Non-existence of time operator in quantum mechanics In quantum mechanics, there is no operator for time instead of this we  we have $\psi(t)$ which is related to $\psi(0)$.

In a book, I read that if there is an operator for time, then it will not be Hermitian".
I am not able to understand how exactly why? What is the reasoning behind this statement.

On the not Hermitian point.

Let $\hat{F}$ be a non-stationary observable (but does not depend explicitly on time $t$, hence $\partial \hat{F}/\partial t = 0)$. in the Heisenberg picture it satisfies the Heisenberg equation of motion for $\hat{F}$, namely
$$i \frac{d \hat{F}}{dt} = [\hat{F},\hat{H}] \tag{1}$$.

Now assume that whenever $d \hat{F}/dt \neq 0$ then it has an inverse
$$(d \hat{F}/dt)^{-1} = i([\hat{F},\hat{H}])^{-1}\tag{2}$$
Now, if
\begin{align}
[(d \hat{F}/dt)^{-1}, \hat{H}] &= 0 \\
\implies [([\hat{F},\hat{H}])^{-1},\hat{H}] &=0.
\end{align}
Now introduce a Time operator, $\hat{T}$ as
\begin{align}
\hat{T} &= \frac{1}{2}\left(\hat{F}\left(\frac{d \hat{F}}{dt}\right)^{-1} + \left(\frac{d \hat{F}}{dt}\right)^{-1} \hat{F}\right) \\
&=\frac{1}{2}\left(\hat{F} ([\hat{F},\hat{H}])^{-1}+([\hat{F},\hat{H}])^{-1}\hat{F}\right)\tag{3}
\end{align}
Now using equations $(1)$ and $(2)$ into $(3)$ one finds that
$$[\hat{H}, \hat{T}] = i \tag{4}$$

That is, $\hat{T}$ and $\hat{H}$ form a canonically conjugate pair. Owing to the fact that the Hamiltonian spectrum is bounded from below, the time operator $\hat{T}$ is not self adjoint, in other words, not Hermitian.


Why is the dz2 orbital so different from the rest? What makes dz2 orbital so special?

Although degenerate with other d orbitals,
It has no nodal planes, instead it has  2 nodal "cones".

Instead of having 4 lobes, it has 2 lobes and 1 ring.

Also, its electron density is prominently distributed in all x,y and z directions unlike others.

I know the wave function is what determines the shape, but what makes this particular orbital different?
Is there any fundamental reason?

The wikipedia is helpful in explaining why radial variations should arise in the density of non-s orbitals:

  The non radial-symmetry properties of non-s orbitals are necessary to localize a particle with angular momentum and a wave nature in an orbital where it must tend to stay away from the central attraction force, since any particle localized at the point of central attraction could have no angular momentum.

What's unique about the $d_{z^2}$ orbital (see the table above, from the wikipedia) compared to the other $l=2$ angular momentum wavefunctions is that the z-component is zero ($m=0$). This further constrains the geometry of the wavefunction.

The functions describing the angular dependence of the hydrogenic wavefunctions are Legendre polynomials $Y_{lm}(\theta,\phi)$, solutions of Legendre's differential equation. In the case of d-orbitals, they satisfy

$$ \hat{L}^2Y_{lm}(\theta,\phi)=\hbar^2l(l+1)Y_{lm}(\theta,\phi)$$

with $l=2$, where $ \hat{L}$ is the angular momentum operator. Since the z-component of the angular momentum is also quantized, the following eigenequation also holds:

$$ \hat{L}_zY_{lm}(\theta,\phi)=\hbar mY_{lm}(\theta,\phi)$$

with $m=0$ in the case of the $d_{z^2}$ orbital, and this last equation leads to the following condition:

$$ \frac{\partial\psi}{\partial y^2} = \frac{\partial\psi}{\partial x^2}$$

which implies that the solutions must be cylindrically symmetric about z. However, the condition $l\neq 0$ implies that the solution is not spherically symmetric. The result is the unexpected shape of the $d_{z^2}$ orbital.


Heisenberg uncertainty principle and lifetime of the excited state A key relation in optical spectroscopy also used to derive natural bandwidth is:

$\Delta E \Delta t=ℏ$

$\Delta E$ is the energy uncertainty of a state, $\Delta t$ should be the uncertainty of the lifetime $τ_b$ of the state. I don't understand why $\Delta t$ can be confounded with the lifetime itself, $τ_b$, and the relation written as

