$

For conversion of pyruvate back to glucose by gluconeogenesis (note that gluconeogenesis usually uses 4ATP and 2GTP - for simiplicity let's consider that a GTP has about the energy equivalence to an ATP):

$$\ce{2C3H3O- + 2NADH + 4[MgATP]^{2-} + 2[MgGTP]^{2-} + 6 H2O ->}$$
$$\ce{C6H12O6 + 2NAD+ + 4[MgADP]- + 2[MgGDP]- + 6HPO4^{2-} +2H+} $$

If you compare the two processes directly, in absence of the source/fate of any of the components, you see that gluconeogenesis consumes 6 ATP/GTP and 2 NADH while glycolysis produces 2ATP and 2NADH. The NADH consumption and production is balanced out.

  Why would we compare these two directly and not worry about coupled processes?

Not all organisms undergo oxidative phosphorylation. Most do, but not all.
More importantly, in the context of your question, NADH can do other things. Not all NADH goes to oxidative phosphorylation.
Most importantly, the direct comparison is carbon balanced with the same organic species: glucose and pyruvate. If we try to include all possible variations and coupled processes we find ourselves eventually trying to do a general chemistry level stoichiometric and thermochemical analysis of an entire life-form. I'd rather not. There are lots of feed stocks into gluconeogenesis besides lactate and pyruvate. An amino acid or lipid that can be catabolized into acetyl-CoA can be fed into gluconeogenesis.

  As for lactate...

True, pyruvate can be generated from lactate $\ce{C3H5O3-}$, which also produces NADH:

$$\ce{C3H5O3- + NAD+ -> C3H3O3- + NADH + H+}$$

If we start with lactate, we have no net generation of NADH in gluconeogenesis. However, we are also no longer comparing the interconversion of the same carbon species. Lactate is produced from pyruvate (consuming NADH), so we are back to square one. Including lactate in the comparison instead of pyruvate leads to no net NADH production/consumption.


Elution Rate - Alkanes VS Alkenes in Gas Chromatography Some references say that alkanes elute faster than alkenes, whereas in my textbook the higher volatility of alkenes means they elute faster than alkanes. This is for gas chromatography.

  Alkenes are more polar than alkanes due to more electron-withdrawing
  $\text{sp}^2$-hybrid orbitals (more s-character draws electrons closer to
  nucleus).

Can someone please clarify - bit confused.

Chromatography in general is a separation method in which the partition equilibrium of a compound between two phases is permanently disturbed since one of the phases is changing/exchanged (mobile phase).

Apparently, the mobile phase in gas chromatography is the carrier gas.
The stationary phase isn't just a hollow tube but a capillary covered with various differently functionalized materials, that may be rather unpolar (squalane), more polar (polyethylene glycol) or polar and shape-selective (cyclodextrines). Other very common materials are alkyl-substituted polysiloxanes, such as polydimethylsiloxane.

Using the latter stationary phase, Alan Katritzky and coworkers published QSPR: the correlation and quantitative prediction of chemical and physical properties from structure, Chem. Soc. Rev., 1995, 24, 279-287 (DOI, PDF). Here, the retention times for dodecane and 1-dodecene were given as 4.59 and 4.45 min, respectively.

Ladislav Sojak published a freely available 35-page review Separation and identification of isomeric hydrocarbons by capillary gas chromatography and hyphenated spectrometric techniques in Petroleum & Coal, 2004, 4(3), 1-35 (PDF). You might want to have a look at this article on the effect of having non-terminal $\ce{C=C}$ double bonds and cis-trans isomers of longer alkenes, which therefore have significantly different shapes.


Why is the rate of a reaction always positive and why? Why rate of reaction can never be negative? And why rate is an experimental derived expression?

No mystery. Chemists write equations so that reactants (on the left) yield products (on the right). So:
$$\ce{H2O -> H^+ + OH^-}$$
and
$$\ce{H^+ + OH^- -> H2O}$$

The reactions are read from left to right and reactivity is defined as positive from left to right - this means the reactivity for the reaction we want to look at is always positive. (great clarification from user Stian Yttervik.


What is the explanation of the structure of H2XeO4? Using the obsolete VSEPR theory I had a hard time predicting the structure of $\ce{H2XeO4}$. So I looked it up. I found the molecule via chemapps.stolaf.edu:

It looks weird to me. I can't explain the molecular structure. Can anyone explain it using more sophisticated theories like molecular orbital theory (MOT)?
I don't know much MOT, yet.

When I think about such structure, I draw all valence electrons around the atoms and connect them somehow reasonably. There are two ways to connect the hydrogens to xenon. Either directly bonded as H-Xe bonds, which for me does not feel right here, or as part of OH groups, which also make more sense in the "acid" part.
By connecting everything that way, you end up with:

This let you end up with a lone pair at xenon which is the reason for the non-tetrahedral structure.

A quick and dirty quantum calculation[1] shows what I mean. By calculating the ELF (Electron Localization Function), you can easily see where the electrons are located. The lone pair is shown by the yellow volume surrounding no atom.

This structure is not clear from the Lewis formula shown on Wikipedia, but can be seen in the ball and stick representation that is shown there (don't care about the "flipped" hydrogens):

[1] ORCA 4.1.2, best functional of the planet B3LYP def2-QZVP // Crest 2.6, Visualization with Chimera, iso value 0.8 ... whatever ... camels? ... what unit has the ELF?


Modified DNA bases in Pymol and Coot When using Pymol or Coot molecular visualization software to solve x-ray crystallography structures, how can I mutate a cytosine to a 5-methylcytosine?

I got the answer on a Pymol listserve:

Yes, it can be done in Pymol.

Choose the residue you want to mutate from cytosine to 5-methylcytosine.
Use Action -  Hydrogens - Add function to add hydrogens.
Change mouse selection to choose atoms.

Click on the hydrogen you want to replace.

Click builder (there is more work to do if you're on a Mac, see below) to open builder window.

Click C

If you're on a Mac, you won't have Builder functionality unless you change your MacPyMOL.app file name to PyMOLX11Hybrid.app


Replacement for optical calcite (Glan Thompson prism)? I notice that Glan Thompson prisms seem to be the preferred optics for doing polarimetry and have been for at least 75 years. One instrument maker says that polaroid films can be used but they tend to degrade and warp over time.

It seems strange to me that no better substitute has been found for Glan Thompson prisms given their expense. I know at one point Polaroid was saying that Sodium Nitrate crystals (which can be grown) could potentially be used, but this effort seems to have gone nowhere.

Why has no cheaper alternative been found? Aren't there growable crystals that have the birefringence necessary?

I have a pair of Glan Thompson polarizers (Karl Lambrecht) that cost a pretty penny many years ago. I also have several other (less expensive) pairs of calcite polarizers. Nothing to date can match high quality calcite in terms of low scattering, low fluorescence and low extinction ratio. In some cases, with the highest quality calcite devices, it is possible to find 'sweet spots' having $10^{-10}$ extinction ratio. (The best I ever got, after hours in a dark room, was about $8 x 10^{-9}$). In contrast, the plastic sheet polarizers are nice and convenient, but nowhere near good enough for serious polarimetry work. Sodium nitrate crystals might have excellent optical properties (I do not know anything about them), but would they need to be protected from water vapor in the air? After all, potassium nitrate is used in gunpowder and I suspect one reason is that it 'keeps the powder dry'!


Meaning of the word 'croton' This is but-2-yne, which is also known as crotonylene.

This is but-2-enal which is also known as crotonaldehyde.

Coming to my query, what does the word croton indicate? Does it refer to $\ce{CH3-CH=CH-R}$? Can this prefix be used anywhere or only for pre-specified compounds? If it can be used anywhere, then can but-2-en-1-ol be called as crotonalcohol or hydroxycroton?

As with many chemical compounds (see the large class of terpenes, for instance) "croton" derivatives take their name from the natural source of their "father" compound. They are a family descending from crotonic acid, which was found in the oil from the seeds of the Croton plant.

"Croton" compounds are derivatives of the father compound "crotonic acid" (2-butenoic acid), so you can guess the names of the derived compounds by the logical chemical modifications of such compounds. For instance, as you might guess, crotyl alcohol is But-2-enol, from the reduction of crotonic acid; crotonylene's structure can be imagined with an analogy between acetic acid and acetylene, and so on.

I don't think that nowadays new names for the derivatives could be accepted, but for historical reasons many derivatives have these funny names.

And, knowing the stucture of the father compound, it's easy to imagine the structure of the derived ones.


Why plastic sheets contract when heated? Why do thin plastic sheets contract when heated, contradictory to the behavior of most other materials ?

What are the things going on at the molecular level ?

When plastic sheets are produced, they are rapidly cooled to keep the polymer chains oriented in a way that makes the sheets nice and flat. This is a relatively high-strain orientation since it is associated with the energy level of the molecules at the casting temperature.

Once the plastic is heated above its glass transition temperature, the polymer chains are no longer locked in that high strain orientation. They relax to a low energy orientation- curled and bending in a way that shrinks the bulk material.

As for the precise mechanism, I'm not exactly sure. I would guess that the shrunken conformation is entropically favorable because there are more arbitrary bends. This would decrease the Gibbs free energy, making it a more stable shape.

Alternatively, it could be that hydrogen bonding between chain elements makes the folded shape more enthalpically favorable.


In a weak acid-strong base titration why is [H+] no longer equal to [A-] after addition of the base? (A2 level chemistry) I understand that before any base is added $\ce{[H+]}$ is equal to $\ce{[A-]}$ but why is $\ce{[H+]}$ no longer equal to $\ce{[A-]}$ when the strong base is added?

