How do the electronic properties of a catalyst affect the rate of oxidative addition? In my project I have a coupling that was planned to be done using Pd (Suzuki or Stille possibly).

Initially, I used an aryl chloride (the SM had already the chloride so I went with it), but the oxidative addition was slow. Changing to iodide helped but still slow (can't use triflate as would be destroyed in an earlier step if carry through).

In my lectures, changing the halide is always given as a way of gaining selectivity, this is also what is taught in Clayden if you look for the information on organometallics. But a postdoc suggested changing the catalyst, as apparently changing the electronics in palladium also helps. Is this the case? Or is OA just controlled by the strength of the C–X bond?

To quote Hartwig:

Oxidative addition to more electron-rich metal centres tends to be more favourable than oxidative addition at metal poor centres [1]

This is generally a thermodynamic issue, and can be rationalised quite easily if one considers that the change in electron density at the metal centre during the oxidative addition process (note, theres a difference between formal oxidation state and the 'actual' electronics about the metal centre).
As you point out, most catalysts systems used are either Pd0 to begin with, or generate Pd0 via a Pd2 pre catalyst (the palladium needs to be in its 0 oxidation state for oxidative addition to occur, usually, though 2 -> 4 is also possible in certain circumstances).
Your confusion seems to have came from the actual nature of the active catalyst, although you'll always end up with Pd0 , the metal centre has ligands attached, and these ligands are able to tune various properties at the metal centre.
The issue with trying to optimise a reaction ( as you seem to be doing) is that there is an almost infinite number of combinations of catalyst + ligand that you could consider, and indeed chemical companies such as Sigma-Aldrich even sell screening kits which provide many of the most common.
My honest advice in your situation would be to have a thorough search of the literature and try a few combinations of ligand + catalyst that you have available in your lab. Doing the optimisation rationally can be extremely interesting (doing it properly requires changing catalysts, ligands, solvents, temperatures, additives etc and is all time consuming, though often done in industry during optimisation before scale-up), but is of little use synthetically if you simply need to drag material through.

[1]: Hartwig, J. Organotransition Metal Chemistry- From Bonding to Catalysis; University Science Books:Herndon , 2010


Possible nonclassical ion from a bicyclic system In this question a bicycling alcohol is reacted with acid to make what appears to be a tertiary carbocation, and the OP asked whether it could become aromatic.  The given answer suggests it could be, but only if a nonclassical ion is formed.  So, does that happen?

The substrate is 8a-methyl-1,3,4,8a-tetrahydronaphthalen-4a(2H)-ol.

While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does not occur but rather the reaction proceeds through the spiro carbocation 5. continued

Applying this approach to the carbocation 7 generated from the alcohol in this question, spiro carbocation 8 forms the predicted tetrahydronaphthalene 10. One way to distinguish between the two mechanisms is via a labeling experiment. All of the carbon label in 7 will retain its location as the red star in 10 if the 1,2-methyl shift mechanism applies. The spiro mechanism will partition the label ~50:50 between the two ring benzylic carbons. For related studies, see reference 2.
1) R. B. Woodward and T. Singh, J. Am. Chem. Soc., 1950, 72, 494.
2) A. J. Waring, J. H. Zaidi and J. W. Pilkington, J. Chem. Soc, Perkin Transactions I, 1981, 1454.


Determining order of reaction when concentration of reactant is not given Suppose reactant $\ce{A}$ decomposes in an aqueous solution to give $\ce{P}, \ce{Q}$:
  $$
\ce{A(aq) -> P(aq) + Q(aq)}
$$

  Suppose further that $[\ce{A}]_0=0.42\text{M}$, and that $[\ce{Q}]$ was measured as follows (in M):

  \begin{align}
t&=0 &[\ce{Q}]&=0 \\
t&=100 \quad &[\ce{Q}]&=0.101\\
t&=200 \quad &[\ce{Q}]&=0.165\\
t&=300 \quad &[\ce{Q}]&=0.206\\
t&=400 \quad &[\ce{Q}]&=0.239\\
t&=500 \quad &[\ce{Q}]&=0.263
\end{align}

  Using the graph of $\ln [\ce{A}]$ and $\frac 1 {[\ce{A}]}$ vs $t$, determine if the reaction is of order $1$ or $2$.

I know how to solve these kind of problems when I'm given the concentration of $[\ce{A}]$, I just plot the graph and check which one resembles a line more, and the slope gives me the velocity constant $k$.

But here $[\ce{A}]$ is not given. I supposed it would be $[\ce{A}](t)=[\ce{A}]_0-[\ce{Q}](t)$, but I was told that was wrong. Could someone help me out?

If the stoichiochemistry is as you present it:

$$\ce{A -> P + Q}$$

Then if at $t=0$, $[\ce{A}]=[\ce{A}]_0$ and $[\ce{Q}]=0$, then because one equivalent of $\ce{Q}$ is formed for each equivalent of $\ce{A}$ consumed, your method of determining $[\ce{A}]_t$ must be correct.

$$[\ce{A}]_t = [\ce{A}]_0 - [\ce{Q}]_t$$

Perhaps the error your instructor noted came in applying your relationship or interpreting your graph.


Why is there a difference between O2 and B2 sigma 2p molecular orbitals in diagrams? The molecular orbital diagram for $\ce{O2}$ says that the  sigma 2p bonding molecular orbital is lower in energy than the pi 2p bonding molecular orbital. Why is this not the case in the $\ce{B2}$ MO diagram?

This topic is really fun. You see in case of boron molecule $\sigma_{2s}$ and $\sigma_{2p_z}$ (being on the intermolecular axis) will have the same symmetry and hence the orbitals will overlap and result in the formation of 2 bonded $sp_z$ orbitals. So the $\sigma_{2p_z}$ orbital will obtain somewhat $s$ character and it will become unstable or in other words will be of higher energy. The $\sigma_{2s}$ MO will obtain somewhat $p$ character. So in a energy level MO diagram $\sigma_{2p}$ bonding orbital is placed above the $\pi_{ 2p}$ orbital.

However, in case of $\ce{O2}$, due to the higher nuclear charge of oxygen, the energy difference of $2s$ and $2p$ is large so there is no such kind of intermixing between the bonded MOs. Whereas in boron, the energy difference of $2s$ and $2p$ is less, facilitating intermixing.

You can find this topic in any higher level chem books. Hope you got my point; thank you.


What exactly does κ stand for in this equation connecting solubility and molar conductivity? I came across the following equation while studying electrochemistry ("Physical Chemistry" by Wallwork and Grant):

  The solubility $s$ of a sparingly soluble salt can be determined from conductivity measurements, provided that the molar conductivities of the ions at the temperature of these measurements are known to enable $\Lambda_0$ to be calculated. Since the saturated solution is dilute, $\Lambda_0 \approx \Lambda = \kappa/c$, where $c = s~(\pu{mol m-3})$. The conductivity $\kappa$ is the conductivity of a saturated solution of the salt in conductance water minus that of the water alone. The solubility $s$ is calculated from:
  $$s = \kappa/\Lambda_0 \tag{8.36}$$

It's mentioned that κ is the conductivity of the saturated solution of the salt in conductance water, minus that of the water alone. Now the thing is, from what I've read, conductance water has NO (or negligible) conductivity.

In that case, what's the point of having to subtract the conductivity of conductance water? Or is there something I've misunderstood here?

Nomenclature changes and authors don't always follow the rules.  The more typical use now is $\Lambda_m$ for the molar conductivity and $\Lambda_m^0$ for the limiting molar conductivity.

The molar conductivity, $\Lambda_m$, is given by the equation

$\Lambda_m = \dfrac{\kappa}{c}\tag{1}$

where:

$\kappa$ is the specific conductance (which is what you measure...)
$c$  is the concentration (which you know because of how the solution was made)

Now typically the specific conductance, $\kappa$, is plotted against concentration. For dilute solutions, there will be a linear relationship where the slope of the line is the limiting molar conductivity, $\Lambda_m^0$.  However at high concentrations the relationship becomes non-linear and the actual conductance is less than what the linear relationship would predict. That is because of the formation of cation and anion clusters. The local effect is thus to make the cluster uncharged (or charged less is the charges don't balance to zero).

Now the author of your book specified a sparingly soluble salt, so at low concentrations $\Lambda_m \approx \Lambda_m^0$, and we can substitute for $\Lambda_m$ to get:

$\Lambda_m^0 = \dfrac{\kappa}{c}\tag{2}$

which can be rearranged to give:

$c = \dfrac{\kappa}{\Lambda_m^0}\tag{3}$

If the conductivity of the solution is much greater than that of pure water then this is fine. The measured specific conductance will be directly proportional to the concentration since $1/\Lambda_m^0$ is a constant.

However the author is talking about a salt of limited solubility, so the conductivity of pure water is not negligible. Going back to the plot of specific conductance, $\kappa$, against concentration, $c$, we can determine that at 0 concentration there is some conductance which I'll call $\kappa_w$ which is the conductance of pure water. So for a salt with limited solubility the equation is:

$c = \dfrac{\kappa - \kappa_w}{\Lambda_m^0}\tag{4}$

If you look back at equation (2) you can see why equation (3) doesn't work for solutions of very low conductivity. When $c=0$ equation (2) blows up since you can't divide by $0$.

So the author used $\kappa$ in two different ways which is very poor writing. We could define $\kappa'= \kappa - \kappa_w\tag{5}$ then $c = \dfrac{\kappa'}{\Lambda_m^0}\tag{6}$


how to determine the volatility of an organic compound using the physical properties of organic molecules How do you tell if a compound has a high volatility?

Also can the boiling point of a compound be taken into consideration when determining if a compound has a high volatility?

Vapor pressure is the physical property that expresses a compound's volatility at a given temperature, and the enthalpy of vaporization describes the temperature dependence of the vapor pressure.

For some compounds, like water, there are tables and formulae readily available for determining the vapor pressure at different temperatures. If you know the compound's vapor pressure at a given temperature and you also know the enthalpy of vaporization for the compound, you can then estimate the vapor pressure of the compound at another temperature.  This method assumes that the enthalpy of vaporization of the compound is constant with temperature, which is only a good approximation over a relatively small temperature range. You would need to know the heat capacity of the compound to correct for the temperature dependence of the enthalpy of vaporization.

Comparison of boiling points is often used to say "compound x is more volatile than compound y" because compound x has the lower boiling point. This statement is only true for the pressure at which the boiling point was given, often 1 atmosphere.  The vapor pressure of a polar compound like a low-molecular weight alcohol has a greater temperature dependence (enthalpy of vaporization) than a non-polar compound like an alkane.  So an alkane and an alcohol with the same boiling point at one atmosphere will not have the same boiling point at another pressure. Correspondingly, an alkane and an alcohol with similar vapor pressures at one temperature can have very different vapor pressures (volatilities) at another temperature.


Types of chromatogram shapes I'm trying to generate a testing set of different chromatograms to check peak detection (whether peak boundaries are detected correctly) and smoothing algorithms. Currently I'm figuring out which features of chromatograms I should vary for my tests. Hopefully someone could help fill my blanks. So far I'm thinking of all combinations of all bullet points:

Peak shapes: Gaussian, ...?
Sharp & wide peaks
Different levels of skewness
Baseline drifts: any types? Should it also resemble Guassian peak?
2 peaks next to each other:

Peak resolution e.g. from .2 to 1. Allows to test valleys & round peaks.
Different peak heights. Together with different levels of resolution allows to test valleys, shoulders, rider peaks.