$\Delta E τ_b=ℏ$

For practical purposes the relevant aspect of the relation between the uncertainties in time and energy is that it is a Fourier relationship that explains how peak widths in a spectrum (with linewidths in frequency units) relate to the decay rates of time-dependent signals. A decay can be considered the statistical result of many independent events with an average probability of occurring described by a time constant $\tau$. It can further be shown via the Fourier transform that when a signal decays exponentially with time constant $\tau$, this leads to a Lorentzian peak shape with a width at half height proportional to $\tau^{-1}$. It is even simpler to show this in the case of a Gaussian shaped decay with linewidth ~$\sigma_t$. The resulting frequency peak is then Gaussian with linewidth ~$\sigma_t^{-1}$. When you multiply the linewidth parameters of the Gaussian functions you get:

$$\sigma_f\sigma_t=1$$

If you use Planck's relation $E=h\nu$ if follows that

$$\sigma_E\sigma_t=h$$

Now - as mentioned earlier - you can think of the width of the Gaussian or Lorentzian time-dependent decay as proportional (or described by) a time constant $\tau$ (which is the default description in the case of exponential decays). It follows that

$$\sigma_E~\tau \approx  h$$


Is there a consensus how to report coupling patterns greater than quartets? When analysing NMR spectra, every student at one time during their chemistry studies learns about vicinal coupling and that it produces the characteristic signal shapes of proton spectra. The following multiplets are usually taught and considered consensus:

singlet (s)
doublet (d)
triplet (t)
quartet (q)
multiplet (m)

But that doesn’t even adequately cover all hydrocarbons; quintets and septets need at least be considered. If, furthermore, there are heteroatoms with a half-integer spin of larger than $1$ present, not even these will suffice.

In my lab experience, I often came across a septet of doublets, which ACD termed sptd. However, I just found an internet source proposing the abbreviation sepd.

Is there a consensus on how coupling patterns with more than four of one coupling should be abbreviated? If so, what are the official abbreviations?

There is no absolute consensus for reporting abbreviations for splitting patterns. Even Magnetic Resonance in Chemistry, a journal that I generally take as the best guide for reporting NMR data, does not have recommended abbreviations for multiplicities above quartets. Further, there are general inconsistencies in reporting formats within this journal, with abbreviations such as s, sept, spt being used for septet. Additionally, the actual term used to describe the splitting varies. For instance, both pentet and quintet are commonly used to describe a 1:4:6:4:1 splitting.

What is consistent, however, is that the abbreviations are defined within the paper.

Commonly used terms and abbreviations that I have come across include:

singlet (s)
doublet (d)
triplet (t)
quartet (q)
pentet (p, pent, pnt, pentet) or quintet (q, quint, qui, qnt, quintet)
sextet (sext, sxt, sextet)
setpet (s, sept, spt, septet) or heptet (hept, hpt, heptet)
octet (oct, octet)
nonet (non, nonet)

While splittings of higher multiplicity are less common within mainstream CHN organic molecules (septet for isopropyl substituents are a common exception, and 'apparent' splittings are excluded), they are more common with heteronuclear substitutions and organometallic systems, where spin numbers greater than 1/2 are frequently encountered. Reporting 'apparent' splittings of, say, (CH3)2CHCH2R as a nonet (where Jab ~ Jac) is not strictly correct according to the 2nI+1 rule, although in non-NMR journals still occurs, and is certainly taught at undergraduate NMR level.

As for reporting multiple couplings (doublet of septets, for example), again there is no strict rule to adhere to, although the commonly adopted practice is to report the larger coupling first. So, a doublet coupling of 10Hz and septet coupling of 2Hz would be a doublet of septets. If it were doublet (2Hz) and septet (10Hz), one would report a septet of doublets.


At what temperature (in kelvin) are most of the elements on the periodic table liquids? This question is out of pure curiosity.

At what temperature are a majority of the elements on the periodic table in a liquid state/phase of matter? For the purpose of this question, assume the standard pressure (100,000 Pa). Is there some sort of formula or theorem that explains this? Please give a detailed explanation on your conclusion.

Now that's a great question indeed! Evidently, at 0K all elements except helium are solids, at 10000K they are all gases, so someplace in between the number of liquids must reach a maximum; what and where might that be?

Well, there is no formula or theorem that says liquid hydrogen must boil at 20K, nor is there such a thing for any other element, so this line of thinking hardly provides any insight. Let's try another way.

Let's take a look at all elements' melting and boiling points. (I used the great site referenced in a comment by ringo, where all data are conveniently stored on line 47 in the file http://www.ptable.com/Static/interactivity-2fcd37b.js). Let's simply make a joint list of all melting and boiling points, sort it (keeping track of which is which), and then walk through it from bottom to top, calculating the number of liquids on each interval. If you know some kind of programming, you can do it with a script.

It turns out that the maximum number of simultaneously liquid elements is 49, and it is reached several times between 2183K (melting of vanadium) and 2792K (boiling of aluminium). By dragging the slider you may actually see which elements are included in these 49. Now, the data are not terribly reliable (for example, in higher actinides they claim some melting points obtained by God-knows-what extrapolation, but no boiling points at all), yet still this is something to think about.


How can I choose the correct mechanism for nucleophilic substitution of aryl halides? I was reading about nucleophilic substitutions of aryl halides, and there are three different mechanisms discussed, namely the SN1, SNAr (addition-elimination), and benzyne (elimination-addition) mechanisms.

However, I am confused as to how to identify which mechanism is operative. How can I tell this from the reaction conditions and reagents?