If all the following reactions below are occurring in solution and both $\ce{CH3COO-}$ an $\ce{H+}$ are able to react in either one of two of these different reactions then why should the $\ce{[H+]}$ no loner equal $\ce{[A-]}$ if both have equal opportunities to be consumed? (I am aware that not all the $\ce{CH3COOH}$ will have dissociated but surely this would not affect the concentrations of either $\ce{[H+]}$ or $\ce{[A-]}$?)

$$\ce{Na+ + CH3COO- -> CH3COONa}$$
$$\ce{CH3COO- + H+ -> CH3COOH}$$
$$\ce{H+ + OH- -> H2O}$$

Is it because the salt $\ce{CH3COONa}$ is much more likely to dissociate than $\ce{CH3COOH}$ and $\ce{H2O}$ meaning more $\ce{CH3COO-}$ remains in solution than $\ce{H+}$?

$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$

$$\ce{NaOH + H3O+ -> Na+ + 2H2O}$$

The base removes the hydrogen ions from solution, and the acid further dissociates to maintain LeChatelier's equilibrium. The $\ce{H3O+}$ ion concentration will decrease as it is continually removed whereas the $\ce{A-}$ concentration will increase until the acid is fully neutralized into a salty solution.

I can't make much sense of your chemical equations, but it looks to me like they are incorrect. There is no reason for formation of the ionic compounds in solution. They should actually be written in reverse. And you have no strong base to react with the hydronium ions.


Why do Grignard reagents form? Why does magnesium insert itself between a carbon atom and halogen atom? Can anyone give any vague idea about why this occurs?

Also does a Grignard reagent only form when there is carbon atom and halogen atom with single bond between them? For example, can magnesium insert between the carbon and oxygen atoms of a carbonyl?

Yes, this is a very interesting question that till today, not everyone agrees the precise mechanism as to how it works. Here is one with homolytic fission:

For your second query, I don't think it is feasible as the $\ce{C=O}$ carbonyl bond is rather stable.


Calculating entropy change: reversible vs irreversible process Since the change in internal energy and enthalpy, which are equal to the heats for a constant-volume and constant-pressure process, respectively, are state functions, the heats for a reversible v.s. irreversible process should be equal. Thus, the entropy change of the surroundings which equals the quotient of the heat and temperature are equal for reversible v.s. irreversible processes.

As Engel's Physical Chemistry states it:

  Because H and U are state functions, the amount of heat entering the surroundings is independent of the path; q is the same whether the transfer occurs reversibly or irreversibly.

However, when the book proceeds to evaluating the change in entropy for the surroundings for a reversible vs irreersible process, the q used for evaluating the change in entropy of surroundings are not equal for the reversible v.s. irreversible process, which contradicts what the textbook previously stated.

The question:

  One mole of an ideal gas at 300. K is reversibly and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath thermal reservoir in the surroundings is very large, T remains essentially constant at 300. K during the process.

Calculation of change in entropy for reversible path:

Calculation for irreversible path:

It seems that the book used the actual heat for the irreversible process. However, I don't understand why the entropy's from the two calculations are not equal. Also, I fail to understand why the change in entropy for the surroundings are not equal for the two process for which the final and initial states are equal --> change in entropy is a state function and should be the same regardless of path.

References

Engel, Thomas, and Philip Reid. Physical Chemistry. San Francisco: Pearson Benjamin Cummings, 2006.

You are confused about how to calculate the entropy change for a system and surroundings that have experienced an irreversible process.  For the irreversible path, you correctly calculated the final state using the first law of thermodynamics.  From this point on, you must totally forget about the irreversible path and, instead, focus exclusively on the initial and final end states.  You must separate the system from the surroundings, and then devise an alternative  reversible path for each of them separately that takes each from its initial state to its final state.  You did that correctly for the system, and determined its correct entropy change (-7.62 J/K).  This is the entropy change for the system both for the reversible path you devised as well as for the actual irreversible path.

In the case of the surroundings, during the irreversible process, its internal energy increased by 3.741 kJ at the constant temperature of 300 K.  So, for its alternative  reversible path, to achieve this same internal energy change at 300 K, the amount of reversible heat it would have to absorb would be the same 3.741 kJ.  So, the change in entropy of the surroundings in the irreversible process was 12.47 J/K.

I know that this might all sound a little confusing.  So here is a more detailed explanation of the fundamentals as well as a cookbook (foolproof) recipe for determining the change in entropy of the system and surroundings for any arbitrary irreversible process on a closed system (including worked examples of the methodology):  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/


How does an additive like aluminum oxide make a material like polyurethane harder/more resistant to wear? I have wood flooring which is coated with a polyurethane finish. According to the marketing, aluminum oxide has been incorporated into the polyurethane to make it more durable, but that's all that is said about it.

I understand that aluminum oxide is a hard substance, but I don't understand how it could be incorporated into a polymer to improve wear resistance. If it occurs as microscopic beads or granules of alumina within a polymer matrix, how does that confer durability? I can understand how a coating of alumina on top of the polymer would provide durability, but how could that even work?

So, my question is: What form does the alumina take in the end product and how, in that form, does it improve durability?

Alumina is extremely hard and very wear resistant, likely harder than anything which would come in contact with the floors. Dirt for example would mainly contain silica based abrasives which are far softer than alumina and hence alumina additives, of sufficient size would provide scratch resistance when present in sufficient quantities.

If you want the exact details, they are going to vary from one manufacturer to another. If you do a search for patents you can see various procedures some of which are pre-mixed, some of which require onsite mixing, for example :

Surface coverings containing aluminum oxide

That kind of post-mixing is very common in concrete, though there the aggregate usually isn't alumina (could be just decorative as well or dual purposed as is common locally) and is far bigger. In general you want the aggregate to be at least as large as, and ideally bigger than the particle which is trying to cut it. If the aggregate is much smaller it won't resist wear strongly as it will be plowed out with the substrate.

Effect of abrasive particle size on abrasive wear
of hardfacing alloys

If you are curious about a specific product then just call the manufacturer and ask to speak to the engineering department. However as they are all likely patented products, you can even search manufacturer specific and turn up results.


XPS/EDS vs FTIR/Raman So I have been studying the four methods the past couple of weeks. To my understanding, XPS is a very surface-sensitive technique. It can give information about the composition of the surface but also of the bulk material via depth profiling. I think it is mainly used for qualitative analysis, i.e. find out the elements of the sample.
The same goes for EDS/EDX (in SEM), it is used to identify samples. Strictly speaking, EDS is not a surface sensitive technique I think, because the spectral resolution is not as good as XPS. XPS looks at the first few layers of atoms, while EDS looks more at the bulk of the sample.
Now my question is: can't we do the same with FTIR or Raman? Except of course that with FTIR or Raman it would be difficult to study metals, whereas with EDS/XPS it is easy.
What are the advantages of FTIR or Raman w.r.t EDS/XPS or the other way around?
I know that there are handheld devices for FTIR and Raman, so they would be easier to use to identify unknown samples and require less sample preparation than XPS or EDS.
To me it seems that all four can be used interchangeably, except for maybe XPS, because it is so surface-sensitive. I know it depends on the applications, but in general, can either of these be used to identify unknown materials?
Summary of the question by the OP from their comments:  I mean to say that either of the methods can be used to identify unknown samples. But of course, it depends on the sample and what you want to know from the sample to know which method works best. While I agree that the methods are different and answer different questions regarding the sample, I don't agree that they are "completely" different. To some extent they do answer the question of what the unknown sample is. Right?

I think your main question is related to the difference? Have you heard a famous poem, The Blind Men and the Elephant? It is a story, I'd rather copy from Wikipedia

It is a story of a group of blind men who have never come across an elephant before and who learn and conceptualize what the elephant is like by touching it. Each blind man feels a different part of the elephant's body, but only one part, such as the side or the tusk. They then describe the elephant based on their limited experience and their descriptions of the elephant are different from each other. The moral of the parable is that humans have a tendency to claim absolute truth based on their limited, subjective experience as they ignore other people's limited, subjective experiences which may be equally true.

So replace these men with these analytical techniques and the sample with the elephant. XPS /EDS can see what Raman/FTIR cannot and vice versa.
XPS/EDS are elemental techniques and they will tell you what elements are present. Sometimes XPS spectrum, for an expert, can also reveal what is the oxidation state of that element, if you do high resolution scans. Each sample may take hours and hours.
Raman and FTIR can used to probe surfaces (they are fancy surface vibrational spectroscopy techniques) and it will tell you the molecular information, and even the angle at which the molecule is standing on the surface.
From the Raman spectra of an unknown it will be very hard to guess what elements are present.
In short, these are complementary techniques. Someone who is trying to study surfaces would use all of them - if needed.


Viscosity of liquid interrupts accuracy of micropipette? Does viscosity of liquid interrupt accuracy of micropipette?

I would like to dilute 85% $\ce{H3PO4}$ acid to $\pu{1.194 mol/dm3}$ so I had to pipette 366 microliter into 4.634 mL of water. Then, I adjusted the pH to 11 by dropping 25% $\ce{NH3}$ solution. I did 4 times but final volume was not equal (about 50, 50,100, 200 mL). I think the problem is viscosity of liquid when I pipetted concentration acid.

I think you try to use glass pipette instead. It may decrease viscosity effect.


Bromination of ethyl substituted benzaldehydes In a material that I use to study High School Organic Chemistry, the bromination of ethyl substituted benzaldehydes is given as follows:

Source: FIITJEE Grand Masters Package - Chemistry (JEE Preparatory Material)

$\ce{-CHO}$ is a meta directing group, and $\ce{-C2H5}$ is an ortho, para directing group. So isn't some of the products given wrong?