Tailing and fronting: are there some known distributions that resemble them?

As for smoothing:

High frequency noise: uniform distribution?

So far this is what I've got for each resolution that I choose: chromatogram-shapes.pdf. Want to add baseline drift next, but haven't decided which distribution(s) it should resemble.

Good to know that someone is working in DSP and chromatography but there are many software out there like PeakFit, why a new one?

Your first test could be peak detection as function of resolution as well their height ratios. Find out under what thresholds your peak detection fails. This is the most important thing.

Peak shapes: Gaussian, Exponentially modified Gaussians are very very common. Tailing is far more common than fronting. There is a very nice equation that covers all three shapes. Search BI-EMG in Google Scholar and you would find it. "BI" here is bi-directional."

Sharp & wide peaks: It is a relative term. You have to check what maximum efficiencies are achievable on instruments on GC, SFC and LC.

Different levels of skewness: Asymmetry of 2 is usually the maximum allowed number. Asymmetry of 1 is highly desirable. In practice it is possible to obtain 1.0 to 1.1.

Baseline drifts: Look more like linear ramps. You have to familiarize yourself with gradient elution.

Peak resolution: Start from peak resolution of > 0.5

Tailing and fronting: There is a very nice single equation that covers all three. See point no. 1.

Smoothing: I have written a couple of papers on smoothing. You can play with wavelets (most complex), Hamming filter, Savitsky Golay, simple central moving average, and Gaussian weighted moving average is the most common. I cannot disclose which instruments use which method. You can assume Gaussian noise rather than uniform. To the best of my knowledge wavelets or FFT filtering is not used in commercial GC, HPLC etc.


Why are tertiary amines less basic than secondary and primary amines in water than when vapourised? In gaseous state:

$$\ce{(CH3)3N} > \ce{(CH3)2NH} > \ce{CH3NH2} > \ce{NH3}$$

However, when dissolved in water:

$$\ce{(CH3)2NH} > \ce{CH3NH2} > \ce{(CH3)3N} > \ce{NH3}$$

My notes mentioned the reason for this is because of the increased steric hindrance from the greater number of alkyl group which makes the lone pair of electrons on the N atom less available. But, if this is the case, why doesn't steric hindrance affect the basicity of tertiary amines when in a gaseous state?

After some research, I found that the conjugate acid of an amine can be stabilised by solvation. But, shouldn't the stabilisation be the same, because of the attraction of electronegative O in a water molecule to the positive charge on $\ce{N}$ in an ammonium salt? (can ammonium salts still be stabilised by hydrogen bonding? I thought ions don't form hydrogen bonds with water, they form ion-dipole attractions)

Unless the effect of steric hindrance comes in here, where the alkyl groups reduce the strength of the ion-dipole attraction between the ammonium ion and thus the stability?

I'm really confused how everything links up; appreciate it if someone can kind of clear this up.

A possible answer from what I've searched on the internet so far..

The strength of basicity of amines depends on two factors: 1) the availability of the lone pair for donation to a proton, 2) the stability of the conjugate acid of the amine.

In methylamine, the conjugate acid, $CH_3NH_3^+$ will attract water molecules and ion-dipole interactions will form between the cation (the conjugate acid of methylamine) and the partial negative charge on O in the water molecule. This ion-dipole interaction or attractive force helps to stabilize the positive charge on the cation.

So, the more water molecules you have solvating your cation, the more stable the conjugate base. However, the more alkyl groups you have, the less stable the conjugate base, because these alkyl groups are hindering water molecules from coming and stabilizing the conjugate base.

[The N atom is "shielded" by the bulky alkyl groups which take up a lot of space and prevent any water molecules from stabilizing the conjugate acid. Water molecules can only come in from where the H atom is where the shape of the conjugate acid is that of a pyramid. Note that there is a positive charge on N.]

As methylamine has the least number of alkyl groups as compared to dimethylamine and trimethylamine, at this point, it would be the strongest base. However, basicity also depends on the availability of the lone pair of electrons on the N atom of the amine itself. As methylamine has the least number of alkyl groups, the inductive effect by the alkyl group is the least out of methylamine, dimethylamine, and trimethylamine.

So, it really is a measure of which factor is more dominant or which factor outweighs the other. In this case, although trimethylamine has the most number of alkyl groups, the three bulky alkyl groups prevent a lot of water molecules from stabilizing the conjugate acid. And, for dimethylamine, although dimethylamine has a fewer number of alkyl groups than trimethylamine, the inductive effect of the alkyl groups outweighs the steric hindrance from the 2 alkyl groups as compared to the three in trimethylamine.

Hence, in an aqueous solution, $(CH_3)_2NH>CH_3NH_2>(CH_3)_3N>NH_3$

However, in the gaseous state, there are no water molecules to stabilize the conjugate acid and basicity depends on the inductive effect of alkyl groups; there are no water molecules to hinder. As trimethylamine is trigonal bipyramidal in shape, the factor of steric hindrance does not apply here. Hence, as trimethylamine has the most number of alkyl groups, the lone pair of electrons on trimethylamine is most available for donation.


Why does the addition of PCl5 increase the rate of dissociation of PCl5? $\ce{PCl5 <=> PCl3 +Cl2} $

It's stated in my book that addition of $\ce{PCl5}$  to the equilibrium mixture increases the rate of forward reaction but no reason is mentioned for it.

However, before this the author derived $K_c = \dfrac{x^2}{(1-x)V}$ (using law of mass action at equilibrium ) where V is the volume of the container and $x$ is the number of $\ce{PCl5}$ molecules dissociated from $a$ number of moles initially.
I couldn't relate the "rate of forward reaction" statement to this equation while earlier in case of the reaction:

$\ce{H2 + I2 <=> HI}$

I was able to relate the equation I obtained through law of mass action with the effect of concentration on rate of forward or backward reaction.

Could someone please explain why addition of $\ce{PCl5}$ to the equilibrium mixture increases the rate of forward reaction?

Imagine a overpopulated kingdom suddenly gaining a large area of land (Start of Reaction). Immediately, groups of people start migrating to the area of land (Forward Reaction). However, as people migrate to this new land, some decide that, for whatever reason, they don't like this area (Reverse Reaction). As more and more people move there, more start coming back until there is an equal rate of people moving to the territory and people moving back (Dynamic Equilibrium).

  To your question: To model this system using my example of immigration, we suddenly add an enourmous influx of refugees into the old kingdom (Increase in $\ce{PCl5}$).

So the land in the outer territory seems more valuable now (less crowding, more open space, better air, etc.). This causes a temporary increase in the forward direction until the reverse reaction increases to compensate. This is called Le Chatelier's principle in chemistry.

References: Le Chatelier's Principle, My Handy Old 2011 Introductory Chemistry Textbook.


How is Raoult's law a special case of Henry's law and not vice versa? I know that Raoult's law can be a special case of Henry's law when the value of the Henry's constant becomes equal to the vapour pressure of the pure state of the liquid. Why can't we say the opposite, i.e. Henry's law is a special case of Raoult's law?

Raoult's law is a proposition regarding the relationship between the vapor pressure of a solvent (component 1 below) in its pure form ($p_1^\circ$) and when a solute (component 2) is present at a mole fraction $\chi_2=1-\chi_1$ (the vapor pressure then $p_1$):

$$\chi_1 = \frac{p_1}{p_1^\circ} $$

It is generally true for very dilute solutions because we know independently that

$$\frac{p}{p^\circ} \rightarrow1 \,\, \text{as} \,\,\chi_1 \rightarrow 1$$

For some ideal binary systems Raoult's law is applicable over a broad range of composition. In fact, sometimes the solute also observes Raoult's law in dilute solutions:

$$\chi_2 = \frac{p_2}{p_2^\circ} $$

This however is exceptional behavior. It is more common that the solute observes the general Henry's law

$$\chi_2 = \frac{p_2}{K_H} $$

where $K_H$ is a proportionality constant for the solute-solvent pair. It should be evident that when $K_H=p_2^\circ$ the solute displays Raoult's law behavior even in dilute solution.

Henry's law is a general statement observed by solutes with a measurable vapor pressure $p_2$, since the following limiting behavior is expected as solute-solvent interactions far exceed solute-solute interactions in importance in the limit of zero solute concentration:

$$\frac{p_2}{K_H} \rightarrow1 \,\, \text{as} \,\,\chi_1 \rightarrow 1$$

However, there is no principle that guarantees that $K_H=p_2^\circ$. Again, for a substance to follow Raoult's law at low concentrations is regarded as exceptionally ideal behavior.


Chemiluminescence vs. Bioluminescence Luminescence in general, is the term of emission of light from specific substances/objects. Chemiluminescence is the emission of light due to chemical reactions. Bioluminescence on the other hand is the emission light from living/biological organisms such as insects ( this is of course the way I understood them ), but I'm a little bit confused because what is the cause that makes insects emmit light? Doesn't that also have to do with chemical reactions but inside living organisms ?

Just a clear explanation of the difference  between these two terms  would be helpful.

Doesn't that also have to do with chemical reactions but inside living organisms?

Exactly! That's why there is one scientific society for both phenomena: ISBC - International Society for Bioluminescence and Chemiluminescence ;-)

The thermal reactions, that lead to an excited species which eventually emits a photon are quite similar. From a reactive compound, typically called a luciferine, a four-membered ring, often a dioxetanone is formed. This decarboxylates and gives rise to a molecule in an electronically excited state. In reality, the decarboxylation is often not just a spontaneous process but requires an activator. W. J. Baader at IQUSP in Sao Paulo, Brazil has examined that in great detail.

The luciferines can be quite different compounds, such as

diaryloxalates (these are used in glow sticks, where the diferent colours come from diferent fluorphores that absorb the initial photon and emit at different wavelength)
benzothiazole derivatives (firefly)
triazine derivatives (marine organisms)


(Effective) number of molecules interacting with light to produce optical activity The question is: How can I get a sensible and concrete (not relative) value for the number of molecules light has interacted with to cause an optical activity phenomenon?

Specifically, my issue in answering the question is that although initially I thought of calculating the number of molecules contained in the volume that the light is occupying, this does not seem correct to me. The reason is that the optical activity is independent of the volume, it only depends on the path length but not the transverse area. So it seems to come down to some kind of "effective" number of interactions, which is invariant to the transverse spread of the light.

Thanks for any help you can give me!

I think the picture above is self-explaining.

Now you must know the diameter and lenght of the lightpath so you can calculate the illuminated volume. Taking this volume and the concentration c you get the mol quantity (6.022 * 10^23) of molecules which have caused optical activity.