Before comparing the three main mechanisms, I will first explain why aryl halides typically do not undergo SN2-type reactions. Take bromobenzene as a typical example of an aryl halide. A nucleophile, like hydroxide ion, cannot simply displace the bromide from bromobenzene. Quoting from Organic Chemistry (1st ed., p 589) by Clayden et al.,

  In the aromatic compound, the C–Br bond is in the plane of the ring as the carbon atom is trigonal. To attack from the back, the nucleophile would have to appear inside the benzene ring and invert the carbon atom in an absurd way. This reaction is not possible!

In rare cases, SN1 reactions can take place; these characteristically occur when there is an excellent leaving group on the arene. One of the few examples is in the case of an aryldiazonium ion, where the corresponding leaving group is nitrogen gas.

Note that the presence of a diazonium ion does not necessarily imply a SN1 mechanism; there are also other possible mechanistic pathways, mainly involving radical chemistry.

The next possibility is that of an addition-elimination reaction, commonly termed SNAr. This is made possible by electron-withdrawing groups (e.g. nitro or carbonyl groups) when they are ortho and/or para to the halogen, as the negative charge in the intermediate can be delocalised into these electron-withdrawing groups:

Finally, the benzyne mechanism. Consider bromobenzene: it cannot undergo SN2 reactions (as explained above), SN1 reactions (bromide is not a sufficiently good leaving group), or SNAr reactions (no electron-withdrawing group ortho or para to bromine). However, under strongly basic conditions, it is possible to eliminate HBr by abstracting a proton ortho to the halide. Examples of such strong bases include sodium amide (NaNH2), LDA/lithium diisopropylamide (LiNi-Pr2), or potassium tert-butoxide (KOt-Bu).

To summarise the three categories:

If you have a very good leaving group - primarily nitrogen gas - on the aromatic compound, then a SN1 mechanism is possible.
If you have good electron withdrawing groups ortho or para to the halide, then it is likely an addition-elimination or SNAr reaction.
If the above two are not possible, and if you have an extremely strong base such as NaNH2 or KOt-Bu, then the benzyne mechanism may be applicable.


Red Squares in RDKit? I'm newish to RDKit and couldn't find an answer to this online.  The following is a minimal representation of how I drew a molecule

from rdkit import Chem
from rdkit.Chem import Draw as Draw
from rdkit.Chem.rdMolDescriptors import CalcMolFormula

# .. read mol_a from file... then
print(repr(mol_a))
<rdkit.Chem.rdchem.Mol object at 0x7fd5c5494dc0>

print(CalcMolFormula(mol_a))
C4H11N

Draw.MolToImageFile(mol_a, 'pic.png')

Here is what I get:

What are those red boxes?

I'm running Ubuntu 16.04, Python 3.7.3

I had three semesters of college chemistry about a decade ago.

Thanks!


What are potential sources of low recovery in this acid-base extraction experiment? I did an acid-base extraction experiment. First extracted a solid mixture containing benzoic acid, p-nitroaniline, and naphthalene into three separate solutions:

NaOH extract
HCl extract
3.organic with the neutral compound.

Then the HCl and NaOH extracts were neutralized and underwent vacuum filtration to isolate crude p-nitroaniline and crude benzoic acid, while the organic solution underwent rotary evaporation to get crude naphthalene.

Crude benzoic acid was purified by recrystallization using hot filtration, and then benzoic acid crystals were isolated by suction filtration. The melting range of the benzoic was relatively pure.

How can I improve this experiment? My recovery of benzoic acid was less than 40%

And as for sources of error, extraction processes always lose some compound during each step. What are other probable sources of error?

I can only assume that you we following some recipe for the purification of the benzoic acid. Frankly if I were doing this I would have done it differently.

Wikipedia shows that the solubility of benzoic acid is 1.7 g/L (0 °C) and 56.31 g/L (100 °C) in water. So I'd added benzoic acid to a test tube and put in a boiling water bath. I'd added just enough water to get the benzoic acid to dissolve and then let the liquid cool in air. When down to room temp, I'd put it into an ice bath. After 10 minutes in ice bath I'd pour off water. I'd wash with a few drops of ice cold water. I'd test for purity and if not good enough I'd repeat.

0.34 g of crude benzoic acid should dissolve in about 6 ml of boiling water.

Each water cycle should have an efficiency of close to (56.31 - 1.7)/56.31 = 97%


Combining Lead pipe with Zinc or Copper pipes in a domestic water installation Grateful if someone assists in explaining this example that appears on a high school text book. The problem can be summarized as follows:
"Old water networks supply water to a residential area via Pb lead pipes. A user wants to fit his own  installation with pipe made out of either Zn or Cu.  The intent is to reduce the lead presence in the water. Condition: the lead pipe of the municipality will remain unchanged up to the water gauge/meter at the house input.
Account for the presence of O2 in the water and give an explanation on your choice Zinc or Copper."
The solution suggested in the book shows Cu pipes, and excludes the use of Zn.
I would chose Zn. Reason would be that O2 would preferentially oxidize Zn instead of Pb. Thus preventing the formation of Pb oxide/hydroxides and consequent leaching in the water.
Thought I overlooked something and looking for an explanation to the book' choice which I assume is correct.. thanks for your support

The focus of the analysis should be first on the anodic index (see, for example, table here), which relates to the electrochemical corrosion potential of the metal serving as the anode. Second, materially important is the current state of the old municipality Lead pipes.
Per the source, the anodic index of Pb is high (see chart) above copper, so historically it has already electrochemically corroded likely inducing a protective Lead oxide coating in the presence of dissolved oxygen.
However, Zn metal is more anodic than the Pb anode, and as such, would readily corrode and may introduce high levels of Zn ions (even levels considered problematic/toxic).
As such, copper metal, which is lower in the anodic index relative to the Pb electrode, is likely a better choice to preserve the current state of affairs, which is in apparent agreement with the reputed cited textbook 'correct' answer.