For example, let's take the bromination of ortho ethyl benzaldehyde. Shouldn't the 2nd product's bromine be in a way that is in position 4 wrt $\ce{-C2H5}$ and position 3 wrt $\ce{-CHO}$?

In the meta ethyl benzaldehyde case, why is the $\ce{-CHO}$ group not affecting the position of $\ce{Br}$? Even though it is meta directing, $\ce{Br}$ doesn't seem to substitute the $\ce{H}$ in the meta position of $\ce{-CHO}$

I searched the internet for more information regarding such reactions, and could not find any helpful results for this case.

Your first query is genuine and the 2nd product for the first case is wrong. It should be, as you noted as well, 5-bromo-2-ethyl-benzaldehyde.

Regarding your second query, you're wrong. Recall the general rule that:

when more than one activators are present on the same benzene ring, the substitution is governed by the best activator alone
when an activator and a deactivator is present on the same benzene ring, the substitution is governed by the activator alone
when more than one deactivators are present on the same benzene ring, the substitution is governed by the weakest deactivator alone

So, here $\ce{Br}$ will go to the o/p position of ethyl group.

Note that in the high school context, I've assumed we're only talking about the major product, because otherwise, all types of crazy products are generally formed, but in smaller ratios


How do foundries prevent zinc from boiling away when alloyed with Aluminum? How do foundries prevent lower boiling point metals such as zinc from boiling away when alloyed in a furnace with higher boiling point metals such as aluminum?

When alloys are made by mixing molten metals (actually an alloy only need contain one metal and at least one other compound, metal or not) the metals only need to be heated to their melting point, not all the way to their boiling point. In the example you've given, the melting point of aluminum is $\pu{660^oC}$, which is $\pu{247^oC}$ below the boiling point of zinc, so the volatilization of zinc is negligible under these conditions.

However, the issue you bring up does present problems in other cases. For example this article states the following:

  One difficulty in making alloys is that metals have different melting points. Thus copper melts at $\pu{1,083^oC}$, while zinc melts at $\pu{419^oC}$ and boils at $\pu{907^oC}$. So, in making brass, if we just put pieces of copper and zinc in a crucible and heated them above $\pu{1,083^oC}$, both the metals would certainly melt. But at that high temperature the liquid zinc would also boil away and the vapour would oxidize in the air. The method adopted in this case is to heat first the metal having the higher melting point, namely the copper. When this is molten, the solid zinc is added and is quickly dissolved in the liquid copper before very much zinc has boiled away. Even so, in the making of brass, allowance has to be made for unavoidable zinc loss which amounts to about one part in twenty of the zinc. Consequently, in weighing out the metals previous to alloying, an extra quantity of zinc has to be added.

Summary, TL;DR:
In your example of aluminum and zinc, each metal melts well below either of their boiling points so that loss via volatilization is not a problem. There are cases however, such as alloying copper and zinc, where the boiling point of one metal is lower than the melting point of the other. One way to minimize (but not eliminate) the loss of the more volatile metal is to quickly dissolve it in the high-melting metal and then cool the solution. Although this does not eliminate losses due to volatilization, it can greatly reduce the problem. And actually, since alloys are frequently composed of predominantly one metal, it is not uncommon to dissolve the lesser components into the primary component as a matter of practice anyway.

I hope the example I gave addresses your question. Please don't hesitate to ask for any clarifications in the comments below.


Why do displacement reactions occur? Suppose we have a compound, say copper sulfate. I know that sulfate is a poly-atomic ion, having valency 2. And, in this case, copper also has valency 2. So, to fullfill their 'octet' requirements, they both react. The problem is: Suppose we take the metal zinc and bring it in contact with copper sulfate. Now, a displacement reaction occurs, in which zinc replaces copper because of its higher reactivity.

Copper sulfate is a compound that is neutral; both the cation and anion have completed their octets; why then does zinc replace copper? Or, in other words, after completing the octet, why it is necessary for sulfate to leave copper and react with zinc?

It is due to that zinc is more reactive than copper which is less reactive metal so, zinc can easily displaces copper from its solution. We can understand it as zinc ion, $\ce{Zn^2+}$ ion form more strong electrostatic attraction with sulphate ion, $\ce{SO4-}$ ion.

Also, we can also consider the electrochemical series or activity series to give a better reason for the same. In activity series, zinc has greater value of standard reduction potential than copper's standard reduction potential value. It means, zinc can easily reduce Copper and it get oxidised.

$$\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}$$


Why does old gasoline seem to have more "kick"? From my personal experience, old gasoline (leftover in a container from the previous year) has more "kick" (power) when I actually use it. Higher idle RPM, higher throttle sensitivity.

It says here that it's not the gasoline itself that burns, but the vapour. And old gasoline is bad because it has vaporised over time. Thus, old gasoline must be weaker.

However, I read somewhere that gasoline vapour has nothing to do with octane rate, so there's no difference in octane between the gas left out in the open for a while and the gas sealed in a container.

I'm confused. Does old gasoline seem to have more "kick" because it needs more compression to combust? Or what is it?

My candidate to explain the increased performance is the removal of water as part of the corrosion reaction occurring in metal storage containers. To quote a reference:

  Since oxygen has different solubilities (higher for the organic and lower for the aqueous phase), anode or cathode regions are formed on the steel or metal surface. This creates favorable conditions for electrochemical corrosion in a given environment.

As such, I would argue that not only does oxygen content impact corrosion rate, but potentially more important, is that the removal of water over time (from corrosion chemistry) also increases O2 solubility (per above). This could be especially favorable when the fuel is vaporized and mixed with air in the ignition chamber.

[EDIT] Further comments, same source:

  Oxidation of biodiesel leads to the formation of various products such as peroxides and hydroperoxides. During degradation, these products transform into shorter-chain compounds such as low-molecular weight acids, aldehydes, ketones, and alcohols [9].

Per another source:

  In the winter, companies produce a gasoline containing lighter hydrocarbons, making the liquid more volatile and therefore easier to ignite.

This source also notes with more volatile fuels, evaporation can leave behind a heavier fuel mix.

Corrosion induced oxidation of biodiesel, in particular, appears to not only impact potential oxygen content of the fuel, but may add peroxides as well. The creation of lower molecular weight organics should also increase flammability.

An interesting point once presented in Wikipedia on Methanol, to quote:

"As a fuel for mud racers, methanol mixed with gasoline and nitrous oxide produces more power than gasoline and nitrous oxide alone."

Which, in the current context, supports my opinion that fuel mix chemistry (whether corrosion based, or otherwise, involving O2 or N2O) CAN IMPACT PERFORMANCE.

So, more oxygen content, less heavy fuel mix, more complete ignition, may explain some of the so-called 'kick' that has been referenced. On the other hand, significant evaporation is likely detrimental.

[EDIT] Yet another path as a consequence of corrosion chemistry, to quote from Wikipedia on Tetraethyllead:

  At the temperatures found in internal combustion engines, (CH3CH2)4Pb decomposes completely into lead and lead oxides as well as combustible, short-lived ethyl radicals. Lead and lead oxide scavenge radical intermediates in combustion reactions. Engine knock is caused by a cool flame, an oscillating low-temperature combustion reaction that occurs before the proper, hot ignition. Lead quenches the pyrolysed radicals and thus kills the radical chain reaction that would sustain a cool flame, preventing it from disturbing the smooth ignition of the hot flame front.

Also on the performance effect of Tetraethyllead (TEL):

  Antiknock agents allow the use of higher compression ratios for greater efficiency[17] and peak power.[18] Adding varying amounts of additives such as low percentage TEL or high percentage ethanol to gasoline allowed easy inexpensive control of octane ratings... Aviation spirits with TEL used in WWII reached 150 octane to enable supercharged engines such as the Rolls-Royce Merlin and Griffon to reach high horsepower ratings at altitude.[19] In military aviation, TEL manipulation allowed a range of different fuels to be tailored for particular flight conditions.

So, the corrosion process on even minor alloy metals, found in the metal storage container, may have also introduced an organometal compound with similar properties to TEL in its ability to scavenge radical intermediates.


Find second acid dissociation constant at a given pH In short, the problem states that:

  We have a molecule that has two ionizable groups
  $\mathrm{p}K_\mathrm{a,1}$ is between $5.0$ and $9.0$; $\mathrm{p}K_\mathrm{a,2}$ is outside of this range.
  $100\ \mathrm{mL}$ of a $0.1\ \mathrm{M}$ solution of molecule, all in the unprotonated form
  added $12\ \mathrm{mL}$ of $1\ \mathrm{M}\ \ce{HCl}$
  final $\mathrm{pH}$ is $7.4$

The goal is to find $\mathrm{p}K_\mathrm{a,1}$. This is how I've approached it:

First, I converted everything to moles:

$0.012\ \mathrm{mol}$ of $\ce{HCl}$ added $\ce{->} 0.012\ \mathrm{mol}\ [\ce{H+}]$
$0.01\ \mathrm{mol}$ of molecule in solution $\ce{->} 0.02\ \mathrm{mol}$ of $[\ce{A-}]$ since the problem states that everything is in unprotonated form I think this is were I am wrong in my thinking
Next, I reason that some $\ce{A-}$ is going to be neutralized by $\ce{H+ ->} 0.02-0.012$. Leaves me with $0.008\ \mathrm{mol}$ of $\ce{A-}$
Then, I have $0.012~\mathrm{mol}$ of $\ce{HA}$ that was formed.
Im plugging this in to the Henderson-Hasselbalch equation $\ce{->} \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log ([\ce{A-}]/[\ce{HA}]) \ce{->}$
leads to the incorrect answer.