Mechanism for the addition of hydrogen iodide to 3,3‐dimethylbut‐1‐yne The following reaction mechanism was given as a solution to a solved problem in my textbook1 for the addition of hydrogen iodide to 3,3‐dimethylbut‐1‐yne:

It can be seen that that 2,2-diiodo-3,3-dimethylbutane (a geminal dihalide) is the product according to this mechanism. I arrived at a different product 2,3-diiodo-2,3-dimethylbutane (a vicinal dihalide) by using the following mechanism:

The difference is due to the methanide shift I did on the second step, converting 3,3-dimethylbut-1-en-2-ylium cation to 2,3‐dimethylbut‐3‐en‐2‐ylium cation. I consider this to be a reasonable carbocation rearrangement because after the methanide shift, the 2,3‐dimethylbut‐3‐en‐2‐ylium cation is tertiary as well as in conjugation with the double bond. Moreover, the initial carbocation has a positive charge on a more electronegative $\mathrm{sp^2}$ hybridized carbon atom. Due to this, I believe, the methanide shift produces a more stable carbocation. But, why did the author proceed without making this rearrangement?
When I discussed this in chat, there was a consensus that the product of this reaction must be a geminal dihalide as obtained by the author. The only way, I could think of, by which I can obtain the geminal dihalide even after the rearrangement discussed earlier is to do another rearrangement during the addition of second molar equivalent of hydrogen iodide as given below:

The problem with this mechanism is the carbon bearing the postitive charge after the methanide shift also has an iodine atom attached to it. Earlier, I learnt that chlorine atom is the only halogen for which the positive mesomeric effect is stronger than the negative inductive effect thereby stabilizing the positive charge. But here, due to the presence of iodine, I think the carbocation is not stable after rearrangement.
Even after neglecting this fact, there seems to be a major difference betweeen the author's mechanism and the modified mechanism, which I've emphasized using a carbon-12 labelled reactant as given below:

It can be seen that even though we obtain geminal dihalides through either of the mechanism, the products formed aren't exactly the same. One has the iodine attoms attached to the normal carbon whereas in the other they are attached to the carbon-12 atom.
In short, what happens when hydrogen iodide is added to 3,3‐dimethylbut‐1‐yne?
Reference

Solomons, et al. Organic Chemistry for JEE (Main & Advanced). Edited by MS Chouhan, Third Edition; Wiley India Private Limited. ISBN 978-81-265-6065-3

There seems to be an issue with both mechanisms on the front that an SN1 reaction would not take place since the carbocation formed is a vinylic carbocation which is highly unstable.
The actual reaction follows a termolecular mechanism where the rate of reaction is given to be:
$$\text{Rate}=[\ce{HX}]^2[\text{alkyne}]$$
Now, according to Advanced Organic Chemistry by Francis A. Carey, the reaction mechanism would be as follows:$^1$

Step $1$:
A concerted termolecular reaction...
This involves an acid/base reaction, protonation of the alkyne developing positive charge on the more substituted carbon. The π electrons act pairs as a Lewis base.
The other part is attack of the nucleophilic bromide ion on the more electrophilic carbocation creates the alkenyl bromide.

Step $2$:
In the presence of excess reagent, a second protonation occurs to generate the more stable carbocation.$^2$

Step $3$:
Attack of the nucleophilic bromide ion on the electrophilic carbocation creates the geminal dibromide.

[$1$]: The reaction mechanism stated above uses $\ce{HBr}$ and not $\ce{HI}$ instead. However this reaction takes place for $\ce{HX}$.
[$2$]: The same reaction using $\ce{HI}$ would have a comparatively lower yield of the geminal product since geminal diiodides are unstable due to the steric hinderance posed by the large size of the iodine atoms.


Placed in equivalent freezers, would a liter of water or a liter of lava turn from liquid to solid first? (question from a 6 year old) Placed in equivalent freezers, would a liter of water or a liter of lava turn from liquid to solid first? (question from a 6 year old)

Based on this page in a “Blaze” book, my six year old asked “which would win?” between water and lava. On further investigation, we refined the question to: which would turn solid first in similar conditions, a liter of room temperature water or a liter of volcanic lava?

You've got the right idea — you want to simplify the problem — but I don't think you're using quite the right simplification.  Take a look at the book's set-up of the situation, and ask yourself how the water is stopping the lava. You'll see the idea is that we're using liquid water, not ice, to solidify the lava.
So your question should really be:  How much heat do we need to absorb from a liter of lava to turn it into a solid, and how much heat can a liter of water at room temperature absorb before it turns to steam?  If the latter is larger than the former, then a liter of water can cool a liter of lava to the point where it solidifies before the water all changes into steam.
[I'm using "heat" when I should really be using "thermal energy", but this is for a 6 yo, so I'm keeping it simple.]
First, let's do the calculation for water.  Here (since it's for a 6 yo), I'm not going to show all the steps in the calculations:
Energy to heat 1 L liquid water at room temp (25C ) to 100C = 75 kcal
Energy to turn 1 L liquid water to steam at 100C = 533 kcal
Total = 608 kcal
According to https://en.wikipedia.org/wiki/Lava, lava is typically 700 to 1200 C, so let's call it 1000 C.
And, using this source (https://ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/12-109-petrology-fall-2005/lecture-notes/Nov3notes.pdf), let's assume it melts at 900 C (there's actually a wide range of lava types, and thus a wide range of lava temps and melting points).
Density of lava = 3.1 g/cm^3, so 1 L lava weighs 3100 g
Energy released when 1 L lava cools from 1000 C to 900 C = 93 kcal
Energy released when 1 L lava solidifies at 900 C: 310 kcal
Total = 403 kcal
So, based on the above, 1 L of water is enough to solidify 1.5 L of lava.
Finally, why does the water win?  The main reason is that the water is going between a liquid and a gas, while the lava is going between a solid and a liquid.  And, in general, it takes much more energy to change liquids into gases than solids into liquids, because in the former case you are pulling the molecules completely apart from each other.
But I said it takes "much more energy" to change a liquid to a gas than a solid to a liquid, yet the difference here is only a factor of 1.5. The discrepancy is because  in this case you're not comparing masses of water and lava, you're comparing volumes (and for the same volume, the mass of lava is 3.1 x greater). Thus, if you do it on a mass basis, 1 kg of water is enough to solidify 6 kg of lava (because, while 1 L of water weighs 1 kg, 1.5 L of lava weighs 6 kg).
[For these quantities, mass is typically a more fundamental basis of comparison than volume, which is why heat capacities, heats of fusion, and heats of vaporization are usually quoted on a mass basis. When you see the term "specific", as in "specific heat capacity, that means "per unit mass".  Of course, one can also use a number basis (moles).]


Why isn't the orbital angular momentum also considered while calculating the magnetic moments 3d transition elements? My textbook (Chapter: The d- and f- Block Elements) makes an interesting assertion, however, without any reason to back it up.

Paramagnetism arises from the presence of unpaired electons, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula, i.e.,
$$\mu = \sqrt{n (n + 2)}$$
where $n$ is the number of unpaired electrons and $\mu$ is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of $1.73$ Bohr magnetons (BM).

Now from what I've learnt from my physics classes last week, the magnetic moment of an electron is calculated using the formula:

$$\frac{m}{L} = \frac{e}{2 m_e}$$
Magnetic moment of revolving electron,
$$m = \frac{e}{2 m_e}L$$
In vector form,
$$\vec{m} = - \frac{e}{2m_e}\vec{L}$$

Where $m$ is the mass of the electron, $e$ is the magnitude of charge associated with the electron and $L$ is the angular momentum of the electron.
Now as I understand it, the angular momentum of an electron in an atom is the resultant of the electron's spin angular momentum and its orbital angular momentum.
But (as you can clearly see) my textbook exhibits an unexplained enthusiasm for only the spin angular momentum. It dismisses the contribution of the orbital angular momentum, as inconsequential, when determining the magnetic moments of electrons in the 3d elements.
As for the 'reason' provided by my textbook:

For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and is hence of no significance.

Well, I fail to see how that even counts as a 'reason'. I did ask my teacher about this, but she expressed her reluctance to go into the details; she's of the opinion that my particular query is “not worth clarifying” since it has no significance from the examination point of view (“Just give them what's in the textbook Aaron, nothing more and certainly not anything less”).
Subsequent internet searches have not yield any satisfying explanation (Perhaps I didn't use the right keywords while searching?).
So I'll just break up my question, point-wise, and list them below:

Is the orbital angular really of no significance when calculating the magnetic dipole moment for 3d elements? If it is significant, is there any 'educational' purpose behind that omission in my textbook, or is it just a mistake?

(As stated in the book) What causes the 'quenching'? How is this 'quenching' effect caused?

Do the transition elements of the other periods also have "insignificant" orbital angular momentum contributions? Is it for the same reason as the 3d transition elements.

Simplistically speaking orbital angular momentum is present when some conditions are satisfied:

A set of orbitals are degenerate;
These orbitals can be "interconverted" by rotation about a certain axis;
The set of orbitals is not empty, half-filled, or fully filled.

Physically speaking, the "rotation" of an electron from one orbital to the next generates orbital angular momentum. The rotation must not have an energy cost, which necessitates degeneracy. It also means that the set of orbitals must have at least one electron, as well as one empty space (a so-called hole) for the electron to "rotate" into. The hole must have the same spin as the electron, which means that a half-filled subshell doesn't allow for orbital angular momentum (since all the electrons have parallel spins).

This is most easily illustrated with the three 2p orbitals in an atom. Rotation of a $\mathrm{p}_x$ orbital by 90 degrees around the $y$-axis, for example, turns it into a $\mathrm{p}_z$ orbital.

Any atom with a $\mathrm{p}^1$, $\mathrm{p}^2$, $\mathrm{p}^4$, or $\mathrm{p}^5$ configuration therefore possesses nonzero angular momentum. This is also reflected in their term symbols where $L = 1$. This is said to be "one unit of angular momentum". An atom with a half-filled shell (for example, $\ce{N}$)  or a closed-shell $\mathrm{p}^6$ configuration (for example, $\ce{Ne}$) does not have any orbital angular momentum.

The five 3d orbitals in an atom also satisfy these requirements. It is less intuitive to see this as compared to the 2p orbitals, because we have taken specific linear combinations of the complex 3d orbitals and converted them into real orbitals. Nevertheless one can use quantum mechanics to prove that the 3d orbitals possess angular momentum. The maths can be found in Figgis and Hitchman's Ligand Field Theory and Its Applications (it is very early in the book, but I do not have it with me now and cannot remember offhand where it is). As such, for free 3d metal atoms or ions, the orbital angular momentum has to be incorporated into the calculation of the magnetic moment. Exceptions are, of course, $\mathrm{d^0}$, $\mathrm{d^5}$, and $\mathrm{d^{10}}$ ions (e.g. $\ce{Sc^3+}$, $\ce{Mn^2+}$, and $\ce{Zn^2+}$).

That is applicable for free ions. However, 3d metals do not commonly exist as free ions, but exist in coordination complexes with a ligand field. For simplicity we'll consider a perfect octahedral ligand field, which causes d orbitals to split into a $\mathrm{t_{2g}}$ ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$) and $\mathrm{e_{g}}$ ($\mathrm{d}_{z^2}$, $\mathrm{d}_{x^2-y^2}$) set. This means that the five d orbitals are no longer all degenerate, breaking the first condition stipulated above.

At a simplistic level, many textbooks will claim that this leads to a "quenching" of orbital angular momentum, which just means that there is no longer any orbital angular momentum, and as such, the spin-only formula applies. [Note that there are some second-order effects (quantum mechanical mixing of excited states into the ground state) that can lead to deviations from the spin-only formula.]

This is true to a certain extent. However, the five 3d orbitals are not split into individually nondegenerate orbitals; the three $\mathrm{t_{2g}}$ orbitals, for example, are still degenerate. On top of that, one can verify pictorially that it is possible to rotate the $\mathrm{t_{2g}}$ orbitals into one another. In this aspect the $\mathrm{t_{2g}}$ orbitals behave quite similarly to the three $\mathrm{p}$ orbitals in a free atom; this is called the "$\mathrm{t_2}$-$\mathrm{p}$ isomorphism". As such, the $\mathrm{t_{2g}}$ orbitals still have orbital angular momentum associated with them.