Why is no bromine liberated in the reaction of potassium bromide and concentrated phosphoric acid? $\ce{KBr}$ reacts with concentrated $\ce{H3PO4}$ to give $\ce{HBr}$ and ($\ce{KH2PO4}$ or $\ce{K3PO4}$) (not sure which one, if someone knows it, please tell).

Why isn't bromine gas liberated?

Firstly, $\ce{H3PO4}$ is not a strong enough oxidizing agent to remove the electrons from the $\ce{Br-}$ ions in order for them to then form $\ce{Br2}$, which exists as a liquid at room temperature. Assuming we are working at room temperature, using a strong enough oxidizing agent such as $\ce{H2SO4}$ or a more reactive halogen like $\ce{Cl2_{(g)}}$ or $\ce{F2_{(g)}}$ would be enough to oxidize the $\ce{Br-}$ to $\ce{Br2_{(l)}}$.

If we desired to evolve $\ce{Br2_{(g)}}$, we would have to heat $\ce{Br2_{(l)}}$ to its boiling point of $58.8\mathrm{^oC}$ and then continue to supply it with enough heat to vaporize it. As docscience stated, however, some oxidizing reactions might be exothermic enough to vaporize $\ce{Br2_{(l)}}$ itself. In solution this would depend on $[\ce{HBr}]$ as to whether or not the heat evolved is enough to vaporize the $\ce{Br2_{(l)}}$, but in the solid form $\ce{Br2_{(l)}}$ is vaporized instantly in the extremely exothermic reaction of $\ce{KBr}$ with either $\ce{F2_{(g)}}$ or $\ce{Cl2_{(g)}}$.


Water extraction of ethanol from petrol We have now 10% ethanol mixed petrol. Seller says that if this petrol comes in contact even 10 ml of water, all ethanol which is mixed in the petrol will be extracted by water, is it true?

Water and ethanol are highly "miscible"; a science term that means virtually any amount of water is fully soluble in ethanol, and vice-versa.  The next thing to consider is that ethanol and gasoline are quite miscible. Think of E-10 (10% ethanol) all the way to the E-85 some petrol engines are designed to function well on (with  some loss in power output).
However, water is very poorly soluble in pure gasoline.
So adding when water to E-10 the water mixes well with the ethanol, but the H2O/ethanol mix looses solubility rapidly as more water is added.
In theory, if you set up a separation flask, basically a clear glass vessel with a petcock at the very bottom, filled it at room temp with a gallon of E-10 and added ~3 tablespoons of water, shook it all up then let it settle to separate, you could drain away the bottom (denser) layer of liquid, which would be the water/ethanol mix, an leave you with ~0.9 gallons of relatively alcohol-free petrol.
My final remark would be to ask why anyone would wish to do this as I can think of no reason in the world why someone would bother.


Bond order of H2O and NH3 using group theory to construct MOs The bond order of H2O is 2, and that of NH3 is 3, which makes sense, when considering the number of bonds they have.

However, if using basic group theory, you get the MO diagrams as presented in this paper:

http://www.scielo.br/pdf/qn/v35n7/v35n7a32.pdf

Upon examination, we see that in the case of H2O, there are two orbitals that are strongly bonding, one weakly bonding orbital, and one non-bonding orbital. If we then tried to apply the bonding order equation to this MO diagram, would we not end up with a bond order of 3 - despite the a1 orbital being only weakly bonding, its occupation still stabilizes the molecule relative to its constituent atoms. Is it the case that the concept of bonding order breaks down using these kinds of MO diagrams?

A further example is given using the NH3 MO diagram. Here, they calculate the bond order as 3, ignoring the fact that the NH3 a1 orbital is weakly bonding. If we were to in theory calculate the MO diagram for NH3 in a trigonal planar geometry (same as for BH3), we would also get the bond order as 3. Of course, the occupied a1 orbital is weakly bonding in the trigonal pyrimidal NH3 geometry, leading to a preference over the trigonal planar geometry, but why is this not reflected in the bond order?

It depends on your definition of bond order. If you are using the simple formula

$$\text{Bond order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}$$

then, yes, it won't be able to capture the subtleties that you describe. After all, when you use that, you're categorising electrons into bonding vs antibonding in a black and white manner. So it's really no surprise that it doesn't reflect "weakly bonding" MOs.

You gave the example of water. Well, for that weakly bonding MO, you need to make a choice: is that considered a bonding MO, leading to a bond order of 1.5 per O–H bond in water (doesn't make much sense)? Or is it considered a nonbonding MO, leading to a bond order of 1 (makes more sense but then you have this problem you brought up)? Or is it somewhere in between, leading to a bond order between 1 and 1.5 – and crucially, if this is the route you want to go down, how do you quantify the bonding character of the MO?

