Do vinyl cations adopt a classical or non-classical structure? Whilst reading this question today, I remembered something that I had seen previously here. In the second linked question, @Martin provided a reference to suggest that vinyl cations actually adopt a non-classical structure with $\ce{sp}$ hybridization and a bridging hydrogen, as shown below:

The non-classical structure can be viewed as a protonated alkyne, in the same way that the ethyl cation, which adopts a non-classical structure, can be viewed as a protonated alkene.

What is the correct structure of a vinyl cation? Does it depend on what the substituents are? For example, in the first linked question, do the two cations (shown below) actually equilibrate to the same non-classical structure?

TL;DR: there have been many theoretical investigations of the relative energies of these two forms for the parent vinyl cation with more recent work indicating that the bridged form is slightly more stable (by about 1–3 kcal/mol)4,5. This prediction has received support from recent experimental work as well6.

Taking such relatively small energy differences into account I would not be surprised if the situation is reversed with a particular substituent. Thus, answering the second part of the question ("Does it depends on what the substituents are?") requires high-level calculations for a wide variety of different substituents which seems to be absent in literature yet.

Classical structure: trigonal (sp2) or linear (sp)?

First note that (give Marko a credit for paying our attention to that), as it is mentioned in Stang's "Vinyl Cations" book, the bent $\mathrm{sp^2}$-hybridized structure with a $\ce{C=C-H}$ bond angle of 120° (OP calls it "classical") is about 50 kcal/mol higher in energy than the linear $\mathrm{sp}$-hybridized form in all of the theoretical calculations.

  These calculations confirm the organic chemist's intuitive and
  qualitative feeling that carbenium ions prefer a trigonal planar
  geometry with an empty p orbital and that it takes energy to bend the
  methyl cation from its low-energy planar form.

Thus, in the studies below the meaning of the classical structure is different from the OP's one: the linear $\mathrm{sp}$-hybridized structure is called classical in almost all the studies.

So, Stang's "Vinyl Cations" references a CI/DZP study of the vinyl cation by Weber, Yoshimine, and McLean 1 as the most recent ab initio correlated calculations published at the time the book was written (the late 70s), so I'll start a review from this study.

Weber, Yoshimine, McLean, 1976

It is was found that the linear and bridged conformations have energies equal to within 0.01 kcal/mol with the later being lower in energy. However, such a small difference is below the accuracy of the CI/DZP level of theory, and thus, the authors didn't claim that the bridged conformation is more energetically favourable, just that it can be such ("probably" is the key word in the quote from the paper below).

  While our CI calculation shows the linear and bridged conformations to
  have energies equal to within 0.01 kcal/mol, we clearly cannot claim
  this absolute accuracy. In fact, taking into account basis set
  deficiencies, both in the one- and $n$-particle spaces, along with
  optimization effects as discussed above, we interpret our calculations
  to predict the relative energies of the two structures to be within
  1–2 kcal/mol with the bridged structure probably having the lower
  energy.

Besides, note that in the study the geometries have been optimized at the SCF level rather than at the correlated while CI calculations were done at the SCF optimized geometries. Taking into account that correlation effect were found to have major impact on relative stability of the two forms as mentioned in the study, it is especially interesting to look at the results obtained by Raghavachari, Whiteside, Pople, and Schleyer2 who optimized the structures not only at the HF/6-31G* but also at the MP2/6-31G* levels as well as did more high-level correlated calculations (MP4) with a bigger basis set (up to 6-311G** of TZP quality).

Raghavachari, Whiteside, Pople, Schleyer, 1981

For the bridged structure the $\ce{C-C}$ bond length was found to increase from 1.21 Å at HF/6-31G* to 1.23 Å at MP2/6-31G* while the distance of the bridging hydrogen to the middle of the $\ce{C-C}$ bond stays the same (1.18 Å). For the classical structure the authors report no significant changes in the geometry. With respect to the relative stability it was found that use of the MP2 geometries results in a slight additional stabilization to the bridged form: the MP4(SDQ)/6-31G**//MP2/6-31G* energy difference is 1.3 kcal/mol while the MP4(SDQ)/6-31G**//HF/6-31G* energy difference is only 0.6 kcal/mol (in both cases in favour of the bridged form).

  With the larger 6-311G** basis, the bridged structure is stabilized
  further, and our final result at MP4(SDQ)/6-311G**//MP2/6-31G*
  indicates 11 [bridged] to be more stable than 10 [classical] by 3.0 kcal/mol.

The authors also mention that these final results are quite in line with the CEPA study by Lischka and Kohler3 done using a comparable basis set: the Lischka and Kohler geometries were very similar to the HF/6-31G* geometries obtained in the study and the bridged form was found to be 4.0 kcal/mol lower in energy than the classical form as well.

A further study done by the Pople group is also is in agreement with the findings quoted above: the bridged form predicted to be more stable by 3.1 kcal/mol (including zero point energies)4. А study by Lee and Schaefer5, though, reports a somewhat lower difference.

Lee, Schaefer, 1986

In a recent CI/DZP study by Lee and Schaefer with the largest CI expansion considered being CISDT the energy difference was found to be 0.68 kcal/mol with the nonclassical structure again lying lower. The authors also considered the effect of the zero-point vibrational energy corrections and found that it further lowers the nonclassical structure by 0.29 kcal/mol.

  Our best strictly ab initio estimate of $\Delta E$(classical-non- classical) for $\ce{C2H3+}$ is obtained by adding the above 0.29 kcal correction to the 0.68 kcal difference from CISDT. In this manner, $\Delta E$ is predicted to be 0.97 kcal/mol. We estimate that a proper variational treatment of quadruple excitations might increase this $\Delta E$ value by perhaps 0.5 kcal/mol.

Weber, J.; Yoshimine, M.; McLean, A. D. A CI study of the classical and nonclassical structures of the vinyl cation and their optimum path for rearrangement. J. Chem. Phys. 1976, 64 (10), 4159–4164. DOI: 10.1063/1.431986.
Raghavachari, K.; Whiteside, R. A.; Pople, J. A.; Schleyer, P. V. R. Molecular orbital theory of the electronic structure of organic molecules. 40. structures and energies of C1-C3 carbocations including effects of electron correlation. J. Am. Chem. Soc. 1981, 103 (19), 5649–5657. DOI: 10.1021/ja00409a004.
Lischka, H.; Koehler, H. J. Structure and stability of the carbocations C2H3+ and C2H4X+, X = hydrogen, fluorine, chlorine, and methyl. Ab initio investigation including electron correlation and a comparison with MINDO/3 results. J. Am. Chem. Soc. 1978, 100 (17), 5297–5305. DOI: 10.1021/ja00485a010.
Curtiss, L. A.; Pople, J. A. Theoretical study of structures and energies of acetylene, ethylene, and vinyl radical and cation. J. Chem. Phys. 1988, 88 (12), 7405–7409. DOI: 10.1063/1.454303.
Lee, T. J.; Schaefer, H. F. The classical and nonclassical forms of protonated acetylene, C2H3+. structures, vibrational frequencies, and infrared intensities from explicitly correlated wave functions. J. Chem. Phys. 1986, 85 (6), 3437–3443. DOI: 10.1063/1.451828.
Crofton, M. W.; Jagod, M.; Rehfuss, B. D.; Oka, T. Infrared spectroscopy of carbo‐ions. v. classical vs nonclassical structure of protonated acetylene c 2 h + 3. J. Chem. Phys. 1989, 91 (9), 5139–5153. DOI: 10.1063/1.457612.


What are the products of the dissociation of sodium bicarbonate in water? What is the relative pH of the solution? I had a recent question on a test that asked what the products would be if sodium hydrogen carbonate were dissolved in water. I had a few candidate answers

$\displaystyle\ce{NaHCO3 -> Na+ + HCO3-}$, but that one doesn't involve water at all
$\displaystyle\ce{NaHCO3 + H2O -> Na+ + OH- + H2CO3}$, but doesn't $\ce{H2CO3}$ decompose into $\ce{H2O + CO2}$?
$\displaystyle\ce{NaHCO3 + H2O -> Na+ + OH- + H2O + CO2}$, but that has water on both sides of the equation.

Part two of the question asked whether the solution would be acidic, basic, or neutral. What is the proper reaction for this?

It's not as intuitive as it seems and your questions are all legitimate questions, but sometimes even good arguments can't be used as evidence in chemistry.

  $\ce{NaHCO3 -> Na+ + HCO3-}$, but that one doesn't involve water at all

Look at this reaction:
$$\ce{NaCl <=>Na+ +Cl-}$$

Even this reaction doesn't "involve" water in the schematics but is right, we assume that is a dissociation of a salt in water.

  $\ce{NaHCO3 + H2O -> Na+ + OH- + H2CO3}$, but doesn't $\ce{H2CO3}$ decompose into $\ce{H2O + CO2}$?

Yes, but in fact this is an equilibrium:
$$\ce{H2CO3 <=> H2O + CO2}$$

And then you have to take into account another important equilibrium many time ignored:
$$\ce{CO2 (aq) <=> CO2 (g)}$$
So the amount of $\ce{CO2}$ present in the solution depends on by the type of system and the pressure exerted. Like a Coca-Cola bottle: if the system is closed, the $\ce{CO2}$ remains dissolved otherwise the equilibrium shift to the right and the $\ce{CO2}$ "escapes" until the rate between the $\ce{CO2}$ molecule dissolved and the $\ce{CO2}$ molecule escaped forms the solution is the same.

  $\ce{NaHCO3 + H2O -> Na+ + OH- + H2O + CO2}$, but that has water on both sides of the equation.

Okay, it is not a net ionic equation but the stoichiometry is still right. Maybe water is used to indicate that we are talking about a reaction that takes place in water.

Which reaction is not right?

All the reactions here are right, I think maybe they put some formal inaccuracies to help you, but in fact they are all part of series of equilibria that normally occur and can be described as:
$$\begin{multline}
\ce{NaHCO3 + H2O <=>[A] Na+ + HCO3- + H2O <=>[B] Na+ + H2CO3 + OH-} \\
\ce{ <=>[C] Na+ + OH- + H2O + (CO2(aq)<=>[C_2]CO2 ^ (g))}\end{multline}$$

The question should be more specific asking for the ultimate products of the reaction because all these are products of the dissociation of sodium hydrogen carbonate and because are equilibria is probable that all the species are present in different amounts when the reaction reach the equilibrium. However, the ultimate products are described by the third answer. And this equation can give you a hint about the acidity of the solution. When $\ce{CO2}$ escape from the solution (equilibrium $\mathrm{C_2}$ goes to the right) the concentration of the manly acid species goes down but the hydroxides from the equilibrium $\mathrm{B}$ are still present so at the end the solution in an open system with an atmosphere with a low partial pressure of $\ce{CO2}$ will have a higher pH, will become mildly alkaline.