Any complex with unsymmetrical occupancy of the $\mathrm{t_{2g}}$ orbitals (i.e. $\mathrm{t_{2g}}^1$, $\mathrm{t_{2g}}^2$, $\mathrm{t_{2g}}^4$, $\mathrm{t_{2g}}^5$) therefore possesses orbital angular momentum and exhibit great deviations from the spin-only formula. These complexes all have a $\mathrm{T_{1g}}$ or $\mathrm{T_{2g}}$ ground state term symbol, so it is commonly said that "$\mathrm{T}$ terms possess one unit of orbital angular momentum". Spin-orbit coupling has to be invoked to obtain a better agreement between the calculated and measured magnetic moments.

As an example of how large the deviations can be, consider the compound $\ce{(NH4)2Co(SO4)2.2H2O}$. This features octahedral Co(II) which adopts a high-spin configuration $(\mathrm{t_{2g}})^5(\mathrm{e_g})^2$. From the spin-only formula, we would expect $\mu_\mathrm{so} = \sqrt{3(3+2} = 3.87$. However, the unsymmetrical population of the $\mathrm{t_{2g}}$ orbitals leads to the generation of orbital angular momentum, and at $300~\mathrm{K}$ the measured effective magnetic moment is actually $5.1$. (Data from here.)

A more detailed discussion can be found in any of the standard inorganic chemistry textbooks. Figgis and Hitchman's book (mentioned earlier) is more advanced and has a very comprehensive treatment of the subject.


Calculating Average Rate A $\pu{3.45 g}$ piece of marble ($\ce{CaCO3}$) is weighed and dropped into a beaker containing $\pu{1.00 L}$ of hydrochloric acid. The marble is completely gone $\pu{4.50 min}$ later. Calculate the average rate of reaction of $\ce{HCl}$ in $\pu{mol/L/s}$. Note that the volume of the system remains at $\pu{1.00 L}$ through the entire reaction.

I'm not very sure why the units are in $\pu{mol/L/s}$ instead of just $\pu{mol/s}$. Here is what I did:

$$\text{Rate} = \frac{\pu{3.45 g}~\ce{(CaCO3)}}{\pu{4.5 min}} = \pu{0.0128 g/s}$$

$$\pu{0.128 g}~\ce{CaCO3} = \pu{1.28e-3 mol}$$

$$n(\ce{HCl}) = \pu{2.58e-3 mol}$$

$$\text{Rate} = \pu{2.58e-3 mol/L/s}$$

However, the answer given is $\pu{2.55e-4 mol/L/s}$. What am I doing wrong?

You haven't accounted for the stoichiometry of the reaction, and I suppose you wrongly converted minutes to seconds. Always start solving problems like this with writing down the chemical reaction:

$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$

By definition rate of consumption of hydrochloric acid over time $\Delta t$ is:

$$r=\frac{\Delta c(\ce{HCl})}{\Delta t}$$

Since all calcium carbonate reacted completely:

$$\Delta c(\ce{HCl}) = \frac{\Delta n(\ce{HCl})}{V} = \frac{2n(\ce{CaCO3})}{V} = \frac{2m(\ce{CaCO3})}{V M(\ce{CaCO3})}$$

where $m$ is mass, $M$ - molar mass, $V$ - volume.

And the average rate is:

$$r = \frac{2m(\ce{CaCO3})}{V M(\ce{CaCO3})\Delta t} = \frac{2\cdot\pu{3.45 g}}{\pu{1 L}\cdot\pu{100.09 g mol-1}\cdot\pu{4.50 min}\cdot\pu{60 s min-1}} = \pu{2.55e-4 mol L-1 s-1}$$

Also, be careful with notations. Use proper capitalization, and don't equate moles to grams! This is not tolerable in natural sciences.


Is it true that human cells grow in levorotation? I remember reading this in a popular science magazine, some 20 years ago: that in most of the living organisms, the proteins (in fact, the amino acid molecules that make the proteins) are levorotatory and very few such molecules are dextrorotatory.

Is that true, at least for humans?

I remember reading this in a popular science magazine, some 20 years ago: that in most of the living organisms, cells grow in levorotation and very few grow in dextrorotation.

There is a kernel of truth in what you wrote, but either the science fiction magazine or your memory has distorted things a fair bit.  Cells do not grow "in levorotation" or "in dextrorotation".  Chirality is not a property of cells, it is a property of the individual molecules that comprise the cell.  Many different types of molecules are essential for life and are found in all living cells.

Two examples of these ubiquitous, essential molecules are carbohydrates and amino acids.  In all organisms on Earth, most carbohydrates are found in what is known as the D configuration.  The "D" stands for "dextrorotatory" but the nomenclature is confusing.  For carbohydrates, the configuration is defined as relative to glyceraldehyde, a simple sugar.  Thus D-fructose or D-glucose have the same configuration as D-glyceraldehyde, but only D-glyceraldehyde is guaranteed to be dextrorotatory.

Amino acids in life on Earth are predominatly "L", meaning they share an absolute configuration with L-glyceraldehyde.

  Is that true, at least for humans?

So humans, like most life on Earth, is made of a mixture of levorotatory and dextrorotatory molecules.  Some types of the molecules in humans are predominantly dextrorotatory and some are levorotatory.


Is the change of the Gibbs free energy always zero at constant pressure? At constant pressure, which is the usual condition for chemical reactions, heat absorbed/released by the system is the same thing as enthalpy change. According to the Gibbs equation,
$$\mathrm dG = \mathrm dH - T\,\mathrm dS = \mathrm dH - đq$$But, $\mathrm dH=\mathrm đq$ at constant a pressure. So is the Gibbs energy change $\mathrm{d}G$ simply zero for any reaction?

You wrote:
$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S = \mathrm{d}H - đq$$
There are several issues with this. I don't wish to co-opt other answers and comments, but allow me to mention these here, since my answer stands as the accepted (which I have no power to change):

You haven't assumed constant temperature, which means that there is an additional term $-S\,\mathrm{d}T$ which you did not take into account;
You haven't assumed constant composition, which likewise leads to another term $\sum_i \mu_i\,\mathrm{d}n_i$.

In fact, these considerations (particularly the second) are more important than my original point, which is retained below. See Kexanone's answer for discussion of these points. In the following discussion, one should assume constant temperature and composition.

In the second equality, where you equate $đq$ with $T\,\mathrm{d}S$, this assumes that the process is reversible. Without going into too much detail (it can be found easily elsewhere), entropy is defined by the following equation:
$$\mathrm{d}S = \frac{đq_\mathrm{rev}}{T}$$
where the subscript $\mathrm{rev}$ indicates a reversible process. It can be shown that for an irreversible process,
$$\frac{đq_\mathrm{irrev}}{T} < \frac{đq_\mathrm{rev}}{T} = \mathrm{d}S$$
This result is the second law of thermodynamics. Therefore, for an irreversible process, $T\mathrm{d}S > đq$.
You are right in saying that at constant pressure, $\mathrm{d}H = đq$. Putting it all together, we have:
$$\mathrm{d}G = \mathrm{d}H - T\mathrm{d}S = đq - T\mathrm{d}S < 0 \qquad \text{(irrev. process; closed system; }p, T\text{ const.)}$$
Therefore, the direct answer to the question is: $\mathrm{d}G$ is, in general, not equal to $0$ because $T\mathrm{d}S \neq đq$.
In fact, you may notice that the condition $\mathrm{d}G < 0$ is the condition for spontaneity under constant $p$ and $T$. This is no coincidence. The concepts of spontaneity (in the context of $G$) and irreversibility (in the context of $S$) are extremely similar, in that they both refer to a tendency for a process to proceed.

For a reversible process, $T\mathrm{d}S = đq$. Following the logic that you used in your question, this means that $\mathrm{d}G = 0$, which is correct. If $\mathrm{d}G = 0$, the forward process cannot be spontaneous (that would require $\mathrm{d}G < 0$) and the reverse process cannot be spontaneous (that would require $\mathrm{d}G > 0$).
This can only mean one thing: the system is in equilibrium. Indeed, the condition for equilibrium at constant $p$ and $T$ is $\mathrm{d}G = 0$. And at this point, it should not come as a surprise that a reversible process is, by definition, one that is at equilibrium throughout.


Active methylene group nucleophilic substitution The first product formed after reaction with $\ce{^-CH(COOEt)2}$ followed by hydrolysis is [2-(naphthalen-1-yl)ethyl]propanedioic acid:

But is the reaction $\mathrm{S_N1}$ or $\mathrm{S_N2}$?

On heating, decarboxylation takes place because of an $\alpha$-carboxylic acid group.

Then $\ce{SOCl2}$ followed by $\ce{AlCl3}$ is Friedel–Crafts acylation because using $\ce{SOCl2}$ gives an acid chloride. After that, an intramolecular electrophilic substitution takes place.

But where will the electrophilic attack take place?

You have given correct suggestions for the first two part of the question. Now, the question is where would be the electrophilic attack of acid chloride taken place under Friedel-Crafts acylation conditions. As Ron correctly pointed out in his comment that there are only 2 possibilities to cyclization: That would be to the 2 or 8 positions of naphthalene nucleus, and it would certainly be the 2 position, which produces a 6-membered ring. The method is a named reaction, which is called Haworth Phenanthrene Synthesis (Ref.1 & 2). According to the abstract of Ref.2:

  Haworth synthesis is a multistep preparation of phenanthrenes from naphthalenes by means of the Friedel–Crafts acylation with succinic anhydride, followed by a Clemmensen reduction or Wolff–Kishner reduction, cyclization, reduction, and dehydrogenation. This synthesis is also referred to as the Haworth phenanthrene synthesis. Additionally, the same procedure used for the preparation of naphthalenes from benzenes is called the Haworth reaction. This reaction has been modified to react with 4‐pentenoic acid ester to form alkyl substituted phenanthrenes. This reaction has a wide application in preparation of fused aromatics, including naphthalenes and phenanthrenes.

The Haworth synthesis is depicted in following Scheme under the same title in top box:

You see after the reduction by $\ce{Zn/HCl}$, you get the same compound (4-(1-naphthyl)butanoic acid) you got after acid hydrolysis and heat in your scheme, as indicated in top yellow box. In Haworth reaction, $\ce{H2SO4}$ has been used instead of $\ce{AlCl3}$ for the convenience, the reaction under goes same pathway to give tetrahydrophenanthrene keto derivative by acylating 2-position of the naphthalene nucleus (see second yellow box). There was no evidence for the formation of 7-membered ring (by undergoing cyclization through 8-position).

However, if you had 4-(2-naphthyl)butanoic acid instead of 4-(1-naphthyl)butanoic acid, you certainly would get anthracene derivative (e.g., compound 4) as the major product with phenanthrene derivative (e.g., compound 5) as a trace product (See second box in the Scheme)(Ref.3).