I don't have any experience using different definitions of bond order, but if you used one of these, I would not be surprised to see an O–H bond order that deviates ever so slightly from 1.


What will be the sign of work done? Consider the following question:

The enthalpy of vapour is $10\ \mathrm{kcal}/\mathrm{mol}$ higher
than liquid at $350\ \mathrm{K}$ and $2\ \mathrm{atm}$. Under given
conditions, the difference in internal energy of vapour and liquid
will be:
(a) $10700\ \mathrm{cal}$
(b) $-9300\ \mathrm{cal}$
(c) $9300\ \mathrm{cal}$
(d) $-10700\ \mathrm{cal}$

My solution is as follows:
Let the internal energy of the liquid be $U_\mathrm L$ and that of vapor be $U_\mathrm G$. We are required to calculate $U_\mathrm G - U_\mathrm L$. As per the question,
$\because H_\mathrm G = U_\mathrm G + W$, where $H_\mathrm G$ is enthalpy of vapor.
$10000 + U_\mathrm L = U_\mathrm G + W$
Where $W$ is the work done due to the expansion of vapor as the liquid changes state. The magnitude of work done $|W| = \Delta n_\mathrm gRT = 1 \times 2 \times 350 = 700\ \mathrm{cal}$.
Now the real question starts, what is the sign of $W$? As per the IUPAC convention, work of expansion is negative, hence $W = -700\ \mathrm{cal}$. So our answer will come out to be,
$10000 + U_\mathrm L = U_\mathrm G - 700$
$\implies 10000 + 700 = U_\mathrm G - U_\mathrm L$
$\therefore U_\mathrm G - U_\mathrm L = 10700\ \mathrm{cal}$
But the terse solution given to this question is:

As per the solution $W = +700\ \mathrm{cal}$, i.e. they have considered the work of expansion to be positive.
Where are we wrong? Or are we right?

Although the IUPAC convention is for work done by a system to be negative for expansion, many physicists and engineers use the opposite convention. What's important is that you use the same convention throughout your calculation. You wrote that H = U + W, which is true only if work done by the system is regarded as positive. To be consistent with the IUPAC convention that you used later in the calculation, this should be instead H=U - W.

On a side note, since work represents a change in energy, not a state function, this equation should be written in terms of changes: $\Delta H = \Delta U - w$. Also note that this equation is only valid for the condition of constant pressure.


What's the actual or official definition of activation energy? In some sources I have seen that activation energy is the minimum energy required to cause a reaction. This implies that activation energy doesn't change with temperature (not considering very high changes).
In other sources it is the difference between the energy required to cause the reaction & the energy of the reactants. This implies that activation energy is temperature dependent since the energy of the reactants change with temperature. So which is correct?
Also another source (khan academy's guardian chemist) says that "The official definition of activation energy is a bit complicated and involves some calculus."

Activation Energy is the minimum amount of energy needed extra than the average potential energy of the reactants to cause a reaction.And it does not change with the change of temperature, and for clarification the energy of the reactant molecules don't also change with the reaction. It is just the rate which changes with the reaction according to the arrhenius equation k=Ae^{-\frac{Ea}{RT}}So,the fraction of molecules having energy higher than activation energy changes with change in temperature. I think the easier definition can be stated like this


Definition of an amide If I asked you to draw me an amide, we'd likely end up with something in the form of:

The exact definition of this is given fairly clearly in the IUPAC gold book.

  Derivatives of oxoacids [...] in which an acidic hydroxy group has been replaced by an amino or substituted amino group. Chalcogen replacement analogues are called thio-, seleno- and telluro-amides. Compounds having one, two or three acyl groups on a given nitrogen are generically included and may be designated as primary, secondary and tertiary amides, respectively
  Source: IUPAC Gold Book, 1995 recommendations

There is however another class of compounds which we call 'amides', for instance LDA (lithium diisopropylamide). Again, the IUPAC gold book has this covered.

  The term applies also to metal derivatives of ammonia and amines, in which a cation replaces a hydrogen atom on nitrogen.

I fully accept this, my question is, why we decided to re-use the word 'amide' to describe these compounds. Is the reason historical (i.e. before we knew better), or is there a logical basis for the grouping of the two classes of compounds.

The latter amide definition in you question did not evolve from the former name of organic amides, but the process was rather opposite, or at least they have a common root. The name "amide" appears to be quite old. There is even a notion of amidogen, not used very often today, but it appears in the book Elements of Chemistry from 1840, along with some inorganic amides like "amide of potassium" or "amide of mercury". Later book Chemistry, Theoretical, Practical and Analytical as applied and relating to the Arts and Manufactures by James Sheridan Muspratt from cca 1860 besides inorganic amides mentions also a peculiar substance "margaramid" that was obtained as a byproduct of ammonium salts of fatty acids preparation by treatment of fats/oils with ammonia. In this time e.g. acetamide was known as well already.