I've tried looking at similar problems but I am just having a hard time with this. I feel that the problem lies with this the neutralization. Also, I know that upper boundary of $\mathrm{p}K_\mathrm{a,1}$ is $9$, which is above the final $\mathrm{pH}$, and I know that at $\mathrm{pH}=\mathrm{p}K_\mathrm{a} 50~\%$ would be protonated/unprotonated, however, I don't know how to account for this information in my calculation.

For starters, I’m going to define your molecule as being $\ce{HA}$ and its two ionised forms to be $\ce{A-}$ and $\ce{H2A+}$. This means, we are observing the following equilibrium:

$$\ce{H2A+ <=>[H+] HA <=>[H+] A-}$$

As you correctly calculated, you have $10~\mathrm{mmol}$ of $\ce{H_nA^m+}$ and $12~\mathrm{mmol}\ \ce{HCl}$ that you add to your solution.

$12~\mathrm{mmol} > 10~\mathrm{mmol}$. Thus, there are two possibilities:

we started with $10~\mathrm{mmol}\ \ce{HA}$. We ended up with $10~\mathrm{mmol}\ \ce{H2A+}$ and $2~\mathrm{mmol}\ \ce{HCl}$
we started with $10~\mathrm{mmol}\ \ce{A-}$. We ended up with $8~\mathrm{mmol}\ \ce{HA}$ and $2~\mathrm{mmol}\ \ce{H2A+}$

$\mathrm{pH}~7.4$ is only consistent with possibility 2; in case of 1 it would mean $2~\mathrm{mmol}\ \ce{HCl}$ controlling the $\mathrm{pH}$ which would give us $\mathrm{pH} < 3$. Thus, we are well within the operating scheme of $\mathrm{p}K_\mathrm{a,1}$ and we can easily use the Henderson–Hasselbalch equation:

$$\begin{align}\mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{HA}]}{[\ce{H2A+}]}\\
7.4 &= \mathrm{p}K_\mathrm{a} + \log \frac{\frac{8~\mathrm{mmol}}{V}}{\frac{2~\mathrm{mmol}}{V}}\\
7.4 &= \mathrm{p}K_\mathrm{a} + \log 4\\
7.4 &= \mathrm{p}K_\mathrm{a} + 0.60\\
6.8 &= \mathrm{p}K_\mathrm{a}\end{align}$$

You did indeed identify the spot where you went wrong. Just because a compound is dibasic does not mean you should double its concentration. As I showed above, $10~\mathrm{mmol}$ of original molecule, whether it was entered as $\ce{H2A+}, \ce{HA}$ or $\ce{A-}$ will remain $10~\mathrm{mmol}$; only the number of attached protons will differ.


What are the effects of heating paraffin wax above it's melting point I'm working with a paraffin embedding station that we recently learned was overshooting it's set point (60C) and heating paraffin to 100C. The problem has been corrected, but I'm curious what the mechanical/physical effects of paraffin being heated to almost twice the melting point. This is for a microtome setup (slicing at 5 microns) so micro features can and would be effected.

In short: A brief heating to 100 C (instead to 60 C) probably did not destroy your paraffin.

It is not too surprising the paraffin accidentally was warmed to 100 C, instead to 60 C. Especially in the solid state it does conduct heat not very well, and has a considerable heat capacity, too.  If it melts around 60 C,  heating to 100 C won't lead to vigorous boiling (like boiling water at this temperature).  In addition, I assume once you noticed the temperature higher than necessary / suitable for the embedding, you rapidly down-regulated the heater, too.

Maybe the manufacturer of the paraffin provided some properties of the batch delivered to you, beside melting point an indication of the decomposition temperature, too; which however should be much higher than 100 C (say 150...175 C).  I think it is unlikely that you observed fumes similar to the ones by an overheated frying pan, which were indicative for decomposition of the paraffin.

The reasoning: Many of the contemporary candles are equally made of paraffin, and when the children make new ones, chunks of old candles and new paraffin are equally thrown into the same melting pot, surrounded by (gently) boiling water, too. And while they do so for hours, I still have to see one new candle that became brown or black because the vessel was heated for so long time.


Energy levels of combined orbitals I saw the MO diagram for a NH3 molecule but couldn't understand why the combined atomic orbitals were in the positions they were, and if these positions were fixed.

Why is the sigma anti bonding lower in energy than the pi anti-bonding? Is it something you have to learn off by heart for each common molecule?

In the strictest sense, the orbital energies are simply eigenvalues of the multi-electron Schrödinger equation; unfortunately, the multi-electron Hamiltonian takes a nasty form and is impossible for us to solve exactly, so we can't get any insight on the energies from that approach...
Instead we inadvertently resort to LCAO-based methods, whereby exploiting symmetry the interacting atomic (or symmetry) orbitals can form molecular orbitals of the correct form. The energy of these is another issue; for some back-of-the-envelope calculations, severe approximations such as Hückel MO theory can be useful for a rough idea, but more accurate estimates just require much more complicated computational methods (e.g. the acclaimed Hartree-Fock).
As these methods resort to a large basis of AOs (and not just the simplified 2 AOs mix in- and out-of-phase to give 2 MOs picture) and other complications such as electron correlation, I'm afraid there is no simple rule to give an accurate enough estimate unless thorough calculations are made. So, to forwardly answer your question, rather impractically we can't know a priori the ordering of the orbitals in the general case.
Certain semi-empirical rules (based on the experience from multiple common systems and "chemical intuition") exist, such as the (often useful) approximation formulated for organic molecules $\sigma < \pi < l.p. < \pi^* <\sigma^*$ though this is very rough and suitable for organic systems, usually.
In the case of ammonia (as given in the question) a particular ordering is found; it should be noted that the two types of bonding interactions are both $\sigma$-interactions (i.e. locally cylindrically symmetric; there is no nodal plane within each individual bond). Despite the double degeneracy, the $e$ set still constitutes a $\sigma$-type MO (with a node in the overall MO shape, rationalising a higher energy than the $a_1$); so the ordering in ammonia is $\sigma_{a_1} < \sigma_e < lp_{a_1} < \sigma_{a_1}^* < \sigma_e^*$ -not so counter-intuitive after all, just different types of $\sigma$ MOs possessing different energies. (As a sidenote, the doubly deg. $\sigma_e^*$ are those allowing ammonia to act as a weak $\pi$-acceptor, placing it higher than simple $\sigma$-donors in the spectrochemical series.)
Summary: MO energy calculation especially in polyatomic species is hard and done with complex computational methods. Unfortunately, then, we can't have a general rule for the energy ordering in advance. However, often we can make educated guesses in analogy to systems we have already seen.
(For a more detailed look into the methods used to obtain the energies see Atkins & Friedman Molecular Quantum Mechanics, 4th ed.; Chapters 8–9)


Is there a flame which doesn't produce smoke? I read a book where it states the following:

  There are many types of flame and there are even flames which don't produce smoke.

After reading it, it just got stuck in my mind. Is there a flame which doesn't produce smoke? If so, then what is it?

Yes there are flames that don't produce smoke, but the reasons are complicated and depend on the conditions of the flame as well as the key fuel creating the flame.

The obvious example of a smokeless flame is hydrogen. Pure hydrogen burns in oxygen to generate water and there is no way the normal things that make up smoke can be created (the smoke in the Hindenburg disaster was the gas-bag covers and the content of the zeppelin so isn't a counter example).

Smoke is, technically, a suspension of particulate matter in the atmosphere. Some things look like smoke but are not (like the steam from a kettle or the steam from a power station cooler tower: these consist of liquid droplets that will evaporate).

The main reason smoke is created is because the reaction driving the flame is incomplete and can produce small particles of something. Flames involving hydrocarbons, for example, have side reactions that can generate small particles of solid carbon or a variety of long-chain or aromatic hydrocarbons. The amount of these side-products (and therefore smoke) depends not just on the fuel but also on the control of the combustion reaction.

Flames where the fuel-air mixture is well controlled may generate very few side reactions and no visible smoke (most gas cookers are set up carefully and don't generate visible smoke, but a badly controlled Bunsen burner using the same fuel may be very smoky.) Candles emit smoke by design as this is what generates most of the light in the flame (small particles of carbon become incandescent in the flame emitting yellowish light just because they are hot). Candles are, essentially, optimised (by the design of the fuel and wick) to burn this way. But small burners using wicks and fuels like ethanol or acetone burn with very clean flames with little smoke (they are optimised for heat not light).

Even fairly well optimised burning can produce some smoke. Modern vehicle engines, for example, need to be as efficient as possible but both petrol (gasoline) and Diesel engines still emit some small particulates (though there are often not very visible unlike old Diesel engines which were notoriously smoky). A badly controlled flame from burning hydrocarbon can be very smoky, but a well controlled flame will emit little visible smoke.

Some things are hard to burn efficiently and smoke a lot. Raw coal is full of a variety of things other than carbon (lots of S and N containing compounds, for example) when burning these tend to promote smoky emissions and also release nasty sulphur and nitrogen oxides into the atmosphere which may react further creating more smoky nasties. "Smokeless" coal is heated before use to drive off the worst offenders and results in something like a halfway house on the way to charcoal, which is almost pure carbon. But the burning of solids is always harder to control than gases or liquids and even charcoal will emit small amounts of smoke though it may not be visible.

In short, smoke is usually a product of the control of a flame. Badly controlled flames smoke because side reactions are common; well controlled flames mostly don't as burning goes to the final products (CO2 and H2O which are gases). Well controlled flames don't smoke visibly (but might emit small amounts that are hard to see for some fuels). Pure hydrogen can burn with no smoke at all as there are no possible side reactions.