Parallel 1st Order Reaction Rates Consider some general parallel first order process, where $\ce{A -> B}$ and $\ce{A -> C}$ (rate constants $k_1$ and $k_2$ respectively)
The overall rate equation of this process is said to be $-\frac{\mathrm d[A]}{\mathrm dt} = k_1[\mathrm A] + k_2[\mathrm A]$
I get the impression that this is derived from the addition of each individual rate equation, i.e. the addition of $-\frac{\mathrm d[\mathrm A]}{\mathrm dt} = k_1[\mathrm A]$ and $-\frac{\mathrm d[\mathrm A]}{\mathrm dt} = k_2[\mathrm A]$ would result in the overall rate equation shown above.
However, if this is the case, then why isn’t the rate equation actually $-\frac{2\,\mathrm d[\mathrm A]}{\mathrm dt} = k_1[\mathrm A] + k_2[\mathrm A]$? I mean, you’re adding two $-\frac{\mathrm d[\mathrm A]}{\mathrm dt}$ terms together; and so considering basic algebra, surely this would result in $-\frac{2\,\mathrm d[\mathrm A]}{\mathrm dt}$?
Or do differentials just not work in this way, and I’m simply confusing myself?

There is no need adding the number $2$ in front of the definition of the rate. So the rate equation is :
$$-d[A]/dt = k_1 [A] + k_2[A] = (k_1 + k_2)[A]$$


How to find the temperature change of a solution? Right now I'm doing a calorimetry lab and for the pre-lab we have been asked to determine the temperature change (in Celsius) of ammonium chloride in water.

The question asks for the expected temperature change ($\Delta T$) of $\pu{8.5 g}$ of $\ce{NH4Cl}$ in $\pu{100 mL}$ (or $\mathrm{g}$) of water, with the molar enthalpy ($\Delta H_\mathrm{sol}$) of the solution being $\pu{0.277 kJ/g}$. The conversion of $\mathrm{kJ/g}$ is throwing me off and I cannot figure out how to solve for $\Delta T$ with the given info.

I know that there are $0.165048...$ moles of solution, which gives me everything I need to solve. We were given the equation

$$n \Delta H_\mathrm{sol} = mC \Delta T,$$

where $m$ - mass of water and $C$ - specific heat capacity of water. I am assuming it needs to be rearranged to

$$\Delta T = \frac{mC}{n\Delta H_\mathrm{sol}}$$

Any help is greatly appreciated and I can explain in further detail if necessary. Sorry for the Celsius, we apparently don't use Kelvin in our calculations.

The main problem here is a simple mistake in the algebra. You rearranged:

$\pu{n\timesΔH_{sol} = m\times C\timesΔT}$

to

$\pu{ΔT = \frac{m\times C}{n\timesΔH_{sol}}}$

rather than

$\pu{ΔT = \frac{n\timesΔH_{sol}}{m \times C}}$

Additionally, the molar enthalpy of solvation is given in the problem as units of kJ/g, not kJ/mol. Presumably, this is an error in the given problem. According to Parker, V . B ., Thermal Properties of Uni-Univalent Electrolytes, Natl .
Stand . Ref . Data Series — Natl . Bur. Stand .(U.S.), No .2, 1965, the molar enthalpy of solution for $\ce{NH4Cl}$ is $\pu{14.78 kJ/mol}$.

There was also a slight miscalculation in the moles of solute. Where you calculated $\pu{0.165 moles}$ of $\ce{NH4Cl}$, you should have gotten:

$\mathrm{8.5 g / 53.49\frac{g}{mol}=0.159~mol}$

These errors corrected, plugging in the values to solve for $\Delta \text{T}$ is trivial and gives:

$\pu{ΔT = \frac{\pu{0.159 mol}\times \pu{14.78 kJ mol-1}}{\pu{100g} \times \pu{4.186J~g-1~K-1}} = 5.6K}$


Is (1R,3aR,4S,6aS)‐1,4‐dibromo‐octahydropentalene chiral or achiral? Is this molecule on the left chiral or achiral?

I believe it is achiral because its mirror image (on the right) is essentially the same molecule. But I was told that it is chiral because it doesn't have an internal plane of symmetry. Could someone tell me which explanation is correct?

It indeed is achiral, because the red dot is the center of symmetry, hence it is superimposible with its mirror image.
Having an internal plane of symmetry is not the only criterion, the presence of a centre of symmetry also leads to the compound being achiral.


Does a polyethene shrinking signify negative or positive entropy? When I think of something shrinking, I would intuitively think it decreases in entropy as there is less movement. E.g. if a gas was to reduce in volume, there would be less movement of the particles and hence less entropy.

How about for as solid?

I have read if polyethene shrinks, it INCREASES in entropy.

Why is this the case?

When stretched, the polymer chains elongate and are held taut. This means there are more crosslinks and fewer opportunities for internal degrees of freedom (movement). A relaxed polymer is not held taut and therefore can freely move, having more degrees of freedom, and an associated increase in entropy.


Uncertainties Calculation Recently I did an experiment on isolating casein and now that I am processing the results, I feel my method of uncertainty calculation is not feasible. The casein was isolated 3 times and the results are as follows: 2.386, 2.628 and 2.164 grams. The scale had an uncertainty of 3 decimal places. Furthermore, the casein was isolated from 100 mL of milk, measured using a measuring cylinder with an uncertainty of 1 mL. The mean uncertainty was found using the following formula:

$$Δx_\mathrm{avg} = \frac{Δx}{\sqrt{N}} = \frac{R}{2\sqrt{N}}$$

Hence, my calculations were:

$$\mathrm{Range} = 2.628 - 2.164 = 0.464$$

Thus the uncertainty for the average would be:

$$\frac{0.464}{2\sqrt{3}} = 0.133945262 = 0.2$$

This would result in a final value for the average amount of casein in the isolated milk of $\pu{2.4 ± 0.2 g}$.

However, this error seems unreasonable considering that my measurements were to 3 decimal places. Furthermore, how would I add the uncertainty of the 100 mL of milk to this value? Would I simply add the uncertainties?

If we have 2 variables
with absolute uncertainties(+)
$A \pm \sigma_A$ and $B \pm \sigma_B$

(+)  expressed as standard deviations, but applicable to confidence intervals as well.

Then $C = A \pm B$ has uncertainty

$$\sigma_C = \sqrt{{\sigma_A}^2 + {\sigma_B}^2}$$

For $C=A/B$ or $C=A\cdot B$,

$$\sigma_C = C \cdot \sqrt{\left(\frac {\sigma_A}{A}\right)^2+ \left(\frac {\sigma_B}{B}\right)^2 }$$

Let $A$ is content of cassein in 100 ml of milk.
Let $B$ is the actual volume of used milk.
Let $C=A\cdot B$ is the mass of  cassein in the volume of milk used in analysis.

Then
$\sigma_C$
is the estimation of the error of the amount of cassein (known as the result error)
$\sigma_A$
is the estimation of the  error of the cassein mass in 100 ml of  milk (unknown)
$\sigma_B$
is the estimation of the  error of the milk volume  (1 ml )

The error introduced by scales can be neglected, or introduced as additive variable $D=0$ with absolute error estimation $\sigma_D=0.001~ \text{g}$.

Another, the biggest  factor is the used analytical method with its bias $E$ and error $\sigma_E$.

$$\begin{align}
C &= D + E + A\cdot B \\
\sigma_{AB} &= A\cdot B \cdot \sqrt{\left(\frac {\sigma_A}{A}\right)^2+ \left(\frac {\sigma_B}{B}\right)^2 }\\
\sigma_C &= \sqrt{{\sigma_D}^2 + {\sigma_E}^2 + {\sigma_{AB}}^2}\\
\sigma_{AB} &= 2.4 \cdot \sqrt{\left(\frac {\sigma_A}{2.4}\right)^2+ \left(\frac {1}{100}\right)^2 }\\
0.13 &= \sqrt{{0.001}^2 +{\sigma_E}^2 + {\sigma_{AB}}^2}\\
\end{align}$$


Writing a organic reaction in different form Can I write this reaction
$$\ce{A + B ->[\ce{X}] C + D ->[\ce{Y}] E + F}$$

as
$$\ce{A + B ->[\ce{(1) X}][\ce{(2) Y}] E + F}$$

Can I write any organic reaction in this way or are there any conditions/rules?

Textbooks sometimes mention this form and sometimes they do not. What I have seen is that they mention mostly oxidation or reduction reactions this way but other reactions like addition are always presented in long form only.
This answer describes a little bit but my question is a little different.

It is depend on the reaction. For example, the given two reactions you have sighted are related to each other. It is a example of alkene to alcohol conversion using oxymercuration followed by demercuration with $\ce{NaBH4}$. The overall reaction is electrophilic addition of water to the alkene following Markovnikov's rule (when the nucleophile is water). The first reaction gives an organomercury intermediate, which is not important here for authors to mention. Yet, it is necessary to have a second reaction to convert carbon-mercury bond of the intermediate to carbon-hydrogen bond to get the product of interest. However, it is important to know (and is a common knowledge) that this intermediate should be isolated before treating it with $\ce{NaBH4}$ for following reasons:

One of the byproduct of reaction (1) is acetic acid, which violently react with $\ce{NaBH4}$.
One of the reactant of reaction (1) is water, which also react with $\ce{NaBH4}$.
The other reactant of reaction (1) is $\ce{Hg(OAc)2}$, which also react with $\ce{NaBH4}$.

Nonetheless, any chemist following this reaction sequence knows what hidden behind the sequence. It is also important to know that most organic reactions are written without byproducts, which are not important to the next step. To my understanding, it is okay to write two reaction sequence (even three or more) in a single reaction using steps (1) and (2) and more, as long as consequent intermediates are not important. For example, one of your intermediates $\ce{C}$ and $\ce{D}$ has to be a byproduct. But, if $\ce{C}$ and $\ce{D}$ give $\ce{E}$ and $\ce{F}$, respectively, reacting with $\ce{Y}$, you should show that in your sequence for readers benefits. I you are interesting only on $\ce{E}$ and $\ce{F}$ (together or you have a better separating method for the mixture), you can write it however you want it. The ball is always at your corner, but more explanatory way is always better.

Notes:
Your first reaction doesnt make sense unless $\ce{Y}$ is a catalyst. If not, you should draw it as tri-component reaction:
$$\ce{A + B + X -> C + D}$$
If it is a catalyst, then you can write it as:
$$\ce{A + B ->[\text{cat. } \ce{X}] C + D}$$
An common example for this is:
$$\ce{CH3CO2H + CH3CH2OH ->[\text{cat. } \ce{H2SO4}] CH3COOCH2CH3 + H2O}$$


An equal amount of hydrogen gas is dissolved in a 1L vessel and a 2L vessel, both filled with water. Which vessel has higher entropy? This question came up in our chemistry class and we could not determine which would have the higher entropy. Clearly, Vessel 2 has a larger volume, but Vessel 1 has a higher concentration of $H_2$. How do we resolve this?

Denote by Vessel I the vessel with volume $V$, moles of water $N$, and moles of hydrogen $N'$. Denote by Vessel II the vessel with volume $2V$, moles of water $2N$, and moles of hydrogen $N'$.

Increasing the volume will increase the number of microstates available to the system, and hence also the entropy. Increasing the moles of water will increase the number of microstates available to the system, and hence also the entropy.

Both statements imply that Vessel II has the greater entropy.


Work done by the System In thermodynamics work is defined as $~-p_{ex}\Delta V~$ for an ideal gas. If the total work is a negative value, that is in case of expansion of gas, we say that the work is done by the system and is negative.

I don't understand that why we use the pressure exerted on the gas in expression of work that is the external pressure and say that the work is done by the system ? If work is done by the system we should use internal pressure and if we use that in case of expansion the work done by the system will be positive which is contradicting. Please provide an explanation

Well, think of a gas that is expanding against a piston shown below:

We want to find the total work which is given by: $$w = F\Delta d$$ So we need to know what is the force that is applied by the gas onto the piston. To do this, we must realise that, the force that is stopping the gas from expanding freely into the atmosphere is the external atmospheric pressure. It is this pressure which is pushing downwards onto the gas, preventing it from expanding. Therefore, the force that is required for the gas to expand must be equal to the total pressure pushing down it and has nothing to do with its internal pressure.