And the definition in Cyclopaedia of Chemistry with Its Applications to Mineralogy, Physiology, and the Arts from 1854 states:

  Amide, Amidogen. The term amide is given to a class of substances which contain ammonia deprived of an atom hydrogen, (NH2.) Hence, when they are treated with potash, ammonia is disengaged, the third atom of hydrogen derived from water again uniting, so as to form the necessary quantity in ammonia. That such a substance as amide or amidogen exists, may be proved by passing dry ammonia over heated potassium or sodium. A green body is formed, as hydrogen is evolved, which is replaced by the metal, NH2H + Na, becoming NH2Na and H. Some of the most familiar of these amides are potassamide NH2K, sodamide NH2Na, oxamide NH2C2O2, benzamide NH2C14H5O2, sulphamide NH2SO2, carbamide NH2CO, platosamine NH2Pt. Some of these amides are formed by heating organic salts of ammonia, by which they lose water, and leave an amide. Thus oxalate of ammonia, when distilled, gives water and oxamide (NH3C2O3 = NH2C2O2 and HO). It is probable that many animal substances, as albumen, belong to this class.

(typos/mistakes included)

Please note that this is not an exhaustive historic research at all :)


Can dimethylmercury originate from cleaning up a broken thermometer with disposable wet wipes? I broke a CFL in my home (so elemental mercury is present in a small amount) and wiped up some of the mess with a clorox wipe.
The wipe in question contains dimethyl benzyl ammonium chloride and dimethyl ethylbenzyl ammonium chloride.

I'm not chemical-phobic, but can someone please tell me, I didn't accidentally expose myself to dimethylmercury? I realize that it is something you don't find outside of very specific lab environments.

Edit: Thanks everyone for the quick responses. It's been a long time since I took a chem class! Can you help me to better understand why this will not methylate the mercury?

For example, I had been cleaning my carpet at the same time with resolve (water, hydrogen peroxide, sodium lauryl sulfate, probably other ingredients too) and threw out paper towels soaked with it to clean the rug into the same bag as the CFL. Am I correct to assume that despite the methyl group on the SDS, it is also stable like the BZK and thus unlikely to methylate the mercury?

According to the US EPA, disposable wet wipes are a recommended tool for cleaning up a broken CFL. Although having a number of adverse health hazards at high concentration BZK is a common ingredient in many sanitizers, wet wipes and disinfecting wipes, where it occurs at very low concentrations; typically much less than 1%. These ingredients are not likely to react adversely with the released mercury.

For future reference to cleaning up broken CFLs, be sure to read this guide.


Calculating concentration of an ionic compound $n$ liters of $k$ M $\ce{NaCl}$ are poured into $m$ liters of $l$ M $\ce{NaCl}$. Determine $\ce{[NaCl]}$, $\ce{[Na^+]}$, and $\ce{[Cl^{-}]}$.

Keep it as simple as possible :)

If both initial solutions are not saturated and all the $\ce{NaCl}$ is dissolved, this will not change upon mixing.

If in doubt with the result of a calculation, have a look at the units!

Concentrations are given as the quotient of moles and volume, $c = \frac{n}{V}$, the unit apparently is $\mathrm{mol \cdot l^{-1}}$. The absolute amount of $\ce{NaCl}$ in the first solution is therefore given as $n = c\cdot V$. (Note that this is not the n in your question ;).

The new $\ce{NaCl}$ concentration is the sum of the moles, divided by the new total volume:
$c = \frac{(k\cdot n)+(l\cdot m)}{m+n}$

Unless given further information, it is safe to assume that $\ce{NaCl}$ still is fully dissociated after mixing.


Electronic effects of substituents in the benzyne mechanism On the topic of nucleophilic aromatic substitution (S$_N$Ar) reactions, the benzyne mechanism is discussed. When considering the position of substitution, electronic and steric factors are mentioned. When considering electronic factors, effects of substituents on the ring are viewed rather differently, compared to how they are viewed in the conventional electrophilic aromatic substitution reaction.

Pg. 525 from Organic Chemistry (2nd ed.) shows the following diagram:

In the discussion of the benzyne mechanism, conventionally electron-donating groups are said to not be able to exert their donating effects and are even viewed as electron-withdrawing via the inductive effect. However, I cannot see why they behave in this way. Why is their donating effect no longer significant? Certainly, their p orbitals are still in conjugation with the rest of the $\pi$ system right?

However, if their electron-donating effects were more important, then we would observe ortho and para attack by the nucleophile, so as to generate the negative charge at the meta position, which is apparently not the case as experimentally observed.

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc

the electron pair that is creating negative charge on ring is sp2 hybridized and its orbital direction is away from the pi orbital electronic cloud of benzene ring so not able to resonate along with pi electrons. OMe inductively stabilize this electron pair as it is electron with-drawing inductively. thats why NH2 group attaches at meta position rather than ortho position.


Concise description of primary organic reaction types mechanisms Long story short, I'm mostly doing calculations in the past 5 years of my scientific carrier and every year I get less and less confident in my knowledge of organic chemistry which besides was never perfect. Yeah, I still remember pretty well some trivial things like nomenclature, classes of compounds, etc., and I'm pretty comfortable with MO theory, hybridization, etc., but all this are, so to speak, just the foundations of the organic chemistry. What I'm really bad at are the applications, primarily, the reaction mechanisms.