Anode and Cathode Understanding I am very much confused: in the cell (Zn, Cu in $\ce {H2SO4}$), Cu is positively charged and is anode, where in this cell (Zn in $\ce {ZnSO4}$ and Cu in $\ce {CuSO4}$), Cu is positively charged but is Cathode. So how can I define it, where both are electric or voltaic cell. Please help me. Please see these picture for better understanding.

reference:A Text Book Of Electrical Technology,Volume 1,B.L.Theraja

Here Cu plate is cathode and Zn plate is anode, it's written wrong in it. In the first case (Zn in  $\ce{ZnSO4}$ and Cu in $\ce{CuSO4}$) the electrodes are connected through an external circuit (You see that bulb glowing?). But what this article is trying to say is that in the cell (Zn, Cu in $\ce{H2SO4}$), if you leave the solution as it is without connecting externally, Zn having higher tendency to lose electrons, will lose electrons and these will appear on its plate surface making it negatively charged, the $\ce{Zn^2+}$ thus formed will combine with $\ce{ SO4^2-}$ and form $\ce{ZnSO4}$. The remaining two $\ce{H^+}$ ions will more towards copper plate where they will combine with the electrons on its plate and make it positive charged, releasing $\ce{H2}$. Now if a wire is connected across this setup then the electrons on Zn surface will leave it and go towards positive charged Cu plate.


Why does a Diels–Alder reaction not occur here? I have  recently come across this organic chemistry problem.
I thought it would undergo simple Diels–Alder reaction
But it didn't turn out to be. The product to me looks like a nucleophilic attack of (I) on (II).
Can anybody say me why exactly this abnormality seen and what is the mechanism of this reaction?

Firstly, Why doesn't Diels-Alder reaction happen here?

On first sight the reaction seem completely feasible as both the diene and the dienophile are good species that could undergo Diels-Alder, but no. It could be proven by just analyzing the products (the actual one and the expected one).

Resonance. The property that literally defines organic stability. The DA product has not got even a trace of the stabilization that the actual product receives. The actual product gets to experience extended resonance due to the benzene ring that is formed.
The second factor being that the actual product has got TWO (yes...two) aromatic rings fused together, each having 6 electron resonance.
Steric hindrance. The DA product experiences Steric hindrance while the reaction happens due to the bulky dithiane group. While the other path is free from this.
Strain. Consider the carbon that is holding the dithiane group in both the products. In the DA product the carbon is $sp_3$ while the other product has $sp_2$ (perfect 120 degrees required, and 120 degrees received). In fact, all carbons in the actual product are $sp_2$, other than the ones in the dithiane group obviously.

All of these factors encourage the reaction to proceed in the second path which leads to an amazing yield of 62%.

P.S. I kinda feel stupid after typing "actual product" like 10 times.

Now, the MECHANISM (ohh yeahh!)

Let's start with para-benzoquinone. It is an $\alpha-\beta ketone$. Ergo, the $\beta$ hydrogen is acidic. So, lets prepare it for the incoming nucleophile.

umm...something's missing. Yeah!

OK. Looks better.

Now that we are done with this, we will get to the other reactant. Observe that the methoxy group on the $sp_2$ makes it a good nucleophile. But there is a problem, the dithiane group is too bulky to be drawn again and again. Therefore,

Properties of 'G' :

*bulky, must be kept away

*observing the product, where it is intact, we can say that it is useless for the reaction.

Coming back to the reaction, this nucleophile attacks our first molecule.

Here the (*) marked carbon is activated due to the methoxy group, hence the attack is at that site.

Now the carbocation can stabilized using proton transfer onto the oxygen.

The encouraging factor for this is the achievement of an aromatic ring in the molecule.

Observe the (*) marked carbon in the above image. We see that it is strained due to the thiane group present on + the methoxy group. Therefore the methoxy could abstract the proton from the adjacent carbon and exit as methanol.

Yet again, another encouraging factor for this exit is the formation of another aromatic ring.

Replacing the 'G" with it's true form we have...(drum roll)


Formulae distinguished only by case A recent question (on $\ce{CI}$ v $\ce{Cl}$) reminded of this question: are there any real (as opposed to contrived) formulae which would be the same if case was ignored?

One certain example is $\ce{Co}$, Cobalt, and $\ce{CO}$, Carbon Monoxide.  Of course, confusing Cobalt and Carbon Monoxide is rather unlikely.

In my search for others, the best example I could find was $\ce{BaS}$, Barium Sulphide, and $\ce{BAs}$, Boron Arsenide.  Are these real substances and is the stoichiometry just $1:1$?  I had trouble verifying that.

Are there any more interesting examples?

One real example that I saw once was a poster in a university chemistry lab showing some complex organic molecules.  $\ce{Ac}$ appeared at many points but it seemed very unlikely that these compounds contained Actinium.  I then noticed $\ce{Me}$ and $\ce{Et}$ and guessed that they were organic units e.g. methyl and ethyl so $\ce{Ac}$ may have been acetyl.

Letter case in chemical names sometimes matters too! E.g. when o-methylphenol (o-cresol, ortho-cresol, 2-methylphenol, I) improperly capitalized to O-methylphenol (instead of o-Methylphenol), it can be interpreted as a different compound, phenol methylated at oxygen (anisole, methoxybenzene, II).


How did scientists discover the concepts of anode and cathode without knowing about electrons? Scientists researched electrons and came to know about electrons after the battery was developed as we can see in the JJ Thomson's model that he used the cathode and anode. My question is that that if the scientists have known already that there are positive and negative charge that they have made the battery they should have known about electrons.

I am confused in this. Research tells that electrons were first discovered by JJ Thomson and that there are negative charges in atoms. My question is that that, without knowing about electrons how were the ideas of cathode and anode developed.

Electricity is a macroscopic phenomenon. It does not depend on the notion of an electron.

In 1733 Charles François de Cisternay du Fay discovered the existence of positive and negative charge. In 1800, after the invention of the voltaic pile, the first electrolysis experiments that showed the effect of electric current on water have been made.

In 1833 Michael Faraday defined the terms cathode and anode:

  The anode is therefore that surface at which the electric current, according to our present expression, enters : it is the negative extremity of the decomposing body; is where oxygen, chlorine, acids, &e., are evolved; and is against or opposite the positive electrode. The cathode is that surface at which the current leaves the decomposing body, and is its positive extremity ; the combustible bodies, metals, alkalies, and bases, are evolved there, and it is in contact with the negative electrode.

His definition does not involve particles of any type. In fact Richard Laming firstly hypothesized the concept of an indivisible quantity of electric charge in 1838.

You may want to study https://en.wikipedia.org/wiki/History_of_electrochemistry for further information.


Can I use fluorescence microscopy with a solution of hydrogen peroxide? I want to use microspheres which have been dyed with a fluorescent dye for fluorescence microscopy in a solution that has a 1-5% concentration of hydrogen peroxide. Is this possible or will the hydrogen peroxide always bleach any particles I put in?

The Hydrogen peroxide will definitively accelerate bleaching of your fluorescent microspheres. In fact, we routinely add glucose oxidase, glucose and catalase to our single molecule fluorescence experiments in order to get rid of oxygen and hydrogen peroxide in our solution and thus reduce photobleaching. However, Fluorescence microspheres are very photostable and whether or not they work for your application depends on several factors:

how long the microspheres have been in the solution before you start imaging (the sooner you start the better, because of less reaction time between dye and hydrogen peroxide).
How long you need to image (the shorter the better, because of less photobleaching)
How large the microspheres can be (the larger the better because large microsphere contain more dye that is less accessible to the hydrogen peroxide in solution).

Basically, I suggest you try (and let us know how it went :) )


How to avoid oxidation of Si using a sub nanometer layer Does anyone know how and which material to use to protect a Si layer from Oxidation by a  small layer of thickness less than 1 nm ? (probably using atomic layer deposition)

Fluorine termination might be worth a look, if you have the facility for it. Sunada et al., in 1990 Jpn. J. Appl. Phys. 29 L24082, write in the abstract:

  The principal role of silicon-fluorine bonds in the chemical nature of HF-etched Si surfaces has been investigated by angle-resolved X-ray photoelectron spectroscopy. The native oxide growth thickness and the fluorine coverage have been systematically measured as functions of HF concentration, pure water rinse time, air or N2+O2 gas exposure time and gas phase H2O concentration. It is shown that the native oxide growth is strongly suppressed by the existence of Si-F bonds of about 0.12 monolayers on the surface. This is explained by a model in which Si-F bonds chemically stabilize the surface reactive sites such as atomic steps as supported by the result of the layer-by-layer oxidation of Si.

(My bolding).
I haven't read the full text, and I suspect that "strongly suppressed" really means that there was a very small amount of oxidation, rather than none at all. And note that while silicon surfaces may be stabilised by fluorination, scientists tend to get violently destabilised by the same reaction. This is a potentially hazardous process and not to be undertaken lightly.

In any case, "stability" really depends on what the surface is going to be exposed to. F-termination may oxidise only very slowly in air, but if you get water in there, it's a different story. In 1993 Jpn. J. Appl. Phys. 32 3125, Nakamura et al. write that:

  The 1.0 monolayer of Si-F bonds was found to be replaced with the 0.7 monolayer of Si-H bonds and the 0.2-0.3 monolayer of Si-OH bonds after dipping into water for 1 min.

So there may not be a "permanent" solution, just one that resists further oxidation for a while. What are you actually trying to achieve?

One other thing you could investigate is growing a self-assembled monolayer (SAM), probably on an H-terminated Si surface. It's been a while, but IIRC some pretty stable coatings can be generated that way. A SAM forest may offer a decent steric barrier to further reaction.


Is there a relationship between boiling point and van der Waals constants Is there a relation of the boiling point of a gas with its van der Waals constants, $a$ & $b$?