Consider this example. I want to push a box across my room which is covered in carpet. To move the box, obviously I need to apply a force to it. The force that I need to apply to must be equal or greater than the resisting force which is the friction between the box and carpet. This force that is required has nothing to do with the speed of the box.

So now that we know the force, we get that $$w = -p_{ex}A\Delta d$$ but the area times the displacement is actually equal to the volume, therefore we get: $$w =-p_{ex}\Delta V$$

So, as you can see, we have derived the equation for work done by a gas. The main thing that you need to recognise is that the internal pressure is irrelevant when considering the force required for the gas to expand. All you need to consider is the force that is preventing it from expanding which is the external pressure.


Cleaning a round-bottom flask, especially after distillation I was wondering, how can I clean my RBF after doing a distillation? Cleaning it with a simple pipe-cleaner or similar cleaner is a headache, due to the roundness of the flask.

Is there a more effective method?

@Permeakra suggests some good cleaning solutions. I have rearranged them (adding a few) in a suggested order from mildest to most evil. The specific thing you want to use will vary based on what you want to clean. If an initial rinse isn't good enough, you may want to let it soak for a few hours. Always take safety precautions (don't add acid washes to basic compounds, for example) and where the appropriate personal protection equipment (PPE).

Detergent + water + "elbow grease" - there is a lab grade soap out there called "microwash" that is pretty good, and I also tell my students that things are more soluble in elbow grease than water.
Acetone - Acetone will dissolve most organics, and it is less harmful to you and the environment than other organics.
Dilute acid (HCl or H2SO4) - good for some "baked"-on stuff or for inorganic compounds. Watch out for vigorous acid-base reactions.
Dilute nitric acid (HNO3) - since most nitrate salts are soluble, nitric acid does a better job with resistant inorganics. It is also an oxidant and will convert resistance organics to more soluble compounds. Also watch for acid-base reactions.
Other organic solvents - Use whatever will dissolve your crud, but use proper personal protective equipment (PPE), fume hoods, and waste disposal.
"Base bath" (saturated KOH in isopropanol) - This solution eats a fine layer off the glass, which also removes the crud stuck to it. Rinse first with water or neutralize any acid present to avoid vigorous reaction with the bath. Immerse the flask into the bath for a few hours to overnight and it should come out clean. Base bath will dissolve silicone grease. Leave it in too long and it will come out etched. Use heavy butyl gloves.
Concentrated HCl - more powerful than dilute acids, but more dangerous. Neutralize any bases present to avoid vigorous reaction. Use proper PPE.
KOH + KMnO4 + H2O - this solution is basic (eats a fine layer of glass) and oxidative (converts many things, like organics, to a more soluble form). Neutralize any acids present and wear appropriate PPE.
Concentrated H2SO4 - has the added bonus of reacting with some insoluble organics, like polyaromatics, to form soluble sulfonates. H2SO4 is more dangerous than HCl. Follow the same precautions, and be more careful. I would use use butyl gloves instead of nitrile or latex. It will cause serious chemical burns.
"No-Chromix" - a non-chromium alternative to chromerge (below), no-chromix is a proprietary cleaner that is added to sufluric acid. The main ingredient is ammonium persulfate (NH4)2S2O8. You can make a reasonable knock-off by dissolving 2 g of (NH4)2S2O8 in 100 mL of water and adding to 2.5 L of sulfuric acid. Feel free to scale down, as this stuff has no shelf-life to speak of.

  The following five items have dangerous reputations, and for good reasons. Use them as last resorts.

Concentrated HNO3 - strongly acidic and powerfully oxidizing. Dissolves many resistant things, including burned-on crud, but may ignite low molecular weight organics (like solvents). Rinse the flask first with water to remove bases, then with acetone to remove organics, and then with water to remove acetone before adding HNO3. Use butyl gloves.
Dilute HF (no more than 5%) - HF eats glass also, and it does so more quickly than base baths. However, the the hazards associated with HF are severe. HF burns are not like other acid burns. The HF continues to bore into you and the damage might not be evident until later. DO NOT USE UNLESS YOU HAVE THE BURN ANTIDOTE HANDY. The burn antidote
is calcium gluconate, which neutralizes the acid and more important sequesters the fluoride ions. It is available as a gel or cream. You should still go overboard on PPE.
Aqua Regia: 1 part Concentrated HNO3 to 3 parts concentrated HCl. This is more powerful than just concentrated HNO3, since HCl is a stronger acid. It also decomposes to form NO2 and Cl2, both more powerful oxidants than Concentrated HNO3 alone. Aqua regia is famous as the only substance that can dissolve gold. It will also dissolve all manner of inorganics and organics, including silicone and hydrocarbon grease. Treat aqua regia with the respect deserved of its two components and keep it in the fume hood. Do not seal the vessels since it gives off gas. Aqua regia has no shelf life, but it remains a strong acid and will need neutralized before disposal. Note that aqua regia will not dissolve metallic rhodium, iridium, osmium, and ruthenium (all less active than gold).
Piranha solution: H2SO4 + H2O2 - this strongly acidic strongly oxidizing solution will eat off a lot of metal, organic, and inorganic residues by oxidizing then to more soluble forms. As a strong acid oxidant, it shares the same incompatibility warnings that concentrated nitric acid has. Piranha solution is good for glassware that needs to remain calibrated (volumetrics) or uniform (NMR tubes) since it does not eat glass. Piranha solution does not have a long shelf life.
Chromerge H2SO4 + K2Cr2O7 - forms chromic acid, a powerful oxidant. This stuff is considered toxic waste in most jurisdictions, meaning it will be expensive to get rid of. You can store it, however, if you store it in teflon. Do not use on NMR tubes, since chromium is paramagnetic. Treat chromerge with a respect due to the sum of everything above it on this list.


Why is gaining hydrogen called reduction when gaining electrons is called reduction? Aren't they opposites Reduction is gain of hydrogen  (Source)
Reduction is gain of electrons (Source)

Is it because a hydrogen has an electron so gaining hydrogen is technically gaining electrons? But that doesn't seem right as oxygens also have electrons and gaining oxygens is oxidation.
What is confusing me is when we usually talk about hydrogens, its a hydrogen ion. And hydrogen ions and electrons are opposite. I realise this situation is not a hydrogen ion but a full hydrogen, but it's still doing my head in.

As a blanket statement the gain of a hydrogen atom cannot be considered a reduction. The IUPAC gold book defines it as follows:

  reduction

  The complete transfer of one or more electrons to a molecular entity (also called 'electronation'), and, more generally, the reverse of the processes described under oxidation (2) and (3).

  oxidation

  The complete, net removal of one or more electrons from a molecular entity (also called 'de-electronation').
  An increase in the oxidation number of any atom within any substrate.
  Gain of oxygen and/or loss of hydrogen of an organic substrate.

  All oxidations meet criteria 1 and 2, and many meet criterion 3, but this is not always easy to demonstrate. Alternatively, an oxidation can be described as a transformation of an organic substrate that can be rationally dissected into steps or primitive changes. The latter consist in removal of one or several electrons from the substrate followed or preceded by gain or loss of water and/or hydrons or hydroxide ions, or by nucleophilic substitution by water or its reverse and/or by an intramolecular molecular rearrangement. This formal definition allows the original idea of oxidation (combination with oxygen), together with its extension to removal of hydrogen, as well as processes closely akin to this type of transformation (and generally regarded in current usage of the term in organic chemistry to be oxidations and to be effected by 'oxidizing agents') to be descriptively related to definition 1. For example the oxidation of methane to chloromethane may be considered as follows:

As you can see, the reverse of (3) is your first statement with one significant addition: organic substrate.

To understand this one only has to look at the electronegativities. With an electronegativity of 2.20 (Pauling) for hydrogen, it is less electronegative than most other non-metals, or in general, elements that organic substrates are made out of.[1] If you follow the electronegativity scheme of assigning oxidation states, then adding a hydrogen atom (one proton and one electron) results in an decrease of the oxidation number of the element that the hydrogen atom was added to.[2]

On the other hand, adding oxygen would increase the oxidation number of the element (except fluorine) it was added to, since its electronegativity is the second highest, i.e. 3.44 (Pauling).[3]

For example, adding dihydrogen to ethene, the carbons are reduced, while hydrogen is oxidised.
$$\ce{\overset{\color{orange}{-2}}{C}_2\overset{+1}{H}_4
  + \overset{0}{H}_2
  -> \overset{\color{orange}{-3}}{C}_2\overset{+1}{H}_6}$$
Adding dioxygen to ethane to form ethane-1,2-diol, the carbons are oxidised, while oxygen is reduced.
$$\ce{\overset{\color{orange}{-3}}{C}_2\overset{+1}{H}_6
  + \overset{0}{O}_2
  -> \overset{\color{orange}{-1}}{C}_2\overset{+1}{H}_6\overset{-2}{O}_2}$$
You can go as far as looking at the addition of water to ethane to form ethanol. One carbon will be oxidised and one will be reduced.
$$\ce{\overset{\color{orange}{-2}}{C}_2\overset{+1}{H}_4
  + \overset{+1}{H}_2\overset{-2}{O}
  -> (\overset{+1}{H}\overset{-2}{O})H2\overset{\color{orange}{-1}}{C}-\overset{\color{orange}{-3}}{C}\overset{+1}{H}_3}$$

Just keep in mind that your first statement is only true when hydrogen is added to to more electronegative elements. The reversal is the case when adding it to a less electronegative element, like a metal.
A second point to keep in mind is that oxidation states are bookkeeping tools only. The world of bonding is not strictly ionic or covalent; it is often somewhere in between. Therefore electrons are almost never completely transferred and the actual charge distribution might be quite different.

References:

Hydrogen: electronegativity. WebElements [http://www.webelements.com/]. [Online]; Dated 30th Nov. 2016; https://www.webelements.com/hydrogen/electronegativity.html
Electronegativity Considerations in Assigning Oxidation States
Oxygen: electronegativity. WebElements [http://www.webelements.com/]. [Online]; Dated 30th Nov. 2016; https://www.webelements.com/oxygen/electronegativity.html


Alternatives to sodium bicarbonate I am searching for a molecule that is edible for human, reacts to liquid by generating $\ce{CO2}$ or a similar gas and that leaves behind a clear liquid, similar to sodium bicarbonate but without the milky whiteness.Any ideas?

Ammonium bicarbonate, "baker's ammonia", is sometimes used as a leavening agent for baking (e.g. cookies) as it decomposes to gaseous ammonia, water, and carbon dioxide which are driven off in the oven heat. You can get it in some food stores. Hope this is helpful.


Regioselectivity of Birch reduction (electron donating groups) Why does Birch reduction on benzene derivatives with an electron donating group happen at ortho and meta positions? When the substitute is electron withdrawing, ipso and para reduction take place because the electron withdrawing group stabilizes the anion in these positions. What's the explanation for electron donating group directivity?

The explanation lies in the distribution of electron density in the intermediate radical anions that appear during the Birch reduction.
I use the following reactions as guiding examples:

Electron-withdrawing groups stabilize electron density at the ipso and para positions through conjugation and so the negative charge will mainly be found in these positions; subsequent protonation occurs para.