References:

R. D. Haworth, “145. Syntheses of alkylphenanthrenes. Part I. 1-, 2-, 3-, and 4-Methylphenanthrenes,” J. Chem. Soc. 1932, 1125-1133 (DOI: 10.1039/JR9320001125).
Zerong Wang (Ed.), In Comprehensive Organic Name Reactions and Reagents – PART TWO; John Wiley & Sons, Inc.: New York, NY, 2010, “Chapter 299: Haworth Synthesis (Haworth Phenanthrene Synthesis),” (https://doi.org/10.1002/9780470638859.conrr299).
Z. Zhang, R. Sangaiah, A. Gold, L. M. Ball, “Synthesis of uniformly $\ce{^{13}C}$-labeled polycyclic aromatic hydrocarbons,” Org. Biomol. Chem. 2011, 9, 5431-5435 (DOI: 10.1039/C0OB01107J).


calculate the pressure at equilibrium Given that $K=18$ for the equilibrium below, calculate the total pressure in a sealed vessel initially containing only $\ce{NH4I(s)}$, that reaches equilibrium.
$$\ce{NH4I(s) \leftrightharpoons NH3(g) + HI(g)}$$
a) $4.2\mathrm{~atm}$
b) $8.5~\mathrm{atm}$
c) $2.1~\mathrm{atm}$
d) $\mathrm{18~atm}$
e) $\mathrm{6.3~atm}$

Using $K = \ce{[NH3][HI]} = 18$, I found $\ce{[NH3]=[HI]=}~\frac{4.24mol}{L}$, and therefore $8.48~\mathrm{mol}$ of gas is produced, but I don't know how to find the equilibrium pressure without temperature and volume.

I think what is given here is $K_p= p(\ce{NH3}). p(\ce{HI})$ and not $K_c$. So, What you have here are the partial pressures of the gases. As you have a sealed vessel, so the sum of the partial pressures of the gases gives the total pressure, 8.5 atm


Saltpetre in the Garage My neighbour warned me that saltpetre (potassium (sodium) nitrate) is danger in the certain conditions. Is it possible that it can explode in my garage?

Potassium and sodium nirate can be used to make black powder (gun powder), but they are not explosives by themselves. It is safe to store bags of salpetre in the garage.

If there's salpetre forming on the brickworks, I'd rather be concerned to keep the bags closed and away from that wall: sodium nitrate is rather hygroscopic and salpetre crystals on and between the bricks indicate that the wall is wet.


Why are dipoles "permanent/induced dipole permanent/induced dipole" and not just "permanent/induced dipole" once? My teacher would always say "induced dipole induced dipole" and while it annoyed my slightly (as you were saying the same word twice) it didn't really bother me, but now looking through some exam papers it's called the same thing "induced dipole induced dipole", so why is it repeated twice? Would I be "wrong" for just saying "induced dipole" once? If so, what's the difference between the two?

(Sorry if this seems like a simple or stupid question, just I'm always curious and the internet doesn't really work for searches like this.)

Because it takes two to tango.

Dipoles interact with each other. A Lone dipole has nothing to interact with (other than an electric field, but if we ignore some externally applied macro field, there is nothing for a lone dipole to interact with).

So molecules with an inherent dipole (like water or chloroform) interact with each other. One molecule's dipole interacts with the other molecule's dipole. So you would never say "dipole interaction" only "dipole-dipole interaction". The repeated word is because it takes two dipoles to interact.

The same pattern applies to non-polar molecules with little or no inherent dipole. For example, benzene. Benzene has no built-in dipole, but the electrons in its bonds are fairly polarisable (which basically means it is easy to induce a dipole in them). So benzene molecules do interact but via London or van der Waals forces which are much weaker than the reactions of molecules with inherent dipoles. But one way to describe those weaker interactions is (simplifying a lot) to see them as forces created by fluctuation and temporary quantum-mechanical induced dipoles. Imagine the electrons in one benzene molecule stray from perfect symmetry in their distribution and create a temporary dipole; this may interact with the electron clouds in other benzene molecules an induce other temporary dipoles (or interact with other existing temporary dipoles). Either way the interaction is an "induced dipole-induced dipole" interaction.

So however you understand the interactions, to describe them properly you have to repeat the phrase twice.


How can I calculate molality from molarity? How can we determine the boiling point of 1 molar aqueous solution (density 1.06 g per ml) of KBr? If I use the colligative property formula than I require molality instead of molarity. Here the density given is the one of the solution not the solvent.

The key here is to find the mass of the solvent. If you take one mL of the solution, it will contain $10^{-3} \mathrm{mol}$ of KBr.

The mass of $10^{-3} \mathrm{mol}$ of KBr is  $10^{-3}\times(39+79,9)=10^{-3}\times 118.9=0.1189 \mathrm{g}$.

Or, the mass of one mL of the solution is $1.06 \mathrm{g}$. This means that the mass of the solvent in one mL of the solution is  $1.06 -0.1189=0.9411 \mathrm{g}=0.9411\times 10^{-3}\mathrm{kg} $.

The molality is:
$$m=\frac{10^{-3} \mathrm{mol}}{0.9411\times 10^{-3}\mathrm{kg}}=1.063\,\, \mathrm{mol/kg} $$


In 2-bromo cyclohexanone,why bromine prefers to occupy axial rather than equatorial position? I read that the substituents generally tend to occupy equatorial position rather than axial because diaxial interactions make the molecule unstable. But why is 2-bromocyclohexanone an exceptional case? I also read that it’s because of the dipolar interactions. What actually are these dipolar interactions and how do they stabilise 2-bromocyclohexanone with bromine being on axial position. How do they outweigh the effect of diaxial interactions.

I think there must have been some misunderstanding. This stabilization is working for all 2‑halocyclohexanones (but does not always cause the axial conformer to predominate), where bromo- one is not a special case, but has been used just as an example.

There is a combination of various effects, but the dipolar one is supposed to cause a destabilization by dipolar repulsion in the equatorial-X conformer. See the difference of partial charges distribution:

Measured or calculated molar fractions (%) of the axial conformers in different phases/solvents (taken from Yoshinaga, Fabiana, et al. "Conformational analysis of 2-halocyclohexanones: an NMR, theoretical and solvation study." Journal of the Chemical Society, Perkin Transactions 2 9 (2002): 1494-1498. doi:10.1039/B204635K):

$$
\begin{array}{c*}\hline
& \text{vapor} & \text{liquid} & \ce{CCl4} & \ce{CHCl3} & \ce{MeCN} & \text{DMSO} \\
\hline
\ce{F}  & 64 &  5 & 28 & 13 &  2 &  2 \\
\ce{Cl} & 86 & 26 & 62 & 42 & 15 & 13 \\
\ce{Br} & 92 & 53 & 81 & 66 & 36 & 33 \\
\ce{I}  & 96 & 87 & 92 & 85 & 65 & 63\\
\end{array}
$$

Overall stabilization effects in axial-X conformation of 2‑halocyclohexanones clearly increases in the order of F < Cl < Br < I.

However, the ‘gauche effect’ participates, which is stabilizing the equatorial conformation in largest extent for F, also for Cl, has almost zero effect for Br, and has a negative effect for I.


How to know which ionic compound have the highest lattice energy without knowing the values of the radii? Given two ionic compounds, for instance:

$$\ce{CaS} \quad \mathrm{or} \quad \ce{KCl}$$

What is the procedure to predict which of the two have the highest lattice energy (in absolute value)?

(Suppose that I do not know the values of radii of the atoms but I can only look at the periodic table).

I know that I must look at the ratio

$$\frac{Z_1 Z_2}{r_1 + r_2} \propto E_\text{lattice},$$

where $r_1 + r_2$ is the distance between the two atoms in each molecule of the compound, $Z_1 Z_2$ is the product of atomic numbers of each element in the molecule.

But, for example, in the case above I have both $r_1 + r_2$ and $Z_1 Z_2$ bigger for $\ce{KCl}$ then $\ce{CaS}$, so I cannot say (without knowing the numerical values) if one of the two ratios would be higher.

Am I missing something?

The Lattice Energy of Ionic Compounds is directly proportion to the Charge Density. As Ca, S have 2 units of charge while K, Cl have one and there is not much of a difference in size ( atleast not enough to overcome the doubling of the charge), the charge density of Ca and S ions would be more. Consequently, the lattice strength of CaS would be more.


Why does tetrachloromethane have a higher boiling point than trichloromethane? London dispersion forces (LDF) are present in all molecules, whether polar or non-polar. Molecules also exhibiting dipole-dipole interactions (in addition to the LDF) must have stronger forces of attraction than those molecules which exhibit only LDF.

Then, why does tetrachloromethane (carbon tetrachloride), which is a non-polar molecule exhibiting only London dispersion forces, have a higher boiling point ($\pu{77 ^\circ C}$) than trichloromethane (chloroform) ($\pu{61 ^\circ C}$) which is a polar molecule, exhibiting dipole-dipole interactions?

You also need to account for the difference in dispersion forces between the two molecules. Chlorine is much larger than hydrogen. Therefore tetrachloromethane has a larger molecular surface area which increases the intermolecular interaction strength. In this particular case, it outweighs the weak dipole interactions present in trichloromethane.


Conditions affecting pseudo rate orders For the reaction:
$$\ce{F
+ Br_2 -> FBr + Br}$$
The gas concentrations as a function of time are given by:
$$kt=\ce{\frac{1}{[Br2]_0-[F]_0}}\ce{\ln(\frac{[Br2][F]_0}{[F][Br_2]_0})}$$
Given that $\ce{[F]_0=4 x 10^{-9}}$ and $\ce{[Br2]=1x 10^{-10}}$ I am meant to show that the rate constant is first order with respect to $\ce{Br2}$. How do I do this?

Fluorine atoms are in excess but is it a large enough excess to assume that the isolation method is valid? I could then cancel terms but would I the have to differentiate to find the rate law?

Maybe it would be best to reorganise that to

$\ce{[Br_2]}=\exp\left[-kt(\ce{[F]_0}-\ce{[Br_2]_0})\right] \frac{\ce{[Br_2]_0}\ce{[F]}}{\ce{[F]_0}}$

and as there is 40 times more $\ce{F}$ than $\ce{Br_2}$, it is safe to assume that $\ce{[F]}\approx\ce{[F]_0}$. I.e. even if all the $\ce{Br_2}$ is gone there is still $\ce{[F]}=3.9\times10^{-9}$. This simplifies the above equation to

$\ce{[Br_2]}=\exp\left[-kt(\ce{[F]_0}-\ce{[Br_2]_0})\right]\ce{[Br_2]_0}$

which is the familiar form of the expression of concentration. It is also easy to see what the pseudo first-order rate constant is.


Baldwin's rules exceptions: cations Why do cations violate Baldwin's rules?

For example, the following acetalization mechanism is plausible even though it contains a 5-endo-trig step:

Image taken from: Clayden, J. Organic Chemistry; Oxford University Press: Oxford, 2013

The acetal formation in question is actually an example of a 5-endo-trig cyclisation, which, according to Baldwin's rules should be unfavourable.

The first thing to realise about Baldwin's rules is that they are largely empirical, that is, based on observation of data rather than theoretical proof (though they can be rationalised based upon orbital overlap arguments).

The second thing to realise is that, as with any rule, there are many exceptions.

  From Wikipedia: Baldwin's rules discuss the relative rates of ring closures of these various types. These terms are not meant to describe the absolute probability that a reaction will or will not take place, rather they are used in a relative sense. A reaction that is disfavoured (slow) does not have a rate that is able to compete effectively with an alternative reaction that is favoured (fast). However, the disfavoured product may be observed, if no alternate reactions are more favoured.