Many times reading a paper on computational studies of a reaction mechanism I caught myself not being able to (relatively quickly) grasp the main idea. I spent a fair amount of time just to recall the very basics of, say, the $S_N2$ reaction. Back many years ago I knew this things. Maybe I knew them not so perfectly and systematically, but even this knowledge of organic chemistry had evaporated almost completely from my head.

So I'm looking for a good resource to refresh and improve my knowledge of organic chemistry primarily (read: almost exclusively) regarding the most common reaction types and their mechanisms. A resource is counted as a good one if:

It is goes straight to the point (reaction mechanism) without discussing the basics of organic chemistry I mentioned, rather assuming that the reader knows them.
It is (relatively) concise and concentrated on the basics. In other words, I have neither time nor interest to read 2000 pages long Advanced Organic Chemistry book with all the details, exceptions, dirty laundries, etc. Just the basics.
On the other hand the basics has to be discussed in sufficient details, i.e. I would not be satisfied with the relevant Wikipedia (or similar) pages.
It is trustworthy. I mean, well established authoritative reference.

I admit I might ask for to much, but who knows, maybe there is something I'm looking for out there.

P.S. The type of a resource does not matter: textbook, lecture notes, MOOC video course, whatever.

The book you are looking for is The Art of Writing Reasonable Organic Reaction Mechanisms by Robert Grossman.

One of the book's publishers offers this description of the book:

  Intended for students of intermediate organic chemistry, this text shows how to write a reasonable mechanism for an organic chemical transformation. The discussion is organized by types of mechanisms and the conditions under which the reaction is executed, rather than by the overall reaction as is the case in most textbooks. The treatment emphasizes unifying principles, showing how common mechanisms link seemingly disparate reactions.
  Each chapter discusses common mechanistic pathways and suggests practical tips for drawing them. Worked problems are included in the discussion of each mechanism, and “common error alerts” are scattered throughout the text to warn readers about pitfalls and misconceptions that bedevil students. Each chapter is capped by a large problem set.
  The author has drawn on his own research and the current literature to ensure that appropriate attention is given to topics across the range of modern organic chemistry. The text is unique in its inclusion of a chapter on reactions mediated or catalyzed by transition metals, an area in which mechanistic understanding is now essential. Relatively new topics such as olefin metathesis and cycloaromatization are covered without giving short shrift to more traditional areas such as carbonyl chemistry. The text assumes a basic knowledge of organic chemistry. It can be used either in a formal course or by students working on their own, and will be particularly useful for graduate students studying for qualifying examinations. It will also be useful to students and researchers in biochemistry, pharmacology, and inorganic chemistry.

Each of the book's chapters are $50 \pm 30$ pages and are packed full of example problems.  Each chapter is organized according to the class of mechanism, not around particular reactants or products.  The chapters are:

The basics.
Polar reactions under basic conditions
Polar reactions under acidic conditions
Pericyclic reactions
Free-radical reactions
Transition-metal-catalyzed and -mediated reactions
Mixed mechanism problems

The author maintains a good web site with additional problems, additional worked examples, and answers to every problem in the book.  You can get a sense of the content of the book by browsing Amazon or by looking some of the answer .pdf files, available for free from Springer.

I found this book very helpful in improving my mechanistic understanding of various types of reactions.


What is the difference between an ion-induced dipole attraction and just an ion-dipole attraction? By what I've read, an ion-induced dipole attraction is a weak attraction that results when the approach of an ion creates a dipole in a non-polar atom or in a non-polar molecule by disturbing the arrangement of electrons in the non-polar species. So this is a dispersion/van-der-Waal force, I believe?

On the other hand, an ion-dipole attraction is a weak attraction that results when the approach of an ion interacts with the dipole in a polar atom or in a polar molecule. And so this wouldn't be a van-der-Waal force would it?

If what I said above happens to be true, does the same apply with dipole-induced dipole forces and just dipole-dipole forces?

In this scenario, a dipole-induced dipole attraction would be a weak attraction that results when the approach of an dipole (polar covalent bond) creates a dipole in a non-polar atom or in a non-polar molecule by disturbing the arrangement of electrons in the non-polar species. So this is a dispersion/van-der-Waal force too, right?

And a dipole-dipole attraction would be a weak attraction that results when the approach of an dipole that already exists  in a polar bond attracts with another dipole in another polar atom or in polar molecule.

It is useful to note the definition of a van der Waal force: weak electric forces attracting neutral molecules to each other in gas thus causing a gas to deviate from an ideal gas.

So, there are three types of these forces

Dipole-dipole force
Dipole-induced dipole force
London dispersion force

It becomes clear that forces involving ions do not fit van de Waals forces as the molecules involved must be neutral.

An ion-induced dipole force is not a dispersion force, either (as dispersion forces are a subset of van der Waals), and with ion-dipole forces, ion-induced dipole forces have a category of it's own

A dipole-induced dipole force is a van der Waal force but not a dispersion force


Why do gases under low pressure conduct electricity? In J. J. Thomson's cathode ray experiment, he used a discharge tube in order to make the gas inside conduct electricity.