Electron-donating groups destabilize a negative charge (when compared to simple benzene) through conjugation and this effect would be strongest at the ipso and para positions.

Thus the negative charge will mainly be found at the ortho and meta position.


Order of positive mesomeric effect of halides I learned that the order of $+M$ effect (mesomeric effect) for halogens is as follows: $\ce{F} > \ce{Cl} > \ce{Br} > \ce I$.

I wonder why this is so?

$\ce{2p-2p}$ orbitals overlaps better than $\ce{2p-3p}$, $\ce{2p-4p}$, $\ce{2p-5p}$. Although $\ce{F}$ is more electronegative, but it is also smaller in size. Both $\ce{F}$ and $\ce{C}$ (next to it) have $\ce{2p}$ orbitals to overlap.


Phenyl vs methyl cis/trans stability Compound I is obviously the Z isomer. I thought it was obviously the less stable one too, given that there's a phenyl and t-butyl on the same side. The answer says that since the phenyl is planar, it actually exhibits less steric hindrance than a methyl group.

Is this because the planar phenyl ring can basically rotate its bulkiness away from the t-butyl, where as the methyl, while also free to rotate, always ends up with a hydrogen in the t-butyl's face?

Thanks!

The molecule contains conjugated system of double bonds, being essentially a substituted styrene. The phenyl cannot therefore freely rotate and its steric requirements are difficult to estimate just by hand-waving arguments.

Correct answer needs some quantitative argument, either e.g. heats of formation or calculation.

From the second, the energies come out using MMFF94 (using Avogadro):

Z: 177.6 kJ/mol
E: 175.6 kJ/mol

Also, both the isomers rotated the styrene substantially from planarity. The difference in results is quite small and can easily change when using other method. According to the numbers I'd agree with your answer, Z is less stable.

I would not consider this a valid question in test, the answer is not straightforward.


Is the compound CH2(OH)(ONa) possible? I was just pondering about the reaction mechanism of Cannizzaro reaction.

$\ce{H2C^{+}-O- + OH- -> H2C(OH)O-}$

$\ce{HCHO + H2C(OH)O- -> H3CO- + HCOOH}$

$\ce{H3CO- + HCOOH -> HCOO- + H3COH}$

I wondered why in 2nd step instead of receiving $\ce{H-}$ from $\ce{H2C(OH)O-}$; $\ce{H2C^{+}-O-}$ wouldn't react with NaOH to form $\ce{H2C(OH)ONa}$.

I don't think this compound will be able to exist. But don't know why! Everything seems right about it.

Of course, you shouldn’t start from $\ce{H2C+-O-}$ unless you note that it is a mesomeric form of $\ce{H2C=O}$ — the more relevant mesmeric structure.

The mechanism as you wrote it and as it is typically written is given without any counterion, i.e. using just $\ce{OH-}$ and not $\ce{NaOH}$. You could instead write the mechanism using a counterion for $\ce{OH-}$ — $\ce{Na+}$, which would result in the first step being:

$$\ce{H2C=O + NaOH -> H2C(ONa)(OH)}\tag{1}$$

This is essentially the compound you are suggesting. Hence it exists, but it is a minor constituent of the overall solution. It’s not part of the second step, it is the first step. For the second step to happen, you need another molecule of $\ce{H2C=O}$, which has not yet reacted to the geminal diolate $\ce{H2C(ONa)(OH)}$.

See this scheme for the transition state of the Cannizzaro reaction and where $\ce{H2C(ONa)(OH)}$ comes into play:


Do enzyme-substrate reactions show reversibility(to a considerable amount)? I was studying about the catalysts of enzyme-substrate reactions. My biology book says that if the density of product gets high the enzyme-substrate reactions get slower. It quite sounded like Le Chatelier's Principle. But Le Chatelier's principle only works for reactions in which backward reaction happens to a considerable amount. But I am not sure if it is the normal case for enzyme-substrate reactions.

So, Does enzyme-substrate reactions show reversibility(to a considerable amount), or my reasoning is somewhere wrong?

There are many enzyme catalyzed reactions that are reversible. In such cases, buildup of product may slow and eventually stop the net reaction as you reach equilibrium conditions. Product binding to the active site or to a secondary allosteric binding site may also inhibit the catalytic reaction. This can occur even in the case of irreversible enzymatic reactions.


Apparent contradiction in the type of deviation of an ethanol-water mixture For a mixture of $\ce{EtOH}$ and $\ce{H2O}$: $\Delta H_{\text{mix}} < 0$ (exothermic) and $\Delta V_{\text{mix}} < 0$ (volume contraction). This means that the unlike interactions are stronger than the like interactions, because the mixture exhibits a lower enthalpy and volume than that would be expected for ideal mixing.

Also, $\ce{EtOH}$ and $\ce{H2O}$ form a minimum boiling azeotrope and the $T-\chi$ graph is below the Raoult's law $T-\chi$ plot. This means that the pair exhibit positive deviation from Raoult's law. This implies that the mixture is more volatile than the pure components. Thus, like interactions should be stronger than unlike interactions.

I am not able to resolve this contradiction. Does the mixture actually exhibit positive deviation? Which interactions are stronger - unlike or like?

As suggested in the comments, there is an entropic contribution to the excess free energy which should not be overlooked, as it is the excess Gibbs free energy that determines whether the deviation from Raoult's law is positive or negative. All of the statements made by the OP are correct, except that negative $\Delta H$ (exothermicity) and $\Delta V$ (volume contraction) do not necessarily imply a reduced volatility (negative deviation from Raoult's law), that is they are not sufficient conditions for a negative deviation, a negative excess Gibbs free energy instead being the sufficient condition. Again, as alluded to in the comments, a simple analysis based entirely on $\Delta H$ is not appropriate here, largely because of the hydrogen bonding (associative) nature of the liquids. In fact, the negative mixing volume and enthalpy might partly explain the negative excess entropy (and vice-verse: they are correlated). The excess entropy of mixing of water and ethanol is highly negative over the entire range of composition at room temperature, leading to a net positive excess Gibbs free energy.

  Does the mixture actually exhibit positive deviation? Which interactions are stronger - unlike or like?

Therefore the mixture does exhibit a positive deviation from Raoult's law, even as interaction enthalpies between unlike molecules are stronger.


How does toluene react at higher temperatures and why? Reaction of toluene at high temperatures gives us (o-m)toluene; whereas at normal conditions of electrophilic attack, it gives us (o-p) directing.

Why does this happen at high temp although as methyl group is activating and o-p directing?

Assuming you're talking about Friedel-Crafts alkylation of toluene, this can be explained by the reaction being reversible and thermodynamic control taking precedence at sufficiently high temperatures. The directing ability of substituents in electrophilic aromatic substitution reactions is a function of the reaction being under kinetic control at lower temperatures. According to Fleming's book on MO theory, a single electron-donating substituent on benzene conjugated to the ring's $\pi$ system disrupts the normal degeneracy of the $\psi_2$ and $\psi_3$ molecular orbitals of unsubstituted benzene. Comparing the intermediates of an EDG-substituted benzene undergoing electrophilic attack, the HOMO of the $\pi$ system, $\psi_3$, is higher in energy when attack occurs at the meta- position than when it occurs at the ortho- or para- positions. Using the concept of resonance, this can be illustrated in an abstract sense by considering the resonance structures of the intermediates resulting from attack at the various positions; an additional resonance structure exists for ortho- and para- attack if the substituent is conjugated to the $\pi$ electron system of the ring.

With alkyl groups, the electrons are donated by hyperconjugation, which is a weaker effect (and not apparent from resonance structures), but nonetheless has a similar impact on regioselectivity and the $\pi$ molecular orbitals. Additionally, the coefficients on the atoms of the HOMO of the $\pi$ system are lowest at the meta positions when an electron donating group is present (including alkyl groups), indicating electron density is higher at the ortho- and para- positions.

(I should clarify, of course, that resonance is not an actual physical process/phenomenon, merely an abstraction which provides a qualitative picture of the extent of electron delocalization and which often correlates well with the actual distribution of electron density in the molecular orbitals.)

Ultimately, at low temperatures, when the reaction is kinetically controlled, the favored pathways are the ones in which the transition states and intermediates are stabilized, leading to ortho-/para- substitution.

At higher temperatures, the preferred product is the thermodynamic one, and the advantage of meta- substitution is that there is less crowding and reduced steric hindrance in, e.g., a 1,3,5-trisubstituted benzene by comparison to a 1,2,4-trisubstituted benzene. Additionally, Friedel-Crafts is actually reversible at high temperatures. Even if the kinetic product forms initially, then it can decompose by dealkylation in the presence of Lewis-Acid catalysts and acid at high temperatures, at which point equilibrium can be established with the thermodynamic product.

According to March, experimental evidence suggests that methyl groups will mainly migrate intramolecularly in the isomerization reaction to give the thermodynamically preferred meta-substituted benzenes, while both intramolecular rearrangements and complete dealkylation with subsequent intermolecular reactions are possible with other types of alkyl groups.


Effect of temperature on equilibrium constant in terms of entropy change This question was asked in the JEE examination of 2017 :

[A]  option is incorrect since the system is losing heat and hence $$ \Delta S\lt0$$
[B]   Lets assume we have an endothermic reaction  $$\ce { A <=> B - Heat} $$
Now if we increase the temperature, the reaction will shift to the product side and hence the entropy of the surroundings will decrease more as compared to the initial conditions. Therefore the Unfavourable change in entropy(decrease in entropy) should increase.
[C]  Though the reaction is endothermic, since we are supplying heat to the system, the net change in entropy of the system is positive.
[D]  Lets assume another exothermic reaction $$ \ce{A<=>B + Heat}$$.
When we increase the temperature, the reaction will shift to the left side and hence less heat will be given off to the surroundings. Therefore, favourable change in entropy(increase in entropy) decreases.
Hence, according to me, the correct answer should only be [D], but [B] option is also given correct. What is the error in my explanation?

The key point is that the entropy of the surroundings changes, not that of the system (at least for a small change in $T$), so only [B] and [D] are relevant, as you correctly determined.
You can write
$$\log(K)=-\frac{\Delta H}{RT}+\frac{\Delta S}R$$
This of course is one form of van 't Hoff's equation. It tells you, assuming $\Delta H$ and $\Delta S$ are constant, that the temperature dependence is due to $\Delta H-$ the heat transferred to/from the surroundings (entropy increase/loss of the surroundings):
$$\log(K)\propto-\frac{\Delta H}{RT}=\frac{\Delta S_\text{surr}}R$$
Now increasing $T$ makes the change in the entropy of the surroundings smaller.
When $\Delta H\gt0$ (endothermic), ie $S_\text{surr}\lt0$, this makes the reaction more likely to happen. When $\Delta H\lt0$ (exothermic), ie $S_\text{surr}\gt0$, this makes the reaction less likely to happen.
So both [B] and [D] are correct.


How are solvents chosen in organic reactions? Just reading through organic chemistry book (Modern Projects in Organic Chemistry by Mohrig, et al.), and it seems like the choice of solvent is usually arbitrary.

One of the experiments designed uses hypochlorite ion as the oxidizer and acetone as a solvent to oxidize 4-tert-butylcyclohexanol, but it isn't clear why this solvent is chosen.

I'm curious if there is a list that gives the suggested solvents for different kinds of reactions, as well as some insight for this particular experiment.