I know if constant $a$ is greater, then the boiling point will also be greater, because of more attraction between the molecules, but does the boiling point depend on $b$ too? If two gases have the same $a$ but different $b$, can I tell which of the two gases will have a higher boiling point or nearly same boiling point?

There will be a relationship, but as the comments on the question indicate, the relationship will not be simple. There are two complications:

The van der Waals equation is an empirical model built to fit experimental pressure/volume/temperature data for real gasses.
The boiling point of a liquid is not a single number but a line on the pressure/temperature phase diagram.

Supposed the van der Waals equation can predict phase behavior (source: 1) However, I made some graphs. The van der Waals constant data are from Wikipedia and the boiling point data looked up from the Wikipedia articles.

  Constant a

The blue line is a cubic polynomial fit, while the grey shaded area is the error. With the exception of the metal outlier (mercury), this fit seems pretty good.

  Constant b

The blue line is a logarithmic fit, while the grey shaded area is the error. Again, with the exception of the metal outlier (mercury), this fit seems pretty good.

  Finally, I decided to plot the van der Waals constants against each other.

The van der Waals constants are fairly co-linear, so any clear relationship between one and boiling points implies a clear relationship between the other and boiling points.

Source:

Terrell L. Hill, 2012 [1960], "An Introduction to Statistical Thermodynamics" [Dover Books on Physics], Chicago, IL, USA:R.R. Donnelly (Courier/Dover), ISBN 0486130908, referenced on the Wikipedia article, but I do not have access.


Finding all Chemical Reactions Given Products and Reactants Suppose only a set of chemical formulas is given. How can you find all mathematically possible chemical equations whose educts and products are only from this set?

Take e.g. the set $\{ \ce{C, H2, O2, N2, CO, CO2, H2O, NH3, NO, NO2, NO3}\}$.

Or consider e.g. the example in MathStackexchange: Finding all chemical equations (Linear Algebra).

To find all combinations and partial reactions between such given species by hand, is possibly incredibly tedious or near impossible for more complex chemical reaction systems.

I guess these methods are topic of Mathematical Chemistry, Computer Chemistry or Chemoinformatics.

The purely combinatorial methods are a first step in modeling chemical reactions. They alone however cannot consider the possibility of the chemical reactions and the chemical stability of the reaction products.
$\ $

1.)

You can build all combinations of at least two formulas, all from the original set of given chemical formulas, and treat this as your set of given formulas. But you can also treat the whole original set of given chemical formulas.

I demonstrate the method with the first example given in the question:

$$\{ \ce{C, H2, N2, O2, CO, CO2, H2O, NH3, NO, NO2, NO3}\}$$

We set up one chemical equation with all chemical formulas that are in the given set. Let the stoichiometric factors in the chemical equation be denoted by $\nu$. Because we don't know which of the given substances will be educt and which of them will be product, we write the chemical equation without reaction arrow. Instead, a negative stoichiometric factor will later mark an educt, a positive a product:

$$\nu_1\ce{C}+\nu_2\ce{H2}+\nu_3\ce{N2}+\nu_4\ce{O2}+\nu_5\ce{CO}+\nu_6\ce{CO2}+\nu_7\ce{H2O}+\nu_8\ce{NH3}+\nu_9\ce{NO}+\nu_{10}\ce{NO2}+\nu_{11}\ce{NO3}$$

We have to obey one of the first fundamental laws of chemistry: the Law of multiple proportions. To fulfill this law, the total sum of each chemical element in our chemical equation has to be $0$. And if ions are among our given chemical formulas, the total sum of all electric charges has to be $0$ also. But no electric charges are involved in our given example.

We will now build a mathematical model of our chemical problem.

For each of the given chemical elements, we have to set up its balance equation from our chemical equation above:

$\ce{C}:\ \ \ \nu_1+\nu_5+\nu_6=0$
$\ce{H}:\ \ \ 2\nu_2+2\nu_7+3\nu_8=0$
$\ce{N}:\ \ \ 2\nu_3+\nu_8+\nu_9+\nu_{10}+\nu_{11}=0$
$\ce{O}:\ \ \ 2\nu_4+\nu_5+2\nu_6+\nu_7+\nu_9+2\nu_{10}+3\nu_{11}=0$

This is a linear equation system. In the general case, too, a linear equation system results. It consists of the coefficients (the numbers above) and the stoichiometric factors $\nu_i$, which are sought. In our example, the equation system has 4 equations (the number of different chemical elements and the electric charge) and 11 unknowns (the number of chemical formulas in the given set).

Linear Algebra says how a linear equation system can be handled. We use here the matrix presentation.

Each of the 11 given chemical formulas is presented by a column vector which contains the frequency of occurence of each chemical element in the chemical formula in a prescribed order. Each of the 4 given chemical elements is presented by a row vector which contains the frequency of occurence of the chemical element in the chemical formulas in a prescribed order.

All 11 column vectors or all 4 raw vectors of the equation system are combined to give the coefficient matrix $A$. Our 11 wanted stoichiometric coefficients build the solution vector $x$, which is sought. The matrix representation of our linear equation system is than:

$$A\cdot x=\emptyset,$$

wherein $\emptyset$ is the zero column vector with 4 rows. It is written out:

$$\left(
\begin{array}{}
1&0&0&0&1&1&0&0&0&0&0\\
0&2&0&0&0&0&2&3&0&0&0\\
0&0&2&0&0&0&0&1&1&1&1\\
0&0&0&2&1&2&1&0&1&2&3
\end{array}
\right)
\cdot
\left(
\begin{array}{}
\nu_1\\\nu_2\\\nu_3\\\nu_4\\\nu_5\\\nu_6\\\nu_7\\\nu_8\\\nu_9\\\nu_{10}\\\nu_{11}
\end{array}
\right)
=\left(\begin{array}{}0\\0\\0\\0\end{array}\right)$$

The solution vector $x$ can be found by solving the linear equation system by methods of Linear Algebra.

The solution vector of our example is:

$$x=\left(
\begin{array}{c}
2\nu_4+\nu_6+\nu_7+\nu_9+2\nu_{10}+3\nu_{11}\\-\nu_7+3\nu_3+\frac{3}{2}\nu_9+\frac{3}{2}\nu_{10}+\frac{3}{2}\nu_{11}\\\nu_3\\\nu_4\\-2\nu_4-2\nu_6-\nu_7-\nu_9-2\nu_{10}-3\nu_{11}\\\nu_6\\\nu_7\\-2\nu_3-\nu_9-\nu_{10}-\nu_{11}\\\nu_9\\\nu_{10}\\\nu_{11}
\end{array}
\right)$$

The different combinatorially possible chemical equations are obtained by choosing suitable values for the free variables $\nu_3,\nu_4,\nu_6,\nu_7,\nu_9,\nu_{10},\nu_{11}$.
$\ $

2.)

Further investigations could be made by searching the solutions that don't contain any other solution. That are the minimal solutions, that means the solutions that are linearly independent of the other solutions.

I found 663 chemical equations and 83 minimal solutions for the example above.

Generating only the minimal solutions in polynomial time needs particular algorithms:

Schay, G.; Pethö, Á.: Über die mathematischen Grundlagen der Stöchiometrie. Acta Chim. Acad. Sci. Hung. 32 (1962) 59-67

Pethö, Á.: Zur Theorie der Stöchiometrie Chemischer Reaktionssysteme. Wissenschaftl. Zeitschr. 6 (1964) 13-15

Pethö, Á.: Algebraic treatment of a class of chemical reactions in stoichiometry. Acta Chim. Acad. Sci. Hung. 54 (1967) 107-117

Pethö, Á.: On a class of solutions of algebraic homogeneous linear equations. Acta Math. Acad. Sci. Hung. 18 (1967) 19-23

Pethö, Á.: Kémiai reakciók egy osztályának algebrai elemzése. Magyar Kémiai Folyóirat 74 (1968) 488-491

Kumar, S.; Pethö, Á: Note on a combinatorial problem for the stoichiometry of chemical reactions. Int. Chem. Eng. 25 (1985) 767-769

Pethö, Á.: The linear relationship between stoichiometry and dimensional analysis. Chem. Eng. Technol. 13 (1990) 328-332

Szalkai, I.: Generating minimal reactions in stoichiometry using linear algebra. Hung. J. Ind. Chem. 19 (1991) 289-292
with a software code at the end of the article

Pethö, Á.: Mathematical discussion of the application of Hess‘s law. Hung. J. Ind. Chem. 21 (1993) 35-38

Pethö, Á.: Further remarks on the linear relationship between stoichiometry and dimensional analysis. Chem. Eng. Techn. Chem. Eng. Tech. 17 (1994) (1) 47-49

Pethö, Á.: Further remarks on the analogy between stoichiometry and dimensional analysis: The valuation operation. Hung. J. Ind. Chem. 23 (1995) 229-231

Laflamme, C.; Szalkai, I.: Counting simplexes in $R^n$. Hung. J. Ind. Chem. 23 (1995) 237-240

Laflamme, C.; and I. Szalkai, I.: Counting simplexes in $R^3$ Electron. J. Combin. 5 (1998) (1) #R40 11. Printed version in: J. Combin. 5 (1998) 597-607

Szalkai, I.: Handling multicomponent systems in $R^n$ I Theoretical results. J. Math. Chem. 25 (1999) 31-46

Szalkai, I.: On valuation operators in stoichiometry and in reaction syntheses. J. Math. Chem. 27 (2000) 377-386

Szalkai, I.: A new general algorithmic method in reaction syntheses using linear algebra. J. Math. Chem. 28 (2000) 1-34

Szalkai, B.; Szalkai, I.: Counting minimal reactions with specific conditions in $R^4$ J. Math. Chem. 49 (2011) 1071-1085

Szalkai, I.; Dósa, G; Tuza, Z.; Szalkai, B.: On minimal solutions of systems of linear equations with applications. Miskolc Mathematical Notes 13 (2012) (2) 529-541

Szalkai, B.; Szalkai, I.: Simplexes and their applications - A short survey. Miskolc Math. Notes 14 (2013) (1) 279-290

Szalkai, I.; Tuza, Z.: Minimum Number of Affine Simplexes of Given Dimension. Discr. Appl. Math. 180 (2015) 141-149

Szalkai, I.: Reakciómechanizmusok algoritmikus és matematikai vizsgálata (Algorithmic and Mathematical Examination of Reaction Mechanisms). PhD thesis, University of Pannonia, Veszprém, Hungary, 2014

Szalkai, I: An algorithmical and mathematical investigation of reactions. English extract of the author's PhD thesis, 2014
See the examples and references therein.