The 5-endo-trig acetal formation is actually one of the most well known exceptions to Baldwin's rules (Baldwin himself acknowledged this).

In an 1981 JOC paper (J. Org. Chem. 1981, 46, 2196), Mike Perkins and Jack Barrett attempted to understand why it was that these acetalizations were so 'favourable'.

Using very basic computational methods (MNDO methods), they studied the energy needed in order to get the molecule into the correct conformation for the hydroxyl to attack the oxonium (this is the ring closing step, and the one which you might expect to he the disfavoured step).

The conclusions of the paper are that the barrier is actually much smaller than expected, making formation of the acetal acceptable, in spite of baldwin suggesting otherwise.


E2 Elimination; Anti-periplanar beta hydrogen I'm confused by this picture I found on masterorganicchemistry.com—a very popular organic chemistry website:

I agree that the product is a product.

However, can't there be another product? Specifically, can't the methyl group by the bromine have an anti and beta hydrogen relative to the bromine?

So shouldn't there be two products?

Specifically, this one?

A reasonable chemist would be on the lookout for two products, especially given that E2 elimination from 1-bromo-1-methylcyclohexane produces some methylenecyclohexane

In the transition state for the E2 elimination the pi bond is already partially developed.  This means that the relative thermodynamic stabilities of the products will play a role in determining the product distribution.  Since more highly substituted olefins are thermodynamically more stable than less substituted olefins, one would predict that the product with the double bond internal to the ring would predominate over the analogue with the external double bond.

Reaction temperature, base size, steric constraints around the various hydrogens that might be involved and the numbers of the various hydrogens all factor into the final product ratio, but these effects are usually secondary unless you change circumstances so much that the reaction mechanism changes and shifts away from E2.


Giving too much energy to an exothermic reaction What will happen in an exothermic reaction if more than required energy is given? I mean, will it produce even more energy which will be highly dangerous or nothing will happen according to Le Chatelier's principle (as we are trying to carry the reaction in an opposite direction)?

There are two separate effects we need to consider here: thermodynamic and kinetic.

Let's assume you are only providing the energy thermally. So providing more energy means increasing the temperature (T).  Thermodynamically, if a reaction is exothermic, and you increase T, the reaction becomes less favorable (assuming it stays exothermic over that temperature range) according to, as you mentioned, Le Chatelier's principle.  I.e., the equilibrium shifts to the left.

However, we also need to consider kinetic effects.  Here, as you increase T, the rate of reaction increases, and thus the rate of thermal energy production increases, thus further increasing T, thus further increasing the rate, and so on, leading to the thermal runaway effect that Buck Thorn mentioned in his comment.  I.e., we get a positive feedback loop.


How are lowercase k and uppercase K used in general chemistry courses? Where do lowercase k and uppercase K make appearances in general chemistry, and what do they signify?

In addition to the original answer I also allowed myself to invite lowercase Greek letter kappa κ to the table as typographically it has a similar-looking appearance and add some more definitions:

Lowercase k

$k$ is one of the Miller indices.
$k$ is a symbol for thermal conductivity.

Uppercase K

$\ce{K}$ is a symbol for potassium.
$\pu{K}$ is a symbol for Kaiser, unit of energy $(\pu{1 K} = \pu{1 cm-1})$.
K is a symbol for aminoacid lysine.
K is used in spectrometry to denote spectral lines.
$K$ is sometimes used as a symbol for decadic absorption coefficient.

Lowercase Greek letter kappa κ

κ is used to denote denticity.
$κ$ is a symbol for [specific] conductivity.
$κ$ is a symbol for molar Napierian absorption coefficient.
$κ$ is a symbol for asymmetry parameter.
$κ$  is sometimes used as a symbol for bulk viscosity.
$κ$ is sometimes used as a symbol for magnetic susceptibility.

Uppercase Greek letter kappa Κ

Κ is a symbol for a kappa number characterizing bleachability.

Both lowercase k and lowercase Greek letter kappa κ

$k$ and $κ$ is a symbol for imaginary refractive index.

Both lowercase k and uppercase K

$k$ and $K$ is in general may refer to a coefficient (German "Koeffizient"); indices are often used to denote which one, e.g. mass transfer coefficient $k_\mathrm{d}$, octanol/water partition coefficient $K_\mathrm{ow}$.


If "Gas" belongs to "State of Matter" then what does "Metal" belong to? My question is, if "gas" belongs to "State of Matter" then what does "Metal" belong to.

Gas can be substituted with any state of matter(in the analogy); metal can be substituted with metalloid or nonmetal(in the analogy). I've researched this and I can't seem to find an answer to this. So, this could read as "if 'solid' belongs to 'State of Matter', then what does 'nonmetal' belong to'"?

I'm not trying to assert that metal is a state of matter.
As most of y'all (hopefully all) know, the elements on the periodic table can be a metal, metalloid, or nonmetal. Would "Metal" belong to "Metal Status"? I'm asking because I'm writing a program to represent the PeriodicTable and I need an appropriate name for my variable. I may just opt for "Metal Status".

"Metal" is a classification used by chemists and usually means something along the lines of "this electrically conductive, usually solid element that is deformable".

The other way around: Chemists use "gas" to refer to a state of matter, not some defined gas (for example the atmosphere).

Putting these two together: "Metal" belongs to the periodic table of elements, while "solid", "liquid", "gas" and "plasma" are merely words to describe what state the element or compound is in. It is possible to have liquid, solid or even gaseous metals.

Edit: If you want a name for a variable that is either in the state Metal, Half-Metal or Metalloid and Non-Metal, I'm not sure such a name even exists. Feel free to make one up that suits your needs, comment your code sufficiently and it will not be a problem. "Metallicness" or something along those lines would be my choice.


Hydrogen Bonding (with oxygen) - restricted to oxygen singly bonded to H or not? I understand the following definition of a hydrogen bond from Merriam Webster:

  an electrostatic attraction between a hydrogen atom in one polar
  molecule (as of water) and a small electronegative atom (as of oxygen,
  nitrogen, or fluorine) in usually another molecule of the same or a
  different polar substance

However, I wasn't sure if this applied to F, N, or O atoms that were not covalently bonded to H in the first place. A question regarding hydrogen-bonding came up on an exam for which the marks scheme read that both element X (with an O-H bond) and element Y (with an O in an ether group). However my teacher doesn't not agree with the marks scheme and, "[believes that the marks-scheme is] incorrect because Y doesn't contain O-H bonds so the O will not carry a large enough δ- to create hydrogen bonds with water."

Is she correct, or is the marks scheme?

Note: I thought the same as my teacher up until I saw the MS.

Hydrogen-bonds require a $\delta^+$ hydrogen atom attached to an electronegative element, i.e. O, F, N, or Cl, and also an "active" lone pair on another electronegative atom.  An electrostatic attraction between these two is an extremely strong kind of intermolecular dipole-dipole interaction.

Compounds which have O-H or N-H groups are able to both form hydrogen-bonds through the $\ce \delta^+$H atom and accept hydrogen-bonds through the lone pair on the oxygen or nitrogen.  These compounds tend to have higher then expected melting and boiling points due to the hydrogen-bonding.  Compounds are even able to form internal hydrogen-bonds with other O or N atoms in the structure.

However hydrogen bonding can occur between substances that cannot form the bonds with their own molecules.  Because compound Y is an ether and can only accept hydrogen-bonds there are fewer bonds made overall between the mixture of substances, so the effect on the physical properties isn't so great.  Nonetheless water can form hydrogen-bonds through its proton and functional groups such as ethers, esters, amines, amides, alcohols and carboxylic acids can all accept those bonds through the relevant lone pairs.

Another example would be a tertiary amine such as $\ce{N(CH3)3}$, which is miscible with water because of the hydrogen bonds it accepts from the water molecules even though it has no N-H bonds itself.  Even $\ce{N(C2H5)3}$ is miscible with water at room temperature, and other tertiary amines continue to be soluble in water for the same reason: they are hydrogen-bond acceptors.

So the mark scheme is correct, and the teacher has shown a poor understanding of intermolecular interactions.


Rate of dehydration of alcohols having keto group The following question was asked in IITJEE-Screening 2000,

Which of the following will be most readily dehydrated in acidic conditions?

I thought that the answer would simply be (c), as its intermediate would be stabilized by resonance with keto group and it is a sec-carbocation (which is common in all the options, so deosn't matter).
But my answer was wrong as per answer key. Am I missing something?

The correct answer as per key was,

 (a)

A carbonyl group normally destabilizes carbocations but stabilizes carbanions.
In this particular case, the rate of dehydration isn't guarded by carbocation stability, but by CH- acidity. In acidic conditions, OH group in alcohols is relatively easily protonated. Then, if there is a relatively acidic hydrogen in $\beta$-position it immediately dissociates, together with a water molecule.
Excercise: Try to draw a mechanism for dehydration of (a) in basic conditions. Hint: it does occur and produces the same product. It does require more heating, though.


How to write correct NMR integrals in a formula? for my masterthesis I want to include the Equation how I calculate my conversion in a polymer reaction.
It is pretty easy, but I still want to write it down.
Basicly the equation is: (Integral A usally 4-5 and Integral B always 1 beacuse its one Proton, Integral B and C are equal)
$$ Conv=\frac{Integral_A-Integral_C}{Integral_A+Integral_B}=\frac{Integral_A-1}{Integral_A+1}
$$
But the formular with "Integral" in it looks not very nice while I thin the $\int{}{}$ sign seems to be a bit over the top.

If someone wants to know how I end up with the equation, here is the explanation:

It comes from the point, that my Monomer has 2 protons on each side of the double bond. When the polymer is formed they shift forming one broad signal. In the polymersolution I can identify one proton of the monomer which gets the Integral 1 while the other big peak contains the polymer as well as one proton from the monomer. Because both monomer protons have the same integral, both get a 1. My "clean" polymersignal is therefore the broad Integral -1. This has to be devided by the overall Integral which is the broad signal (which already includes one monomer proton)+ the other proton.

You assume the total number of protons summed over reactants and products is constant during the course of the reaction, which is perfectly reasonable, and that all monomer protons are incorporated into the polymer (perfectly reasonably), but also that the total signal is a constant. Note that because polymer signal will relax more rapidly by $T_2$ relaxation, usually, signal ratios are not expected to be strictly proportional to concentration, but we'll ignore that point, assuming this is a solution experiment and the polymer is highly dynamic.

Therefore, you could write

$ f_{conv}= (s_A-s_C)/(s_A+s_C) = (s_{A,norm}-1)/(s_{A,norm}+1)$

where $s_C$ is the isolated signal from one type of monomer proton and  $s_B=s_C$, the other monomer signal, overlaps with the polymer signal to generate $s_A$. Then you define units for the area $s_B=s_C=1$ according to the expression on the right. This is explicit in the way I have written the equation. You should explain the meaning of the symbols in the text, of course.


Osmium tetroxide production I know osmium powder will produce osmium tetroxide when exposed to air,but what about solid piece of osmium,such as pellet or ring,finely polished to minimize surface area,will this produce significant tetroxide at 24 degrees celsius air?

How much tetroxide will 1 cm2 surface area of polished osmium produce in 24 hours when exposed to air at 24 celsius?