There is no list per se, but there is certainly a process to determine appropriate solvents for a given reaction. For the specific reaction you mention, here are some criteria you would consider to select a solvent:

Because this is an oxidation reaction, choose a solvent that cannot be oxidized, known as an aprotic solvent.
The solvent needs to dissolve the reagents in the reaction: hypochlorite and 4-tert-butylcyclohexanol. Hypochlorite is a polar molecule, and 4-tert-butylcyclohexanol is a little polar thanks to the alcohol group. The alcohol group will be oxidized to a ketone by the reaction, so the solvent should be able to dissolve that reasonably well. Conclusion: choose a moderately polar solvent.

So a polar aprotic solvent is needed for this reaction. There are two more considerations: cost and safety

Obviously, the cheapest solvent with the lowest potential for harm should be selected.

A list of some common polar aprotic solvents for consideration might include:

tetrahydrofuran
N,N-dimethylformamide
acetonitrile
dimethyl sulfoxide
acetone

Tetrahydrofuran could be a reasonable choice, but it forms peroxides over time, which are quite explosive.

N,N-dimethylformamide has some health concerns: GHS07 and GHS08

Acetonitrile also falls into the GHS07 category: it causes cyanide to form in the body at high concentrations.

Dimethyl sulfoxide does not have health or safety concerns.

Acetone also falls into the GHS07 category: it can cause nausea and irritation at high exposure levels.

The health concerns of acetonitrile, dimethyl sulfoxide, and acetone are largely not going to pose a hazard for a laboratory experiment. All three of these can likely be found in any organic teaching lab.

Cost comparison, from Sigma Aldrich:

Acetonitrile: 2 L = $172.50

Dimethyl sulfoxide: 2 L = $208.50

Acetone: 2 L = $94.80

Therefore, choose acetone for this reaction!


What will be the order of acidity in the following compounds? What will be the order of acidity in the following compounds ? From My point of view it should be 3>1>2>4. But the answer given is 3>2>1>4. Can anyone explain how ?

I think you are correct. The $\mathrm{p}K_\mathrm{a}$ of 4-methylphenylacetic acid is quoted as 4.37 here. The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is quoted as 4.19 by the same source here.


"4K-metal chelated" A chemical term used in Sagan's Cosmos Maybe this question doesn't belong here. It's about a hypothetical xenobiology imagined by a prominent scientist in a seminal work of popular science. So I dunno. Tell me to go away and I'll see if some other SE can help me.

I'll never forget the races Sagan made up in his book Cosmos. He gave descriptions of them in the form of excerpts from his imagined Encyclopedia Galactica, a compendium of all the races in the galaxy.

There was one term that never made sense to me: "4K-metal chelated". What does the "4K" part mean? Here is the context:

Civilization Type: 2.3 R
Society Code: 1H1, "We Who Became One".
Interstellar civilization, no planetary communities,
  utilizes 1504 supergiants, OV, BV, AV stars and
  pulsars.
Civilization Age: 6.09 * 10^15 s.
First local initiated contact: 6.09 * 10^15 s ago.
Receipt first galactic nested code: 6.09 * 10^15 s ago.
Source civilization, neutrino channel.
Local Group polylogue.
Biology: C, H, O, Be, Fe, Ge, He. 4K metal-chelated
  organic semiconductors, types various. Cryogenic
  superconducting electrovores with neutron crystal
  dense packing and modular starminers; polytaxic.

(By the way, if anyone knows what "Local Group polylogue" is supposed to mean, I care about that too.)

These guys are "cryogenic superconducting electrovores" who apparently have "no planetary communities" so my guess is that "4K" should probably be taken to mean "4 Kelvin" and "metal-chelated organic semiconductors" as that they are composed of various types of semiconductors whose molecules are organics chelated with a metallic ion. About the "Local Group polylogue" thing, it's probably fictional stuff like a sort of interstellar Commonwealth or something.


How does fluorouracil inhibit thymidylate synthase? I'm curious to know how just by adding a fluorine atom to uracil (to form 5-fluorouracil) permanently inhibits the enzyme thymidylate synthase when they bind together. This molecule is widely used as an antimetabolite in chemotherapy as it stops the synthesis of thymine.

What changes undergoes the enzyme-inhibitor compound so it is permanently bonded to the fluorouracil? Probably the chemistry is too complicated to be explained in this Q&A format, if that is the case, references like books or papers would be great.

The inhibition comes about because:

Uracil and $5$-fluorouracil are sufficiently similar that both moieties have a strong binding affinity for the active site of thymidylate synthase.
The presence of the fluorine prevents $5$-methylation of the uracil core by a folate cofactor, which otherwise would turn it into a thymine moiety.
This inability of the enzyme to convert the $5$-fluorouracil moiety into a thymine moiety (which has a much lower affinity for the active site) means that the enzyme has no way to reject the $5$-fluorouracil or the methylating folate species, once bound.

The particulars of this interaction are nicely spelled out in Mishanina et al.$^\dagger$, conveniently available in preprint:

  [5-fluorodeoxyuridylate] reacts [as normal] with the active site cysteine and undergoes condensation with [N,N-methylene-tetrahydrofolate] but prevents proton abstraction and elimination of the tetrahydrofolate [product], as a consequence stalling the reaction at the covalent ternary complex [of enzyme, inhibitor, and folate species].

Scheme 4A from the preprint linked above illustrates well the specifics of the chemistry.  (I refrain from reproducing it here due to uncertainty regarding its copyright/license status.)

As well, Figure 2 from the Mechanism Description section of the Wikipedia page on thymidylate synthase shows very nicely the enzyme interactions with the normal deoxyuridylate substrate (click image for larger version):

$\hspace{20pt}$
Image by CJerc (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0 )], via Wikimedia Commons

$^\dagger$ Mishanina et al. Bioorganic Chemistry 43: 37 (2012). doi:10.1016/j.bioorg.2011.11.005


Role of tautomerism and its difference from hyperconjugation I would like to know the role of tautomerism, if any, in increasing stability of an organic compound. Moreover, how is it different from that of hyperconjugation?

I know that hyperconjugation involves shifting of protons between adjacent
atoms in different hypothetical structures to form a practical hybrid structure.

I am not clear with tautomerism, so kindly deliver a precise explanation.

Tautomerism basicially means that you have different isomers that are able to convert into each-other. This is not really similar to hyperconjugation which is an electronic effect in a given molecular geometry. The main difference between them is that hyperconjugation is thought to be a property of a given molecular geometry while tautomerization is an equlibrium between separate, distinct, stable configurations.

There are many types of tautomerizations, as explained for example on the Wikipedia. Probably the most well-known type of tautomerisation is the keto-enol tautomerisation, which is an equilibrium between a ketone molecule and an enol (an alcohol connected to an $sp^2$ carbon). The equilibrium between two isomers is the explanation why enols are only stable under special scenarios: they tautomerise into ketons! Another important example for tautomerism is the equlibrium between the chain and the ring isomers for sugars.

So again: tautomerism is an equilibrium between different (real, existing) isomers as opposed to hyperconjugation which is just defined within a single structure.


What is the reaction between plain soda and sugar? Today I was having lunch and I ordered plain soda. And some amount of sugar was added, suddenly all the liquid fell of the glass. I thought there should have been a reaction between plain soda and sugar. What is it?

Soda is carbonated water - it has carbon dioxide dissolved under high pressure. Carbon dioxide is a gas that would like to escape. Shaking the bottle of soda, for example, provides a suitable mechanism to allow release of the gas.

Adding dry sugar also provides a mechanism for this to occur through nucleation. The sugar crystals give the carbon dioxide a surface via which to escape the solution. This is not a chemical reaction per se, but a physical reaction. It is essentially the same process as the coke-mentos reaction. Sugar crystals are great for this, as they have a rough surface and therefor a very high surface area available.


For the half reaction HSO3- -> S2O6^2- why is it that 2e- are on the right hand side when balancing and not one electron? $$\ce{HSO3- -> S2O6^2-}$$

The oxidation number of $\ce{S}$ on the left is +4 and +5 so when adding electrons to balance the oxidation number why is $\ce{2e-}$ added to the right instead of one?
Why is 2 placed in front of the $\ce{HSO3-}$?

Why is $2$ placed in front of the $\ce{HSO3}$?

This is, of course, the elemental law of the conservation of mass. You have $x$ atoms that you put into your equation, you need to get $x$ atoms out again. While hydrogen and oxygen can be taken care of with water, you need to account for the sulphur immediately. Hence you need two sulphurs.

  The oxidation number of S on the left is +IV and +V so when adding electrons to balance the oxidation number why is $\ce{2e-}$ added to the right instead of one?

The oxidation numbers are, of course, correct, but remember that in the first step you wrote a $2$ before the sulphur. Therefore you now have two sulphurs that you need to treat with the appropriate number of electrons — either by stealing or by gifting.

Since you have two sulphurs and they each lose one electron, you need to add $\ce{2 e-}$ to the right.

Note: The OP already understood what went wrong through hints in the comments. Unfortunately, OP decided not to write up a self-answer.


Epoxidation with peroxyacids In this particular epoxidation/hydroxylation reaction, why does the epoxide only form at one of the double bonds and not both?

The compound initially reacts with the peroxyacid to form an epoxide, follow by an acid catalyzed epoxide opening resulting in syn-stereochemistry. My initial guess was that the epoxide would only form by the double bond that is not sterically hindered. I don't understand why the epoxide only forms at the double bond flanked by two methyl groups and not both.

This source states that MCPBA gives preferential epoxidation of the more substituted alkene. This source explains that due to intra-molecular hydrogen bonding in solution, the high degree of polarisation results in an electrophilic oxygen atom that adds to alkenes. The most substituted alkene will be the most electron-rich alkene and hence the most nucleophilic. This is supported by the data shown here and here 4 on the relative rate of reaction of unsubstituted/mono/di/tri/tetra subsituted alkenes.


SN2 reactions and inversion of configuration in compounds with more than one chiral centre Suppose I have a compound with 2 chiral centres which are at (S,S). Now, suppose I have an iodine atom attached to the back carbon, and I wish to carry out an $\mathrm{S_N2}$ type substitution in a non-polar solvent. Obviously, the iodine atom would get replaced by the nucleophile, and there should be an inversion of configuration. But is this inversion only for the back carbon or the compound as a whole? Should the product have a configuration (R,S) or (R,R)? Someone please explain this to me briefly.

Here I have a compound with 2 chiral centers. There is only inversion at Carbon with iodine undergoing $\mathrm{S_N2}$ reaction. The second carbon does not undergo $\mathrm{S_N}$ reaction. There is no change in its configuration.

Example 1

In the example below ,

[Ref : https://en.chem-station.com/reactions-2/2016/05/neighboring-group-participation.html]

before a nucleophile attacks externally, OH of carboxylic group attacks internally via $\mathrm{S_N2}$. The 3 membred ring opens via another $\mathrm{S_N2}$ attack by hydroxide ion. The net result is retention of configuration .

Example 2

In this example , two chilral carbons are at neighboring carbons. As explained earlier , Sulphur attacks internally via $\mathrm{S_N2}$ attack. Again bromide ion attacks via another $\mathrm{S_N2}$ leading to retention of configuration ( Anchimeric assistance ).

[Ref : http://www.chem.ucla.edu/~harding/IGOC/A/anchimeric_assistance.htmll]

Example 3

This example also behaves similarly to one discissed above.