Tóth , J.: Reaction Kinetics: Exercises, Programs and Theorems. Springer, 2018
$ $

3.)

The answer above treats overall reactions. The underlying elementary reaction steps (elementary reactions) can be generated by working with the Ugi-Dugundji model: The chemical formulas are presented as graphs together with their graph matrices (Bond-Electron matrices (BE-matrices) and Reaction matrices (R-matrices)). The single elementary reaction steps will be generated successively by stepwise shifts of electrons or bonds in the set of given chemical formulas by stepwise shifting of single edges in the reaction matrix.

Ugi, I.; Gillespie, P.; Gillespie, C.: Chemistry, a finite metric topology - synthetic planning, an exercise in algebra. Trans. New York Acad. Sci. II 34 (1972) 416-432

Dugundji, J.; Ugi, I.: An algebraic model of constitutional chemistry as a basis for chemical computer programs. Top. Curr. Chem. 39 (1973) 19-64

Dugundji, J.; Gillespie, P. D.; Marquarding, D.; Ugi, I.; Ramirez, F.: Metric spaces and graphs representing the logical structure of chemistry. in: Chemical applications of graph theory. Balaban, A. T. (Ed.), Academic Press, London 1976, 107-174

Brandt, J.; Friedrich, J.; Gasteiger, J.; Jochum, C.; Wolfgang Schubert, W.; Ugi, I.: Computer programs for the deductive solution of chemical problems on the basis of a mathematical model of chemistry. in: Wipke, W. T.; Howe, W. J.: Computer-Assisted Organic Synthesis. American Chemical Society 1977, chapter 2, 33-59

Ugi, I.; Brandt, J.; Friedrich, J.; Gasteiger, J.; Jochum, C.; Lemmen, P.; Wolfgang Schubert, W.: The deductive solution of chemical problems by computer programs on the basis of a mathematical model of chemistry. Pure Appl. Chem. 50 (1978) 1303-1318

Ugi, I.; Brandt, J.; Friedrich, J.; Gasteiger, J.; Jochum, C.; Lemmen, P.; Schubert, W.: Computer Programs for the Deductive Solution of Chemical Problems on the Basis of a Mathematical Model of Chemistry. Pure Appl. Chem. 50 (1978) (11-12) 1303-1318

Behnke, C.; Bargon, J.: Computer-assisted topological analysis and completion of chemical reactions. J. Chem. Inf. Comput. Sci. 30 (1990) (3) 228-237

Ugi, I; Dengler, A.: The algebraic and graph theoretical completion of truncated reaction equations. J. Math. Chem. 9 (1992) (1) 1-10


How do the fields of medicinal chemistry, pharmaceutical chemistry, pharmacy, and pharmacology differ? I would like to know what the difference between medicinal chemistry, pharmaceutical chemistry,  pharmacy and pharmacology. All of them seem to involve the development of drugs, so I am confusd as to how to distinguish them.

Also, does specialising in medicinal chemistry require a biology background? It seems to lean more towards biology.

I'm in a medicinal chemistry department, the floor below is the pharmaceutics department. We are both located in our university's college of pharmacy. The pharmacology department is under the college of medicine in same building as biochemistry, microbiology, etc. There is little interaction between departments here and that's not good.

But we can break the differences in each discipline down into some order based on drug development. Medicinal chemistry and pharmacology come first, with the discovery of molecules that bind to receptors and other proteins and have biological effect. This used to be based almost entirely on natural products, and those still play a large role, but high throughput screening and computer modeling are becoming more important for discovery of drug candidates. While pharmacology focuses on the effects of drugs and other molecules on cells and the body, medicinal chemistry has traditionally focused on the chemistry behind them. A medicinal chemist might look at a natural product with desirable activity, and ask "how can I make this better?" and apply organic synthesis to produce new versions of that drug, with better activity, reduced toxicity, and longer half-life. Of course the biology cannot be ignored, and has become more important in recent decades. Our department used to be pure synthesis, but now only 2 of our 8 labs are pure synthesis, and 2 almost entirely biological.

Once the medicinal chemists have developed a drug candidate to the point they think it can be used in the real world, it moves on to pharmaceutical chemists. They not only have to refine the synthesis of the drug to produce it efficiently and reliably on large scale, they have to develop the formulation the drug is delivered as. If it's a tablet, they have to determine what to make the bulk of the tablet from, how to press the tablets, how to store them, etc. A large part of pharmaceutics is stability, how long can the product sit on a shelf and not degrade? How does temperature and humidity affect degradation? What does it degrade into and are those products dangerous? Pharmaceutics is also concerned with drug absorption and excretion. They often try to develop "extended release" formulations so that a tablet takes a longer time to break down in your gut, releasing the drug over a longer time and keeping an effective concentration of drug in your blood longer, thus requiring fewer pills and less variation in concentration. Pharmaceutics is a lot of organic chemistry, physical chemistry, analytical chemistry, and quality control.

Pharmacy is the end product of all of these. After a drug is developed and approved the pharmacist is responsible for distribution. While doctors prescribe drugs, they usually have very little training in how the drugs work and interact. The pharmacist needs to know more than just how to sell pills, they must have at least a working knowledge of the pharmacology behind the drug, be aware of potential interactions and side effects, and advise the patient on the best practices for using the drug. They must also be aware of the rules and regulations, and make sure the controlled substances under their care are not stolen or lost or abused. Substance abuse is actually the biggest workplace hazard for pharmacists.

I have some knowledge on these things because I'm in a med chem department, my father is a chemist for a pharmaceutical company, and my brother just got his pharmacy license, so we see all parts of the process.


Calculate the third and fourth nearest neighbours in bcc My textbook has given data for third and fourth nearest neighbours to be 12 and 8 with distances $\sqrt{2}a$ and $\frac{\sqrt{11}a}{2}$.

I have been able to calculate for  the first and second nearest neighbour but it has become difficult to visualise for the other two to calculate. Can you help me with hints on how to proceed preferably with a diagram.

With $\sqrt3\over2$ being that close to $1$, BCC packing is better not looked at in terms of coordination spheres. But if you insist...

Say you are sitting in the center of a cell. Then:

Your first neighbours are at the corners of the same cell.
Second neighbours are at the centers of the nearest adjacent cells.
Third neighbours: centers of the next adjacent cells (those which share two corners with your cell).
Fourth neighbours: far corners of the nearest adjacent cells.

On a side note, there are more than just 8 of the latter.


Origin of electrode potential and its connection with electric double layer Recently, I was learning about electrode processes and I came across the topic of "electrode potentials".
I understood that at equilibrium, a potential difference was generated at the interface between the electrode and the electrolyte because of the formation of an electric double layer.
I was reading
this
article which says that on dipping metals into their respective salt solutions, positive ions readily form and move into the solution leaving behind electrons in the electrode and a dynamic equilibrium is achieved soon.
Across different metals, only the position of the equilibrium (magnitude of negative charges on the metal electrode) differs due to their differing tendencies to oxidize (or reduce).
(I understood this part)

On the other hand,
I’ve also seen in some other texts that at equilibrium, the charge acquired by the metal might be either positive or negative depending on its tendency to oxidize (or reduce)
(unlike the $1$st source where they say the charge will be always negative).
Examples of
$\ce{Zn}$ ($\ce{Zn}$ has greater tendency to form $\ce{Zn^2+}$ ions hence metal acquires negative charge)
$$\ce{Zn<=>>Zn^2+ + 2e^-}$$
and
$\ce{Cu}$ ($\ce{Cu^2+}$ ions have greater tendency to get reduced hence $\ce{Cu}$ metal acquires positive charge)
$$\ce{Cu^2+ + 2e^-<=>>Cu}$$
are usually given.
(I also understood this explanation)

But what is confusing me is which of the above explanations is correct (or more accurate)?
Because if we go by the $2$nd view,
we have that $\ce{Cu}$ ions are more likely to get reduced so let’s say the $\ce{Cu}$ electrode becomes positively charged when ions from the solution stick to the metal, but at the same time few other $\ce{Cu}$ atoms in the metal will also get oxidized (because there are limited electrons in the $\ce{Cu}$ electrode) and dissolve so there’s no net oxidation or reduction. If this indeed happens then at equilibrium the $\ce{Cu}$ electrode will have no net charge (which isn't possible).

Note-I've seen a few other answers here on this site like this, this and this but none of these seem to address the confusion I'm having.
Any suggestion will be of great importance.

Let's see if we can tease out the main points here.  The images below are taken from Lance and Cole's Analytical Chemistry POGIL workbook.
Microscopic processes at a metal surface exposed to an aqueous solution containing the corresponding metal ion(1) will result in a slight negative charge on the metal surface due to oxidation of the metal and a slight positive charge in the region of the solution close to the metal surface owing to the sudden generation (and dissolution) of positively charged metal ions.