A block of Osmium is quite immune to atmospheric attack, due to the low surface area of the osmium, but when heated it can form Osmium tetroxide. I wouldn't like to risk wearing an osmium ring just incase Osmium tetroxide is formed. The oxide has a pungent smell so if you do wear a ring of Osmium, at least you'll have some form of warning to take it off. So really, I wouldn't recommend it, but do it at your own risk.


In orbital hybridisation, what determines the number of orbitals that hybridise to be at the same energy level? For example, in phosphorus pentachloride five orbitals hybridize, but why not 6, as in the hydrated Al ion? I have read that the reason six orbitals hybridize is that a maximum of six oxygens can stably fit around it. Why is it not the same case for phosphorus?
Secondly, why does water $sp^3$ hybridize even though it has enough lone pairs in its p orbital to bond with 2 oxygens. Surely it would take more energy to pull up the lower $s$ orbital to bond than to allow the $p$'s, already elevated to that energy, to do the bonding?

The difficulty in talking about hybridization lies in the fact that it tends to confuse atomic and molecular orbitals, or (equivalently?) one- and many-electron wave functions.

Mixing atomic orbitals generally only becomes possible when we do a mathematical approach which treats more than one electron at once. That, however, is mathematically rather complex, and the resulting wave functions are molecular orbitals, which are entirely different from atomic orbitals. Molecular orbitals are constructed as combinations of atomic orbitals, which means that it is possible to e.g. add s and p wave functions together to obtain something that looks like sp hybrids. But, and this is important, there are actually many different possibly ways for such combinations, and none of them should really be considered physical reality; and furthermore, this "mixing together" has to follow very strict mathematical rules which in the end affects the entire set of molecular orbitals at once. Describing molecular orbitals in terms of hybrid orbitals (or something similar) is possible if one does it thoroughly, in terms of a full and correct mathematical treatment. Still, even when applying the concept in this way, it is somewhat arbitrary, since it represents only one of many possible solutions (and depending on the problem, not even the best one).

In the simple and qualitative terms often introduced in chemistry (especially organic chemistry), it is a very crude concept which should be regarded with great caution and especially not as something that would be some sort of reality. Hybridization is a mathematical procedure for many-electron wave functions; nothing more, nothing less.

In more technical terms: The role of one- and many-electron wave functions in atoms is crucial here. Atomic orbitals are assumed to be hydrogen-like wave functions; in the one-electron atom, the orbitals are $\ell$-degenerate, so linear combinations of wave functions for different orbitals would mathematically still be solutions to the Schrödinger equation. In many-electron atoms, the degeneracy breaks down, so combining different orbital wave functions would be invalid, since they belong to different eigenvalues. We do, however, employ a many-electron Hamiltonian such as the Fock operator in this case, and unless I am mistaken, the electron-electron interactions lift this restriction. So, by some unitary transformation of the many-electron wave function of a single atom, I guess one might arrive at something like hybrid orbitals. Then again, this does not reflect physical reality.


As adjectives, not in a math context: 'different' vs 'differential' I looked up  differential, but still don't savvy why the books below use 'differential' rather than 'different' and vice versa.

Advanced Organic Chemistry: Part A: Structure and Mechanisms by Francis A. Carey, Richard J. Sundberg. p 211.

Electroanalytical Chemistry: A Series of Advances
edited by Allen J. Bard. p 195.

Tietz Textbook of Clinical Chemistry and Molecular Diagnostics - E-Book by Carl A. Burtis, Edward R. Ashwood, David E. Bruns. p 519.

"Differential" is a compact way of saying "difference between the" (or its equivalent).  For instance, in your first example, where the author writes "Chromatographic separations result from the differential interactions of the enantiomers...", what's meant is  "Chromatographic separations result from the difference between the interactions of the [two] enantiomers...".

I.e., chromatographic separation results from the fact that there is a difference between how the two different enantiomers react with the stationary phase (the solid column packing material).

The first definition given in your link applies here: "Of, showing, or depending on a difference; varying according to circumstances or relevant factors."


Can I use sodium hydroxide (NaOH) to remove oxides from metals? Sodium hydroxide dissolved in water is a very strong base. I think this could mean it could de-oxidate the metal oxides during re-ox reaction. Basically, I ask whether the following happens:
$$\ce{NaOH + Fe2O3 -> Fe + something}$$

I have read that it does reduce carbon dioxide producing sodium carbonate and water:

$$\ce{2 NaOH + CO2 -> Na2CO3 + H2O}$$

Would that work with metals? If yes, which ones?

Your second reaction isn't a redox reaction. When carbon dioxide is dissolved in water, two equilibriums are forms:

\begin{eqnarray}
    \ce{H_2CO_3 + H_2O &<=> HCO_3^- + H_3O^+} \\
    \ce{HCO_3^- + H_2O &<=> CO_3^{2-} + H_3O^+}
  \end{eqnarray}

Remember that water also is in equilibrium:

$$\ce{2H_2O <=> OH^- + H_3O^+}$$

Adding $\ce{NaOH}$ will — according to Le Chatelier's principle, the equilibriums are disturbed, and more carbonate ($\ce{CO_3^{2-}}$) is formed.

If you're interested in a nice redox reaction involving iron(III) oxide, aluminium is much better choice.


High concentration of ferric ion catalyst decreases rate of decomposition of hydrogen peroxide in the Fenton reaction I did a project on the Fenton reaction and rate of decomposition of $\ce{H2O2}$. I used colorimetry to measure the rate of decolourisation of methylene blue when I mixed $\pu{5e-6 mol}$ $\ce{FeCl3}$ and $\pu{0.2 mL}$ of different concentrations of $\ce{H2O2}$ (1, 3, 5, 10, 15, and 20%).

After processing the data, I found the following graph:

I was wondering why with higher concentrations of $\ce{H2O2}$ the rate decreases. I initially wanted to analyse this according to the Michaelis-Menten model (as $\ce{Fe^3+}$ ion can be regarded as a catalyst in this reaction). However, it seems my data goes in the opposite direction as the Michaelis-Menten plot. What surprised me most that it was decreasing.

I came up with a few hypotheses for this, and searched for papers, but I could not find any with results similar to mine. My main hypothesis was that because $\ce{Fe}$ ion and $\ce{H2O2}$ form $\ce{OH.}$ radicals and $\ce{OOH.}$ radicals (as reported here and here), and these radicals are what cause the decomposition of methylene blue, I thought that maybe if there was too high a concentration of $\ce{H2O2}$, there would be too many radicals, and this would in turn terminate the free radical reactions (when 2 free radicals meet). At lower concentrations, the radicals would continue propagating, hence decolourising methylene blue faster.

Moreover, I know that this is not a problem with the colorimeter, as I could see this decolourisation outside the colorimeter too. The following is a screenshot of a video I took; each reaction in the cuvette was started at the same time, and concentrations of $\ce{H2O2}$ used in order from left to right are 1, 3, 5, 10, 15, 20%.

I was wondering if anyone knew why this happened.

In this paper, a similar reduction in rate is observed at concentrations of H2O2 above 0.03 M final H2O2 concentration (which is about 0.1 wt %). They attribute it to $\ce{HO.}$ reacting with H2O2 (to yield $\ce{HOO.}$) more quickly than it reacts with MB. The result is that $\ce{HOO.}$ is much more abundant than $\ce{HO.}$ and its lower reactivity means a slower rate of reaction with MB.

Based on their data, I would recommend repeating the experiment with much lower H2O2 concentrations.


Enthalpy of Reaction and Efficiency of Electrochemical Energy Devices For electrochemical energy devices such as batteries or fuel cells working reversibly, efficiency is defined as: $$ \eta = \frac {\Delta_rG^⦵}  {\Delta_rH^⦵} $$ Since cell operates reversibly Gibbs energy change of the system is equal to electrical work done by the cell, so we can instead just use standard Gibbs energy of reaction as energy output. What I am not clear about is why is enthalpy of reaction used as energy input? Since electrochemical energy devices transform chemical energy of electroactive materials to electrical energy, energy input should be chemical energy, but I am not sure why enthalpy of reaction represents chemical energy? Enthalpy of reaction is given by this equation: $$ \Delta_rH^⦵ = \Delta_rG^⦵ + T\Delta_rS^⦵$$
Since cell operates reversibly, $\Delta_rG^⦵$ represents reversible electrical work (maximum non-PV work done by the cell) and $T\Delta_rS^⦵$ represents heat exchanged with surroundings to keep cell at constant T in reversible operation since change in Gibbs energy as measure of non-PV work is only applicable for closed systems at constant p and T.
Why is chemical energy sum of these two quantities physically speaking since I do know how to derive this equation from 1st and 2nd law of thermodynamics, but I am not sure in its physical interpretation in context of electrochemical cells?

The efficiency is  $\eta=\displaystyle \frac{w}{\Delta H}$ where $w$ is the useful work output, and $\Delta H$ the reaction enthalpy. This can be written as $\displaystyle \eta=\frac{(\Delta H-Q_{in})}{\Delta H}$ and $Q_{in}$ is the heat added where $Q_{in}=T\Delta S$ when operated reversibly giving the efficiency equation you quote. In effect the efficiency accounts for heat losses or gains relative to the reaction energy available. Gibbs was first to realise that because heat from the surroundings can be added to or removed in a reaction the work done can be greater (or less than) than $\Delta H$. In your definition the efficiency can therefore be $\gt 100$%.


Temperature & time necessary to carbonize all organics Suppose that I have a cast-iron pot which my awful roommate has vigorously scrubbed with soap and left to soak in soapy water over the weekend. The seasoning has been destroyed and replaced with awful-tasting soapy contaminants.

It seems to me that the simplest and best way to deal with this is to heat the pot over a campfire until only black carbon remains, then clean and reseason.

Not knowing exactly what is in my awful roommate's dish soap, and being a very fussy foodie, I want to be conservative here... so how hot would I have to heat the pot, and for how long, to be quite sure that virtually nothing but carbon remains?

Edit: Compare using the self-cleaning cycle of an oven for similar purpose, reported to work well...

There's a lot of exciting, and not well defined, chemistry that happens with seasoning cast-iron cookware.  Essentially, you are creating a cross-linked polymer film that is hydrophobic, providing a "stick-free" surface.  There are many many websites claiming the "best way" to season (or re-season) cast-iron cookware.  Few of these sites (and I'll let the OP and visitors search for those on their own) are based on science.  (Although I am not going to question tried and true family secrets, passed down from generation to generation.)  One site in particular makes a fairly convincing argument for seasoning that is science-based.  I suggest this approach can help restore your pot.

I suspect what has happened is that the scrubbing did remove the seasoning and exposed base metal to water and oxidation - your pot rusted.  The "soapy contaminants" is probably iron oxide rather than something from the dishwater, although there are such products as metal soaps which are metals chelated to long-chain fatty acids.

As for heating the pot to burn off all the residue - I doubt this is the most effective use of your time.  Scrubbing with acidified water to remove the iron oxide and a long seasoning session with an easily cross-linkable oil should get your pot back to (or better than) original condition.


Why add H2O when calculating the pH of an acid? My question is very easy.

Case 1: I have to calculate the $\mathrm{pH}$ of $0.01\ \mathrm{M}$ of $\ce{HNO3}$ solution.