[Ref: Molecules 11(4):212-8 • April 2006   DOI: 10.3390/11040212 ]

In conclusion,

If there exists no lone pair donating group at neighboring carbon to leaving group, then $\mathrm{S_N2}$ attack takes place with inversion of configuration (Example 1).
If there exists a lone pair donating group (Example 2 and Example 3)at neighboring carbon to leaving group, then retention of configuration is seen.


Crystal Lattice Defects Why do defects in crystal lattices of various molecules and compound occur ?
Is it due to the continuous contact or it arises due to impurities and human error during the formation?

First of all, there are no such thing as a "perfect" crystal.

Defects occur from both human error and the experimental setup.

Defects from human error are primarily due to handling issues, which can never be perfected, especially not if the crystals are solvated.

Experimental defects can occur from either the diffraction experiment or the crystal growth experiment. Impurities can affect crystal growth by displacing the molecule of interest in the lattice and unless you are attempting neutron diffraction, the crystal will be significantly damaged during the data collection.

In my field of protein x-ray crystallography, the typical crystals can have diffraction ranging from nothing at all to ~0.8 Angstrom range (best cases) and as such you can have beautiful crystals that have no diffraction and ugly crystals that diffract very well.

Crystal defects can be perfectly acceptable and are a so common that we usually don't think about them too much.


Why does the energy gap for π - π* transitions shrink with the size of the pi-conjugated system? Quoting from this site:

  As conjugated pi systems become larger, the  energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer.   The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol.

It's also quite easy to see that this is true glancing at various dye absorption tables.

But why is this true?  I can understand why the optical cross-section would increase, implying a higher quantum efficiency at some maximum absorbance wavelength (is this generally true?), but why would the energy gap between the ground state and the excited state correspondingly shrink?

Klaus Warzecha's answer pretty much answers your question. But I know that this subject is easier to understand if supported by some pictures. That's why I will take the same route as Klaus at explaining the concept behind why the absorption in conjugated systems is shifted to higher wavelengths but I will provide some pictures on the way.

In a conjugated carbon chain or ring system you can think of the $\ce{C}$ atoms as $\text{sp}^{2}$-hybridized. So, each carbon has 3 $\text{sp}^{2}$ orbitals which it uses to form $\sigma$ bonds and 1 $\text{p}$ orbital which is used to form $\pi$ bonds.
It is the $\text{p}$ orbitals that are responsible for the conjugation and their combinations according to the LCAO model are the interesting part since the HOMO and LUMO of the system will be among the molecular orbitals formed from the conjugated $\text{p}$ orbitals.

For a start take ethene, the simplest $\pi$-system, being comprised of only 2 carbon atoms.
When you combine two atomic orbitals you get two molecular orbitals.
These result from combining the $\text{p}$ orbitals either in-phase or out-of-phase.
The in-phase combination is lower in energy than the original $\text{p}$ orbitals and the out-of-phase combination is higher in energy than the original $\text{p}$ orbitals.
The in-phase combination accounts for the bonding molecular orbital ($\pi$), whilst the out-of-phase combination accounts for the antibonding molecular orbital ($\pi^{*}$).

Now, what happens when you lengthen the conjugated system by combining two ethene fragments?
You get to butadiene.
Butadiene has two $\pi$ bonds and so four electrons in the $\pi$ system.
Which molecular orbitals are these electrons in?
Since each molecular orbital can hold two electrons, only the two molecular orbitals
lowest in energy are filled.
Let's have a closer look at these orbitals.
In $\Psi_1$, the lowest-energy bonding orbital, the electrons are spread out over all four carbon atoms (above and below the plane) in one continuous orbital.
There is bonding between all the atoms.
The other two electrons are in $\Psi_2$.
This orbital has bonding interactions between carbon atoms 1 and 2, and also between 3 and 4 but an antibonding interaction between carbons 2 and 3.
Overall, in both the occupied $\pi$ orbitals there are electrons between carbons 1 and 2 and between 3 and 4, but the antibonding interaction between carbons 2 and 3 in $\Psi_2$ partially cancels out the bonding interaction in $\Psi_1$.
This explains why all the bonds in butadiene are not the same and why the middle bond is more like a single bond while the end bonds are double bonds.
If we look closely at the coefficients on each atom in orbitals $\Psi_1$ and $\Psi_2$, it can be seen that the bonding interaction between the central carbon atoms in $\Psi_1$ is greater than the antibonding one in $\Psi_2$.
Thus butadiene does have some double bond character between carbons 2 and 3, which explains why there is the slight barrier to rotation about this bond.

You can construct the molecular orbitals of butadiene by combining the molecular orbitals of the two ethene fragments in-phase and out-of-phase.

This method of construction also shows why the HOMO-LUMO gap of butadiene is smaller than that of ethene.
The molecular orbital $\Psi_2$, which is the HOMO of butadiene, is the out-of-phase combination of two ethene $\pi$ orbitals, which are the HOMO of ethene.
Thus, the HOMO of butadiene is higher in energy than the HOMO of ethene.
Furthermore, the molecular orbital $\Psi_3$, which is the LUMO of butadiene, is the in-phase combination of two ethene $\pi^{*}$ orbitals, which are the LUMO of ethene.
Thus, the LUMO of butadiene is lower in energy than the LUMO of ethene.
It follows that the HOMO-LUMO energy gap is smaller in butadiene than in ethene and thus butadiene absorbs light with longer wavelenghts than ethene.

If you continue to lengthen the $\pi$ system by adding more ethene fragments you will see that the HOMO and LUMO are getting closer and closer together the longer the $\pi$ system becomes.


How can I calculate the equilibrium constant for this reaction? Silver(II) oxide partially decomposes into solid silver and oxygen gas according to:
$$\ce{2 AgO(s) <=> 4 Ag(s) + O2(g)}$$
with enthalpy $H^\circ = \pu{62.0 kJ}$ and entropy $S^\circ = \pu{133.6 J K-1}.$
a) Calculate the change in standard Gibbs free energy. (Answer: $\pu{22.2 kJ}$)
b) Calculate the equilibrium constant for this reaction.

This is one of the exam questions I got, but I don't understand how I'm supposed to answer question b without the concentration. I also didn't get a mass or volume so I can't calculate it using the molar mass.
I tried using the formula $G^\circ=-RT \ln K,$ but I don't get the right answer with this formula. Could someone help me find the right way and formula to answer this question?
$$\frac{ΔG^\circ}{-RT} = \frac{-RT}{-RT}\ln K$$
$$\ln K = \frac{ΔG^\circ}{-RT}$$
$$
\begin{align}
K &= e^{-ΔG^\circ/(RT)} \\
  &= e^{-22.2/(8.3145\cdot 298)}\\
  &= \pu{8.18545e-9}
\end{align}
$$
The answer I'm supposed to get is $\pu{1.30e-4}.$ The answer I get according to my calculations is $\pu{8.1854e-9}.$

You might want to check your units.

The energy should be in J/mol, according to the values you put in, and not in kJ/mol.

$e^{-22200/(8.314*298)} = 0.000128397 \approx 1.3*10^{-4}$


Can a solvent be a solid? Can a solvent be a solid? I do not think so. But is it possible for a solvent to be a solid?

When most people think of the term solvent, a liquid medium comes to mind; however, in the technical sense of the definition, this does not have to be the case. Here is the definition of a solution present in the IUPAC gold book:

  A liquid or solid phase containing more than one substance, when for convenience one (or more) substance, which is called the solvent, is treated differently from the other substances, which are called solutes. When, as is often but not necessarily the case, the sum of the mole fractions of solutes is small compared with unity, the solution is called a dilute solution. A superscript attached to the ∞ symbol for a property of a solution denotes the property in the limit of infinite dilution.

The definition of a solvent is incorporated into the definition of a solution. This definition states that the substance present in the larger quantity is the solvent while that present in the smaller quantity is the solute. Based on this definition, it is entirely possible to have a solvent that is in the solid phase. For example, brass is a mixture of 64% copper and 34% zinc. According to the IUPAC definition, a solution can be in the solid phase. Because copper is present in a larger amount, it is considered the solvent -- a perfect, everyday example of a solvent in the solid phase.


Why do you use catalysts if they increase the rate of forwards and backwards reactions equally? So I wanted to know why catalysts didn't affect concentration; since if they increased rate of reaction, presumably the concentration of the products would increase.

I did read online, however, that catalysts increase the rate of the forwards and backwards reactions equally; and then the concentration not changing makes sense, but there's another question now.

I thought the whole point of a catalyst was to increase the rate of reaction so you can get more of something in less time. If it increases only the rate of reaction and not concentration, the net output of the products would be the same as if you didn't use a catalyst — and in that case, what's the point of even using a catalyst?

In a simplified representation, a catalyst lowers the overall activation energy while going from initial state A to the of the products B along the reaction coordinate. The interference of the catalyst on the reaction alters the reaction mechanism, renders alternative pathways accessible, which then may be  more favourable than for the reaction in absence of the catalyst. Catalysts are deployed regardless if the overall reaction is exothermic, or endothermic. And, because of the relationship described in the Eyring equation, changes in the free activation enthalpy influence rate constants.

For the instance of a heterogeneous catalysis, the picture is simplified as assuming energy chances for adsorption of the starting material(s) to and subsequent desorption of the product(s) from the catalyst were non-significant in the overall picture.

You are right that it may ease both the forward, as well as the backward reaction.  However, in absence of catalyst, overcoming the activation energy at all were just too expensive, even if the reaction $\ce{A -> B}$ drawn here happens to be an exothermic one.

As examples, think about the hydrogenation of benzene; in absence of Nickel there were practically no reaction at all, even at elevated temperatures.  Numerous biochemical reactions were unable to proceed in absence of a catalyst at ambient temperature, too.


What effect does diluting a buffer have on its buffer capacity? What effect does diluting a buffer have on its buffer capacity? I am not worried about the effect on the pH, just the buffer capacity.

The buffer capacity β is a direct function of the concentrations of all protolytes in the solution. Thus, if the concentration of the protolytes decreases you should then expect a decrease in the buffer capacity.


Why is the methyl radical planar? Why is the methyl radical planar? The VSEPR theory would predict an angle between 120° and 109.5°, while it is actually 120°.

This question unfortunately illustrates the significant shortcomings of the VSEPR model. This model is based upon classical electrostatic arguments. The molecular geometry will seek to maximize the distances between the repulsive electrons in the valence (outer) shell, that are completely localized in their chemical bonds and their lone pairs/singly occupied molecular orbitals.
This model works well in some cases. However, when quantum mechanics comes into play, it falls pretty flat, as it totally neglects (the really important) quantum effects. A better way of finding the molecular geometry is to find the geometry that minimizes the total energy of the molecular orbitals. In this case, quantum mechanical calculations find that a set of three $\mathrm{sp^2}$-hybridized molecular orbitals arising from a $D_\mathrm{3h}$ symmetry has a lower total energy than breaking the symmetry to form the pyramidal $C_\mathrm{3v}$ geometry.
However, this does not mean that all $\ce{\overset{.}{C}X3}$ radicals follow this pattern. Put an extremely electronegative substituent on in place of $\ce{H},$ e.g. in $\ce{CF3},$ and the situation changes. The lowest energy geometry is now the pyramidal one.
Sadly, quantum chemistry doesn't always care to follow general rules.