The amount of negative charge generated is metal specific and as of now, unknowable in an absolute sense.  Therefore, we are required to compare the extent of this process at one metal electrode to that of another.  If we compare the Zn electrode above to a Pb electrode, we would find that the Pb electrode has less negative charge on its surface relative to Zn.  Both of the electrodes, however, have a slightly negative charge at their surfaces.  Because the charge is greater at the Zn electrode, we say Zn is easier to oxidize than Pb.
To answer your question which of the above explanations is correct, the answer is both (although the clarity of the explanations is up for debate).  A negative charge is generated at both electrodes, but the the amount relative to one another differs, and that gives rise to the notions of easier/harder to oxidize/reduce.
(1) In my experience, microscopic theory of electrochemical processes typically assume the presence of the metal and metal ion in part to avoid issues which identifying what in the system is being oxidized and reduced.


How do you calculate molar specific heat? I know how to calculate the specific heat, but not molar specific heat. What is molar specific heat and how do you calculate it?

Specific heat has the units of $\mathrm{J/(K\cdot kg)}$.

Molar specific heat is in units of $\mathrm{J/(K\cdot mol)}$, and is the amount of heat needed (in joules) to raise the temperature of $1$ mole of something, by $1$ kelvin (assuming no phase changes).

So, the conversion factor you need, from dimension analysis, will have unit $\pu{kg/mol}$.

$\pu{kg/mol}$ is the SI unit for molar mass.

Multiply the specific heat by the molar mass to get the molar specific heat.

For example, the molar mass of water is $\approx \pu{0.018 kg/mol}$.

The specific heat of water is $\approx 4186\ \mathrm{J/(K\cdot kg)}$.

So the molar specific heat of water is $4186\ \mathrm{J/(K\cdot kg)} \times \pu{0.018 kg/mol} \approx 75\ \mathrm{J/(K\cdot mol)}$


Purification of isoborneol How can I convert isoborneol powder to isoborneol flakes?

I have made the isoborneol powder from reduction of camphor using sodium borohydride.

iso-Borneol is a fairly common reagent, and methods for its preparation and purification are readily accessible (even without journal access). For information, the entry in Purification of Laboratory Chemicals suggests recrystallising from ethanol or petroleum ether, melting point data is also provided as a way of assaying the resulting material  (though NMR would be better).

  dl-Isoborneol [124-76-5] M 154.3, m 212° (sealed tube). Crystallise isoborneol from EtOH or petroleum ether (b 60-80°). It sublimes in a vacuum. The 4-nitrobenzoyl derivative has m 153°. [Yager and Morgan J Am Chem Soc 57 2081 1935, Beilstein 6 II 80, 6 III 299, 6 IV 281.]

  Taken from Purification of Laboratory Chemicals, 7 ed. Armagero and Perrin

Its worth noting that a powder and flakes are essentially just two amorphous forms, and there isn't really any benefit of having one over the other (if anything, the powder is more convenient to handle).


Michaelis-Menten rate law for enzyme which catalyzes two reactions: steady state? Suppose an enzyme $\ce{E}$ can catalyze two reactions:

\begin{align}
\ce{S1 + E &<=> S1E ->  P1 + E} \tag{R1} \\
\ce{S2 + E &<=> S2E ->  P2 + E} \tag{R2}
\end{align}

I want to derive a rate law. Can I assume that

\begin{align}
\frac{d[\ce{S1E}]}{dt} &= 0 \tag{1}\\
\frac{d[\ce{S2E}]}{dt} &= 0 \tag{2}
\end{align}

like in the derivation of the Michaelis-Menten rate law?

In the steady-state reaction, the intermediate concentration [ES] is assumed to remain at a small constant value. So in this case only if k2 >> k1 and similar for the second reaction. ES is now a reactive intermediate and there is no stable equilibrium between S, E and P.

\begin{align}
\frac{d[\ce{S1E}]}{dt} &= \ce{k1}[\ce{S1}][\ce{E}] - k_{-1} [\ce{S1E}] - \ce{k2}[\ce{S1E}] = 0\\
\end{align}

Therfore:

\begin{align}
[\ce{S1E}] &= \frac{\ce{k1}}{k_{-1}+\ce{k2}}[\ce{S1}][\ce{E}] = K_a[\ce{S1}][\ce{E}]
\\
\end{align}

Similarly for the second reaction:

\begin{align}
[\ce{S2E}] &= \frac{\ce{k3}}{k_{-3}+\ce{k4}}[\ce{S2}][\ce{E}] = K_b[\ce{S2}][\ce{E}]
\\
\end{align}

The enzyme E is involved in both reactions and its total concentration(bound and unbound) is constant. This total concentration of enzyme $[E]_{0}$ is equivalent to the concentration of the free enzyme before adding the substrates. The concentration of the free enzyme at a certain time t is [E]:

\begin{align}
[E]_{0} &= [E] + [\ce{S1E}] + [\ce{S2E}]
\\
\end{align}

if you substitute $[\ce{S1E}]$ and $[\ce{S2E}]$ with the previous expressions then:

\begin{align}
[E] &= \frac{[E]_{0}}{1 + k_{a}[\ce{S1}]+k_{b}[\ce{S2}]}
\\
\end{align}

Eventually:
\begin{align}
\frac{d[P_{1}]}{dt} = k_2[\ce{S1E}] = \ce{k2}(K_a[\ce{S1}][\ce{E}]) = K_p[\ce{S1}][\ce{E}] = K_p[\ce{S1}]\frac{[E]_{0}}{1 + k_{a}[\ce{S1}]+k_{b}[\ce{S2}]}
\\
\end{align}

and similarly for the rate of production of $P_2$


Is it valid to calculate concentrations given equilibrium constant and initial concentration? Suppose we have the following reaction:
$$\ce{2A <=> B + C}\tag{1}$$
which can be thought the sum of the following reactions:
$$\ce{A <=> B}\tag{2}$$
$$\ce{A <=> C}\tag{3}$$
with equilibrium constants $K_1$ and $K_2$ respectively. The $K$ for the first reaction will be the product of $K_1$ and $K_2$ i.e. $K=K_1K_2 $. Now if we are given that initial we have $x$ mol of $\ce{A}$ and no amount of $\ce{B}$ and $\ce{C}$, one can calculate the amount of B and C present at equilibrium using the equilibrium constant $K$:
$$K=\frac{y\cdot y}{(x-2y)^2}\tag{4}$$
If we solve (4) for a given $x$ (initial amount) then concentrations of $\ce{B}$ and $\ce{C}$ must be equal. But how can this be possible if $K_1$ and $K_2$ have different values? I mean the following equality must holds:
$$[\ce{C}]= [\ce{B}] \Longleftrightarrow K_1{[\ce{A}]} = K_2\ce[{A}] \tag{5}$$
Edit I am asking if it is valid to use the equilibrium constant $K$ to find the concentrations of $\ce{B}$ and $\ce{C}$ because every reaction can be thought of as sum of other reactions. It is a common exercise in many general Chemistry textbooks where you are given a reaction with an equilibrium constant and you must find the concentrations at equilibrium. Well the substitutions into equilibrium constant that they make is the one that I have also did (assuming 0 initial concentration for both $\ce{B}$ and $\ce{C}$. So what is going wrong?

The equation
$$\ce{2A <=> B + C}$$
implies a fixed stoichiometry between B and C. If the two separate reactions (2) and (3) were happening, the correct net equation would be
$$\ce{(x + y) A <=> x B + y C}$$
Here is an example that would be well-described by reaction (1):
$$\ce{2 H2O <=> H3O+ + OH-}$$
One water molecule cannot make a hydronium ion (goes against the conservation of mass and charge). The same goes for a water reacting to form a hydroxide. Instead, for every hydronium that forms, exactly one hydroxide is formed. So the separate reactions do not exist, and the separate equilibrium constants do not either.
Of course, there are other reactions where two independent products exist, such as linear glucose forming the alpha or beta anomer of the pyranose ring. In this case, there are two equilibrium constants, but there is no combined equilibrium constant because the two reactions are not the steps of an overall reaction, but rather two distinct reaction paths.


Why does automotive glass shatter into tiny pieces but glass used in our houses into shards? Image Source : Bigstockphoto

Image source : pinterest

The first picture depicts the glass that is found from an accident site, while the second picture depicts the glass that we use in our home.

What is the difference between these two glasses? Why doesn't the first glass, which is used as a wind shield in vehicles, breaks like the glass that we use in our home?

Let's start out by covering the
Why?
Glasses used in automobiles are designed to break in tiny pieces.
Imagine you're driving down a dark highway hitting dangerously high speeds and then suddenly you lose control and crash.
If the glass used was not to shatter in tiny pieces rest assured you would not remain in one piece. Essentially it shatters for safety reasons.
Household glass on the is generally not involved in such drastic accidents so there is no need for it to shatter into tiny pieces .
How is it done?
From How Stuff Works

Tempered glass is created by heating and then rapidly cooling the glass to room temperature by ushering it through a system of blowers.
The surface of the glass cools much faster than the center of the glass and contracts, causing compressive stresses, while the center of the glass expands because of its temperature, producing tensile stresses. What does that mean? Imagine a piece of glass that could be pulled or stretched to a certain length (tensile stress), while being pushed down and compressed (compressive stress) simultaneously. Both the pulling and pushing stresses achieved through the heating and cooling process give tempered glass its tensile and compressive strength. The differences between these two give the glass 5 to 10 times the amount of strength it originally had.

This system holds a lot of energy due to the stress so when it breaks all of it is released causing the glass to shatter into tiny pieces
Tl;Dr
Automotive glass contains more energy than normal household glass. When it breaks all that energy is released which causes the glass to break into tiny pieces.