This is my solution:

Case 2:I have to calculate the $\mathrm{pH}$ of a $0.01\ \mathrm{M}$ $\ce{H3PO4}$ solution.
This is the provided solution:

My question is why do we add the watermolecule in our calculations in the second case and not in the first? (or vice-verca)

Based on what I can make out, you did not use the concentration of $\ce{H2O}$ in either of your calculations.

In the second case, you wrote it as a part of your reaction equation. Now, the reason for doing this becomes clear when you realise that when you are talking about a $\ce{H+}$ ion, you are in fact referring to a "proton"-- a bare positive charge. Moreover, a proton is extremely tiny, and thus consequently the charge density is quite high.

This makes it extremely reactive (in a sense), and thus in a chemical system of any sort would immediately seek out and associate with the electron clouds of a surrounding molecule.

In an aqueous solution an obvious, readily available target is the water molecule, and you end up with hydronium ($\ce{H3O+}$) ions.

In these particular problems, the solvent i.e water doesn't have a significant effect on the equilibria, however, this may not always be the case; which is why I recommend writing your reaction equations as you did in the second case, as it allows you to identify the acid/base, and conjugate acid/conjugate base in the reaction.


How to Determine Acid Strength? The correct answer is E

But I do not understand why?
Shouldn't (v) have the lowest $\mathrm{p}K_\mathrm{a}$ since it does not have a carboxyl group making it a strong acid? How do we determine the strength of acids?

First of all, let's make something crystal clear: The more acidic is a compound, they smaller will be its $pK_a$.

To evaluate the acidity of these compounds you need to analyze the stability of the corresponding conjugate base of each one, in effect, the stability of the same compound but without a proton and a negative charge on an oxygen atom.

First of all, you have two kinds of compounds:

Two alcohols: V and II
Two carboxylic acids: I, III and IV

From those compounds, alcohols will always be less acidic than carboxylic acid. Why? Because the negative charge left by a proton leaving the molecule can be delocalizaed by resonance troughout the carbonyl, but in the case of the alcohols it will be localized on the only oxygen atom, which makes the conjugate base less stable, and therefore, the compound less acidic.

That being said, from the two alcohols, V will be more acidic, because the presence of the chlorine atoms substract charge from the carbon they are bond to (chlorine is more electronegative than carbon), and helps stabilizing the negative charge that is left in the molecule after the proton is substracted.

In regard to the carboxylic acids, you can use a similar reasoning for the carboxylic acid.
The less acidic is the one in which R is just a hydrocarbon chain, which has no effect in contrast to I, which has 1 chlorine atom that stabilizes the negative charge, and to III, which has 3 chlorine atoms that stabilize the negative charge of the conjugate base even more.

Therefore, the order presented in option E is justified.


Is trans-1,2-dimethylcyclopropane optically active or not? According to my book it is optically active, but no valid reason is given. But what I found is there exists an alternating axis of symmetry, and so my conclusion is the compound would be inactive or has a meso isomer. Am I wrong?

First off ...
There is a difference between "chiral" and "optically active".  Something that's optically active has to have chiral molecules.  However, something can be chiral but if you have equal amounts of the mirror-image isomers, or if molecular motion can make an equal number of isomers like an umbrella-like inversion of a lone pair or an exchange of protons, then you do not see optical activity.  All we are looking at here is whether the molecule is chiral, not whether a sample of the stuff is in general optically active.
With that cleared up:
As Weijun says in the comments you have to have a mirror plane, or an improper rotation axis, or else the molecule is chiral.  A (proper) rotation axis does nothing in this regard.
Each of the carbon atoms with a methyl group on it is, in fact, a chiral center.  The four different groups attached to it are:
No 4.  A hydrogen atom
No 3.  The methyl group that's attached outside the ring
And...
No 2, No 1.  The ring going clockwise and going counterclockwise, as viewed from "above" the C–H bond, are counted as two different attached groups, as if the ring were "cut off" in either direction before rejoining the carbon atom you're looking at.
If you look at the ring attachment that way, you see that the two different directions rank 1 and 2, in some order depending on where the second carbon with the methyl group is, hence my numbering above.  So you can figure out an R or S configuration for each chiral center.  In the cis isomer you end up with one R and one S (thus mirror-image configurations), and a mirror plane does go between them making a meso rather than a chiral molecule.  But the trans isomers give both R or both S, which never allows a mirror plane or any improper rotation axis, and so they are chiral.


Why does the stability of ascorbic acid decrease (increased rate of oxidation to dehydroascorbic acid) when pH increases? I am doing a Chemistry IA on the stability of ascorbic acid in different pH, and found that the stability decreased (there was a lower final concentration of ascorbic acid) as pH increases. Why is this so?

The decrease of ascorbic acid is probably mostly due to oxidation.

Oxidation is loss of electrons.

Increased pH means more ascorbate, thus negative charge and higher electron density.

Higher electron density means easier loss off electrons, thus more oxidation and less ascorbic acid at the end.


A IChO-2014 problem involving reaction mechanism This is a problem of IChO-2014(&solution).But I am having trouble working out reaction mechanism of the second reaction or even how the product B formed. This reaction is somewhat new to me. And googling didn't help much.

Just need a little hint on this regard.

A to B is a classic hydroboration of the least hindered double bond followed by oxidation of the trialkyl borane.


How to find and use the Clausius-Clapeyron equation I know how to get the equation from the Clapeyron equation but I have a question regarding a the integration along a phase boundary and a small step in the derivation that I will make clear when I reach that step. Firstly, the Clapeyron equation:

$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta S}{\Delta V}$$

Or alternatively, by recognizing that $\Delta G=0$ when two phases are in equilibrium; $\Delta G =\Delta H - T\Delta S$ can be rearranged to give:

$$\Delta S = \frac{\Delta H}{T}$$

Substituting this into to the the first equation gives:

$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta H}{T\Delta V}$$

At a solid/gas or liquid/gas phase phase boundary, it is a reasonable to approximate that $\Delta V \approx V_{m,\text{gas}}$ as $\Delta V = V_{m,\text{gas}}-V_{m,\text{condensed}}$ since $V_{m,\text{gas}} \gg V_{m,\text{condensed}}$

Consequently, using the equation of state for a perfect gas and substituting for $\Delta V$ in the second form of the Clapeyron equation, the following result is obtained:

$$\frac1p \frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\mathrm{d}\ln p}{\mathrm{d}T}=\frac{\Delta H}{RT^2}$$

However, why is $\frac1p \frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\mathrm{d}\ln p}{\mathrm{d}T}$?

The following step is the step that I don’t understand In my lecture handout this step is shown as:

$$\int_{p_1}^{p_2}\frac1p\mathrm{d}p=\int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\mathrm{d}T$$

Why can the left hand side be integrated with respect to $p$, yet the right hand side be integrated with respect to $T$. I cannot make sense of this. Other instances in thermodynamics when integration is used like this, both sides are integrated with respect to the same variable. For example, finding how $H$ varies with $p$. I will quickly go through what I mean without much verbal explanation:

$$\mathrm{d}H=T\mathrm{d}S+V\mathrm{d}p=\left(\frac{\partial H}{\partial S}\right)_p \mathrm{d}S+\left(\frac{\partial H}{\partial p}\right)_S \mathrm{d}p$$

Consequently, for one mole of a perfect gas:

$$V=\left(\frac{\partial H}{\partial p}\right)_S=\frac{RT}p$$

Thus:

$$\int_{p_1}^{p_2}\left(\frac{\partial H}{\partial p}\right)_S\mathrm{d}p=\int_{p_1}^{p_2}\frac{RT}p\mathrm{d}p$$

Since $\left(\frac{\partial H}{\partial p}\right)_S\mathrm{d}p=\mathrm{d}H$ at a constant $T$ (I think this is right – please clarify)

$$\int_{p_1}^{p_2}\mathrm{d}H=RT\int_{p_1}^{p_2}\frac{1}p\mathrm{d}p$$

This can obviously be integrated quite easily but this only illustrates my point regarding the variable that the integration is carried out with respect to. Surely it must be the same on both the RHS and the LHS of the equation?

See Integration by substitution, see also this question and another question at math.SE. This is the rule we have to apply here:

$$\int_{x_0}^{x_1} f(u(x))u'(x)\,\mathrm dx = \int_{u(x_0)}^{u(x_1)} f(u)\,\mathrm du $$

So we go from this equation:
$$\frac1p \frac{\mathrm dp}{\mathrm dT}=\frac{\mathrm d\ln p}{\mathrm dT}=\frac{\Delta H}{RT^2},$$
and integrate over temperature:
$$\int_{T_1}^{T_2}\frac1p \frac{\mathrm dp}{\mathrm dT}\,\mathrm dT = \int_{T_1}^{T_2} \frac{\Delta H}{RT^2}\,\mathrm dT,$$
Now we use the first equation, with $f=\frac{1}{p}$, $p=u$, and $x=T$:
$$\int_{T_1}^{T_2}\frac1p \frac{\mathrm dp}{\mathrm dT}\,\mathrm dT = \int_{p_1(T_1)}^{p_2(T_2)}\frac1p\,\mathrm dp=\int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\,\mathrm dT$$

Bonus question: Its just the application of the chain rule, namely
$$\frac {\mathrm d}{\mathrm dx}z = \frac {\mathrm dz}{\mathrm dy} \frac {\mathrm dy}{\mathrm dx} $$
which, if applied leads to
$$\frac{\mathrm d}{\mathrm dT}\ln p = \frac{\mathrm d\ln(p)}{\mathrm dp} \frac{\mathrm dp}{\mathrm dT} = \frac{1}{p}\frac{\mathrm dp}{\mathrm dT} .$$


Why do Grignard reagents react with epoxides but not THF? THF is used as a solvent for Grignard reagents, but epoxides react with Grignard reagents. There is only a small difference between epoxides and THF, what causes the difference in reactivity towards Grignard reagents?

The difference between THF and epoxides is perhapse bigger than you might think. Due to the three-membered-ring of an epoxide, a lot of energy is 'stored' in ring strain. This destabilizes the stucture and is a driving force for a Grignard reaction (or any substitution on an epoxide).

The THF is a five-membered-ring, which is far more stable. This bond is simply an ether, that happens to be in a cyclic structure. Therefore it also reacts like an ether: not easily attacked by nucleophiles.


To which block of the periodic table does helium belong? According to quite a few books I have read, helium belongs to the p-block with reasoning along the lines of "it's an inert gas, so let's clump it together with other noble gases which are in the p-block."

However, on Wikipedia, helium is classified as an s-block element which, to me, makes sense because the last (or so-called "differentiating") electron that is added to helium will occupy an s sub-shell.

But my teachers are adamant that helium is a p-block element, and the fact that our textbook covers helium under the chapter "The p-block Elements" just gives them more reinforcement of that idea.

So: is helium an s- or a p-block element? Could someone provide concrete reasons for and against the block selection?

Blocks in the periodic table should not be mixed up with groups (like noble gases). The reason why Helium is considered a noble gas is because its outermost (and at the same time only) shell is fully occupied by its 2 electrons. When you look at the electron configurations in the PTE you can see that the first element which happens to have an occupied p-orbital is boron. Due to this fact boron also has to be the first p-block element, since the block definition is as follows:

  An element belongs to the block x, if the electron, which is added in comparison to the previous element (by atomic number), is in an orbital of type x in the ground-state electron configuration of that element.

Or the fast explanation: Why should helium be considered a p-block element, if it doesn't even have any electron in a p-orbital or any shell including a p-orbital?