Why would this rearrangement happen? So, I am a first year chemistry student and I saw this chemistry reaction on a previous year's test. I attempted to figure out the mechanism, but I'm rather confused as to how the cyclopentane somehow donates a carbon to the cyclobutane? Why would such a rearrangement occur?

That is how far I get to before I am lost. Is methanol even a strong enough reagent for such a reaction?

Is methanol even a strong enough reagent for such a reaction?

Methanol is a very poor nucleophile, so the rate of the $\mathrm{S_N2}$ pathway will be extremely slow. This is much likelier to be a $\mathrm{S_N1}$ reaction. The starting material is a benzylic bromide from the point of view of the alkyl halide, both pathways are quite possible, but the lack of a good nucleophile suggests the predominance of the $\mathrm{S_N1}$ route.

So, you go through a carbocation intermediate, which undergoes an alkyl shift to generate a more stable carbocation.

Both carbocations are benzylic and secondary, so hyperconjugation isn't the driving force for the rearrangement, as your comment hints at. Instead, it's simply related to the relief of ring strain as you expand a four-membered ring to a five-membered ring. Four-membered rings are pretty unstable to begin with and fusing one to a benzene ring increases the strain even further since you impose more geometrical constraints on the ring.

Interception by methanol and deprotonation gives the desired product.


Number of atoms in NaCl unit cell Consider this picture of an $\ce{NaCl}$ unit cell:

It appears to show 14 $\ce{Cl-}$ ions and only 13 $\ce{Na+}$ ions.  Given that discrepancy, how is table salt balanced for charge?  Why isn't there a charge excess?

The picture you showed does have an unequal number of sodium cations and chloride anions.  However, the picture shows only part of a crystal.  Every atom that is on a boundary of the shown cube, whether on a face, edge, or vertex of the cube, is shared with other "cubes" in the crystal that aren't shown in the picture.

Each of the 8 corner Cl atoms in your picture is shared with 8 cubes (7 not shown). The 6 face-centered Cl atoms are shared with 2 cubes. Each of the 12 edge Na atoms is shared with 4 cubes (3 not shown). The center sodium atom is not shared. Thus there are 8/8 + 6/2 = 4 Cl atoms per unit "cube" in your picture, and 12/4 + 1/1 = 4 Na atoms per unit "cube" in your picture. 4=4, so the charge balances.

You might be thinking that this math checks out only insofar as a crystal is actually infinitely sized.  And you may have noticed that no salt crystals are infinitely big in the real world.  These things are both true.  But even tiny specks of salt crystals are gigantic relative to atoms.  The surface of a salt crystal could involve imperfections that mean that the count of sodium atoms and chloride atoms aren't precisely equal.  But instead of 14 vs. 13, the difference is more like 100,000,000,000,000,000 vs 99,999,999,999,999,999.  And since the imperfections are at the surface, on the outside of the crystal, any charge imbalance can be corrected if an opposite-charged particle from outside the crystals floats by and neutralizes the extra charge from the extra atom.


What are the half reactions for the redox reaction between SO2 and iodine? What are the half reactions for the redox reaction between $\ce{SO2}$ and iodine? The relevant chemical equation:
  $$\ce{ SO2 + H2O + I2 -> H2SO4 + 2HI}$$

This is what I think it is, not sure if it's correct.

Oxidation: $\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-}$

Reduction: $\ce{I2 + 2e- -> 2I-}$

Your initial comment exchange made me realize that you have some difficulties in balancing redox equations. Thus, I decided to include some clues for your benefit.

An important part of writing half-reactions is making sure they're balanced by mass and charge. Since most redox reactions are done in aqueous medium, you can always balance $\ce{O}$ by $\ce{H2O}$. By doing so, you contribute extra $\ce{H}$ to the equation so you can balance them by $\ce{H+}$ since it is mostly an acid medium reactions (if they are in base or neutral conditions, you may cancel $\ce{H+}$ by adding the same amount of $\ce{OH-}$ to both sides of the reaction). Finally, cancel the net plus charge by adding $\ce{e-}$s to the appropriate side.

Let's see the easy one first, in your case reduction half reaction where $\ce{I2}$ reduce to $\ce{I-}$: $\ce{I2 -> I-}$. Balance its mass, which gives you: $\ce{I2 -> 2I-}$. Now, balance the negative charges by $\ce{e-}$s. So, you got balanced reduction half-reaction ($\ce{e-}$s are in LHS):
$$\ce{I2 + 2e- -> 2I-} \qquad \mathrm{E^\circ = \pu{0.536 V}} \qquad \text{(1)}$$

Now, see the more difficult second equation, the oxidation half reaction where $\ce{SO2}$ oxidizes to $\ce{SO4^2-}$: $\ce{SO2 -> SO4^2-}$. Its $\ce{S}$ is already balanced, but $\ce{O}$ is not. So, balance it with $\ce{H2O}$, which gives you: $\ce{SO2 + 2H2O -> SO4^2-}$. Now, balance additional $\ce{H}$ by $\ce{H+}$, which gives you a mass-balanced equation: $\ce{SO2 + 2H2O -> SO4^2- + 4H+}$. Now, balance the negative charges by $\ce{e-}$s. So, you got balanced oxidation half-reaction ($\ce{e-}$s are in RHS):
$$\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-} \qquad \mathrm{E^\circ = \pu{0.157 V}} \qquad \text{(2)}$$
If you add (1) and (2) together in order to cancel $\ce{e-}$s, you get the redox reaction you are looking for:
$$\ce{SO2 + 2H2O + I2 -> SO4^2- + 4H+ + 2I-} \qquad \mathrm{E^\circ_{cell} = \pu{0.693 V}} \qquad \text{(3)}$$

The value of $\mathrm{E^\circ_{cell}}$ is positive means the reaction is spontaneous.

(Note: The value of $\mathrm{E^\circ_{\ce{SO2/SO4^2-}}}$ is from Ref.1)

Reference:

J. A. O’Brien, J. T. Hinkley, S. W. Donne, “Electrochemical Oxidation of Aqueous Sulfur Dioxide II. Comparative Studies on Platinum and Gold Electrodes,” J. Electrochem. Soc. 2012, 159(9), F585–F593 (DOI: 10.1149/2.060209jes).


Difference between the terms “analyte” and “analyate” and their applicability “Analyte” is the term that is usually universally understood and is listed in  IUPAC Gold Book:

  analyte The component of a system to be analysed.

However, recently I stumbled upon the word “analyate”.
Initially I thought of it being a typo or an erroneous translation (I'm not a native English speaker), but upon inspection of the literature sources it turned out be used in cited publications in the context similar to the term “analyte” (emphasis mine):

McMaster's LC/MS: a practical user's guide (323 citations) [1, pp. 2, 16]:

  The mass analyzer is quickly overwhelmed by the signal from the solvent if the HPLC output is introduced directly into the mass spectrometer. The analyate signal is buried beneath this solvent signal avalanche. […] The HPLC solvent gradient used to resolve closely eluting HPLC peaks and decrease HPLC run times also produces solvent composition changes that further complicate the solvent-masking effect of analyate signal.

  […]

  The purpose of an interface is twofold. It must remove as much solvent as possible without losing the analyate, and it must volatilize and ionize its components for submission to the analyzer.

At the same time in the textbook the term “analyte” is utilized on p. 61:

  The ionized analyte ions formed in the interface between the HPLC and
  the mass spectrometer cannot survive collision with air molecules in the analyzer.

… and on p. 151:

  Gas chromatography. A separation technique in which the volatile analyte is swept by a carrier gas down a column packed with packing coated with an absorbing liquid.

Zhao and Swager (272 citations) [2, p. 9379]:

  It is informative to investigate the fluorescence quenching activities of P1-3 in solution. Conducting these studies in chloroform, a good solvent, allows us to evaluate the responses of individual polymer chains and the relative
  analyate affinities to the polymers under these conditions.

Patent US20050137471A1 (182 citations) [3]:

  The Signals are processed in a processing unit to calculate the concentration
  of the analyate.

DeLouise et al. (146 citations) [4, p. 3224]:

  It also simplifies quantitation of the optical response relative to competitive techniques based on evanescent waves where the signal response decays exponentially from the surface causing sensitivity to depend on the surface-analyate distance.

Hart et al. (130 citations) [5, p. 461]:

  It was found that as structural similarity between the imprint molecule and the analyate molecule diminishes, both retention and selectivity decrease.

  […]

  Making a relatively small perturbation to the analyate by removing the aromatic hydroxyl group results in a lower, but still significant, separation factor of $α = 2.5.$

… and many more other less cited papers, over 250 in total over 1887–2019 time span according to Google Scholar (I used Harzing's PoP software for query).
So, is there a difference between the terms “analyte” and “analyate”?
If they are not fully interchangeable, then what exactly “analyate” is, or in what context its usage is more appropriate?

References:

McMaster, M. C. LC/MS: A Practical User’s Guide; John Wiley: Hoboken, N.J, 2005. ISBN 978-0-471-65531-2.
Zhao, D.; Swager, T. M. Sensory Responses in Solution vs Solid State:  A Fluorescence Quenching Study of Poly(Iptycenebutadiynylene)s. Macromolecules 2005, 38 (22), 9377–9384. https://doi.org/10.1021/ma051584y.
Haar, H.-P.; List, H.; Baader, F.; Harttig, H.; Meacham, G. Continuous Glucose Monitoring Device. US20050137471A1, June 23, 2005. (PDF)
DeLouise, L. A.; Kou, P. M.; Miller, B. L. Cross-Correlation of Optical Microcavity Biosensor Response with Immobilized Enzyme Activity. Insights into Biosensor Sensitivity. Anal. Chem. 2005, 77 (10), 3222–3230. https://doi.org/10.1021/ac048144+.
Hart, B. R.; Rush, D. J.; Shea, K. J. Discrimination between Enantiomers of Structurally Related Molecules:  Separation of Benzodiazepines by Molecularly Imprinted Polymers. J. Am. Chem. Soc. 2000, 122 (3), 460–465. https://doi.org/10.1021/ja9926313.

The unabridged Oxford English Dictionary, available by subscription, is the ultimate source to confirm the usages of English words. I checked the word "analyate" and it does not exist in the entire dictionary. The word analyte is also very recent (mid 1950s). I am not surprised that wrong spellings are published in high-end journals. If you study the statistics of wrong references, that is even more shocking. How many times it has happened that you check a citation and it turns out that the citation never mentioned anything like that.

Here is an excerpt from the OED

  Origin: Apparently formed within English, by compounding. Etymons: analyse v., -lyte comb. form.
  Etymology: Apparently < ana- (in analyse v.) + -lyte comb. form, perhaps after electrolyte n.
  Chemistry.
  Categories »

  A substance that is analysed or determined, esp. quantitatively.

  1955   C. M. Crawford in Chem. & Engin. News 5 Dec. 5262/2   There is
  not, as far as I am aware, any short word denoting ‘the substance
  being analysed for’ in use... My proposal is that the word ‘analyte’
  be understood to convey the above meaning.


Is there any software that can do geometry optimization using machine learning? DFT is a computational tool that is used in optimizing and calculating the electronic structure properties of molecules.

Are there any machine learning codes that can do something similar in a shorter time? Is there any software that does this?

There is one Python package that I know: AMP (Atomistic Machine-learning Package). It is based on ASE (Atomic Simulation Environment), a homogeneous interface to a lot of computational chemistry packages.

Basically, after you obtain geometry trajectories through ASE (using any computational chemistry package you like), an approximation to the potential energy surface (PES) is machine-learned. It can be used as an ASE interface.

You can theoretically do anything ASE is capable of with this, from saddle point optimization to quantum dynamics.
I am not an expert in this field, but they seem to apply ideas from convolutional neural networks to molecular structures for describing the chemical environment surrounding atoms.
Their method is described in Khorshidi, A.; Peterson, A. A. Amp: A Modular Approach to Machine Learning in Atomistic Simulations. Computer Physics Communications 2016, 207, 310–324.


How to calculate the Gibbs energy for the vaporisation of solid ammonia? Given that $P_{\ce{NH3}}=\mathrm{10~atm}$ at $298\ \mathrm K$, calculate the $\Delta G^\circ$ of the reaction $$\ce{NH3(s) <=> NH3(g)}$$

I'll begin solving the problem...

I'm using the formula $$\Delta G=\Delta G^\circ+RT\ln Q.$$ The vaporisation of $\ce{NH3}$ is an equilibrium so  $\Delta G=0$ and $Q=K_c$ . Also $K_c=\left[\ce{NH3}\right]$. So $\Delta G^\circ=-RT\ln K_c$.

How do I continue? Is it $K_c=K_p=10$ or $K_c=K_p/\left(RT\right)$?

Actually for this reaction $\Delta G=\Delta G^\circ+RT\ln Q$ holds only when $Q$ is used in terms of partial pressures. This is because the activity of an ideal gas is measured by its pressure, not its concentration. This equation can be derived from the definition of the Gibbs energy.


Can I eat the lemons after I have used them as a battery? After completing the lemon battery experiment, can I eat the lemons when I'm done?

More broadly, what will happen to the lemons? What are the chemical reactions, what chemicals might be present/missing after running my lemons as a zinc/copper battery? Would it be poisonous, or just unpleasant?

In a lemon battery, the zinc from the galvanized nail is giving up electrons and transitioning into an aqueous state:

$$\ce{Zn → Zn^2+ + 2e−}$$

While zinc is entering the electrolyte, two positively charged hydrogen ions ($\ce{H+}$) from the citric acid with two electrons at the copper electrode's surface and form an uncharged hydrogen molecule $\ce{H2}$:

$$\ce{2H+ + 2e− → H2}$$

While zinc is a mineral our bodies need, I wouldn't recommend using this as a source of zinc. Also, zinc toxicity is a thing, so without knowing the dose you're taking, you risk poisoning yourself.

Lemon Battery Wikipedia


Shielding effect- specifically defined? In all the sites I have searched, shielding effect has been defined as the reduction of the nuclear attraction upon the outer electron/electrons, due to the core electrons that repulse the outer electron.

So, is the shielding effect specifically defined for outer shell electrons? Can't the shielding effect be defined for inner shell electrons?

If it can be defined, do the electrons in outer shells than that shell have any impact on the shielding effect?(They too must repulse the electron in the inner shell, possibly helping the nuclear attraction)
If not, why?

Can someone help me out? Tks

Generally, inner electrons shield outer electrons. In order to shield you have to be closer to the nucleus, but it's harder for an electron in a higher energy orbital to be closer to the nucleus than an electron in a lower energy orbital. Therefore the shielding effect from inner on outer is a lot more pronounced than outer on inner.

EDIT (after better understanding of OP's question):
One thing to remember is that the inner electron can be on the same side of the nucleus as the outer electron, or it can be on the other side of the nucleus as the outer electron. In one case, the inner electron is being pushed towards the nucleus but the other case results in the inner electron being pushed away from the nucleus. A hand-wavy application of Gauss' Theorem suggests that these effects might cancel out over time. So, the net effect is that the outer electrons have minimal impact on the inner electrons. This argument fails if the outer electrons have greater nuclear penetration, but that doesn't happen by definition.


Origin of the term "millamolecule" I came across a term "millamolecule" to represent macrocycles with a molecular weight between 500 and 1,000 daltons. Google Scholar show several resources which use this term e.g., Google Scholar Results

What could be origin of this term or is it just a buzzword invented by pharma companies?

In their 2013 Annual Report titled Evolving to a Specialty Care Biopharma Company, a well-known pharmaceutical company, Bristol-Myers Squibb has introduced the following statement under the title, Developing Innovative Drug Platforms:

  As R&D evolves its focus, it is also investing in technology platforms that concentrate on new ways to affect disease targets, including antibody drug conjugates, which combine the targeted benefits of biologics with the cancer-killing ability of traditional small-molecule chemotherapies. R&D also will further expand the potential use of millamolecules, which are larger than small molecules but smaller than biologics. These millamolecules may be able to better exploit novel targets and mechanisms by retaining the desirable properties of small molecules with the high degree of selectivity, especially against antigens, that biologics and small molecules have had difficulty targeting.

And then, Bristol-Myers Squibb Company has filled a patent under World Intellectual Property Organization (WIPO) claiming Pet-imaging immunomodulators, abstract of which states that:

  The invention relates to the synthesis and use of $\ce{^{18}F}$-labeled millamolecules for imaging various processes within the body, for detecting the location of molecules associated with disease pathology, and for monitoring disease progression are disclosed.

Thus, to my understanding, millamolecules is the word invented by Bristol-Myers Squibb Company's scientists to describe special set of molecules with different activity other than those exhibited by small molecules and higher polypeptides.

Finally, Medicinal & Pharmaceutical Chemistry glossary & taxonomy describes it as:

  Millamolecules: Mid-range compounds that fall in size between small molecules and biologics. Size is not the only consideration for this class, as millamolecules should also be orally available and able to interrupt protein-protein interactions.

Unsurprisingly, their reference has also directed to Bristol Myers Squibb: Areas of Focus.


Can a cyclic amine form an amide? This is a question from the Cambridge International Examinations October/November 2017 (pdf from papers.gceguide.com, pdf via the Wayback Machine).

I need to understand why the secondary amine in the serotonin molecule does not undergo condensation reaction with $\ce{CH3COCl}$.
Because, as far as I know, there should be an amide formation for this reaction.

Here's a picture of serotonin:

I must be lacking some information making that $\ce{NH}$ unsuitable for this reaction. And I know that the other amine and the phenol reacts with $\ce{CH3COCl}$.

The heterocycle in this question is indole and is aromatic. This means that the N lone pair is delocalised and not readily available for nucleophilic attack. Think of it as similar in reactivity to a secondary amide nitrogen RCONHR. Generally you need to formally deprotonate to functionalise, though there are some interesting techniques using carbonyl azoles catalysed by DBU[1] and others using aldehyde and alcohol substrates.[2] Note that 3-unsubstituted indoles react with acyl halides by F-C acylation at the 3 position.[3]

References:

Heller, S. T.; Schultz, E. E.; Sarpong, R. Chemoselective N-Acylation of Indoles and Oxazolidinones with Carbonylazoles. Angew. Chem. Int. Ed. 2012, 51 (33), 8304–8308 DOI: 10.1002/anie.201203976.
Maki, B. E.; Scheidt, K. A. Single-Flask Synthesis ofN-Acylated Indoles by Catalytic Dehydrogenative Coupling with Primary Alcohols. Org. Lett. 2009, 11 (7), 1651–1654 DOI: 10.1021/ol900306v. PMID: 19320508 (with free text available).
Okauchi, T.; Itonaga, M.; Minami, T.; Owa, T.; Kitoh, K.; Yoshino, H. A General Method for Acylation of Indoles at the 3-Position with Acyl Chlorides in the Presence of Dialkylaluminum Chloride. Org. Lett. 2000, 2 (10), 1485–1487 DOI: 10.1021/ol005841p.


Can sp2 hybridised carbon show chirality? I know that any carbon to show chirality has to be bonded to 4 different atoms (groups of atoms). So a double bond is
considered as that the carbon is achiral. I learnt from my lecturer that dissubstituted allene is an exception. If it is true that sp2 hybridised carbon cannot show chirality, are there any other exceptions?

As correctly stated in a comment by @Ivan Neretin, chirality is not bound to stereogenic centers (e.g. the textbook examples of carbon atoms with four different substitutents, or chiral sulfoxides) only.

Returning to your question, an other option to build chirality is of course axial chirality, where the allenes -- formally built from $sp^2$-hybridized carbon atoms -- offer enantiomeric forms.

(source)


How are ionic formulas written? Why is $\ce{Ba(OH)2}$ Barium hydroxide not Barium dihydroxide?
Why is Lithium carbonate $\ce{Li2CO3}$ instead of $\ce{LiCO3}$?

Is  there a formula or is just the case for these two?

There is no other Ba(OH)x so there is no need to note a difference between several forms.

This happens with any metal ion that only attains precisely one positive charge, whether it is +1 or +10.

Lithium ions have +1 charge. Carbonate ions have -2 charge. Therefore to balance the charge in the formula we need 2 lithium to 1 carbonate.

This happens with all formulae, and it is a simple mathematical check to decide what the correct ratios should be.


Does hydrogen bonding contribute in solubility of a substance If a compound can't hydrogen bond with itself but can hydrogen bond with the water, why does it still remain immiscible?

Take for example ethoxyethane. It can't hydrogen bond with itself but its oxygen atoms can participate in hydrogen bonding with the slightly charged H atoms of the water.

So why does ethoxyethane still remain immiscible with water?

Ethoxyethane, better known as "diethyl ether" or even just "ether", can form hydrogen bonds with water.  But fitting the ether molecules into the water solvent means you have to break up some water-water hydrogen bonds, and the ether molecules cannot replace enough of those hydrogen bonds to dissolve freely.

Of course, ether does form some hydrogen bonds, which makes is more soluble than it would be without the hydrogen bonds (compare ether with, for example, diethyl sulfide).  But most (or at least the most common) freely water-soluble organic compounds have either ions to generate ion-dipole interactions with the water, or hydroxyl groups that readily fit with the hydrogen bonding structure of the water (think of ethanol or sugars).


Non-Ideal Gas Behavior I obtained measurements of temperature and pressure at various temperatures ($-10, 0, 20, \text{ and } 50^\circ \mathrm{C}$) and plotted them on a graph. Projecting backwards, the best-fit line predicts 0 pressure at significantly above $-273.15^\circ \mathrm{C}$(0 Kelvin). I am wondering why this is.

My understanding is that a true linear relation assumes "Ideal Gas Behavior", which is predicated on the false assumptions that molecules have no attraction for each other and that they take up no space. I know that at high/low temperatures and pressures the realities about volume and attraction begin to make a difference. But I don't understand how these account for my results.

The pressure for all measurements fell between $680$ and $820$ torr. My understanding is that that is not a significant amount of pressure. In terms of predicting pressures at lower temperatures, I understand that the less kinetic energy of the gas the more relevant the attractive forces are...but I would think that would lower the pressure which makes things worse--if the pressure actually falls faster and faster as temperature drops, then the predicted zero-pressure point would be even higher than the predicted values I got. I need it to be lower!

I am under the (possibly incorrect) impression that both the realities of volume and attraction are supposed to be causing my incorrect prediction...but I don't understand how either does...

Thanks for any help!

There is little reality to such an extrapolation. Attaching much significance will just lead you astray. There are a number of real gas equations which purport to correct for this or that "problem" with the ideal gas equation. The greater truth is that the equations just add extra fitting coefficients which make the equation more accurate. Different equations will work for different gases in different ranges.