Why do gases under low pressure conduct electricity?

The cathode ray tube doesn't conduct electricity per se. It liberates free electrons that are simply moving through a vacuum from one end to the other. In practice, you can't create a real vacuum, so a low pressure gas (as low as possible) is used instead.

Air has a high enough dielectric strength that it will dampen the effect of the applied voltage. That is, it'll be harder to liberate free electrons to shoot down the tube.


Obtaining number of functional groups in a compound from SMILES, SDF or MOL2 format I'm looking for a tool that can give how many methyl groups, hydroxyl groups etc are in a compound's structure presented as SMILES, SDF or MOL2 format.

There are numerous programs, but it's pretty easy to use the obgrep tool in Open Babel to do this using SMARTS for the various functional groups:

obgrep -c '[CX3]' file.sdf # count the number of methyl groups
obgrep -c '[#6][OX2H]' file.sdf # count the number of hydroxyl groups

etc. Examples of functional groups using SMARTS can be found on the Daylight SMARTS examples page.

I've also written scripts in Python for these purposes - depends on how you want to run the SMARTS searches.


Determining which compound is more 'ionic' I was going through my chemistry textbook (IB Pearson), and it explicitly stated that the higher the absolute difference between the electronegativity of elements in a binary compound, the more 'ionic' that compound is, which does make intuitive sense given that a higher electronegativity difference leads to the compound being more 'polar.'

However, the book came up with the following example:

  Which of the following pairs will form the most ionic compound?

  1) Be and F (the electronegativity difference is $|4-1.6| = 2.4)$
  2) Si and O (the electronegativity difference is $|3.4-1.9| = 1.5)$
  3) N and Cl (the electronegativity difference is $|3.2-3| = 0.2)$
  4) K and S (the electronegativity difference is $|2.6-0.8| = 1.8)$

  $\ce{K2S}$ will be the most ionic compound as the difference in electronegativity is the greatest.

I don't get how this works, the difference in electronegativity between beryllium and fluorine is visible greater, so why does the book say that the binary compound formed by K and S is more ionic?

Tl, dr -- upon further review we can't really tell whether potassium sulfide or beryllium fluoride is more ionic without some quantitative details.

There is more involved in the balance between ionic and covalent bonding than electronegativity differential.  You have to consider molecular orbital structure, too.  One of the answers to this question describes how molecular orbital structure can drive highly ionic bonding when electronegativity differences are modest (as in salts with aromatic ions) or even the "wrong way" (as in alkalides).

With single atoms, as pointed out in the other question, you cannot expect 100% ionic bonds even when molecular orbitals favor it.  Yet there can still be an impact which we can see by comparing organolithium compounds with organic fluorides.  The former acts as a ready source of anionic carbon, while the latter are notoriously unreactive towards the displacement of fluoride ions, as if the carbon-lithium bond had much more ionic character.  Yet the electronegativity differentials in the carbon-lithium and carbon fluorine bonds are about the same (oppositely directed with respect to carbon, of course).  The difference comes from lithium having a diffuse valence orbital that overlaps poorly with carbon, whereas fluorine is more compact and overlaps well, thus promoting electron sharing between both atoms in the bonding orbital.

Similarly, potassium has a diffuse valence orbital, even more so than lithium, which overlaps poorly with most nonmetals.  The ionicity of potassium sulfide, already considerable based on electronegativity, is thus enhanced even more.  Beryllium and fluorine, with good covalent overlap, would have a more covalent interaction than what might be expected from their electronegativity difference.

Since potassium sulfide has poor overlap and beryllium fluoride has a somewhat higher electronegativity difference, the two effects oppose each other when the compounds are compared.  We cannot tell from these qualitative arguments which one is more ionic.


Which chemical is used in drunken driver test? Which chemical is used in drunken driver test?

Acidified K2Cr2O7
KMnO4
Chromyl Chloride
H2O2

On searching on Google, I got the answer as option 1, acidified potassium dichromate, but my questions are:

How is the test administered? How much time does it take for the results?
On searching, I got a hunch that it had something to do with oxidation of alcohol, so, why can't we use the chemicals of the other options mentioned?
Is my hunch correct?

The test is administered by breathing through a tube, and the results become apparent immediately. Indeed, this has something to do with oxidation of alcohol. As soon as the alcohol in the exhale reaches the contents of the test tube, the latter changes its color, and the police guys put you in the handcuffs. Or it doesn't, and they don't. That's a wonderful example of chemistry improving our life.

Now to the point of comparing different reagents.

Permanganate itself is way too dark. When looking through its solution, you won't be able to detect any trace of any other color, let alone the products of reduction. Besides, the solution is rather sensitive and won't last long.
Chromyl chloride is too dangerous. It is liquid, and the suspect is going to breath in some tiny droplets (and/or possibly vapors) of it, and that's not something you would want to happen to a human being.
Hydrogen peroxide... have you ever seen it, really? It has the almost same color as water (that is, no color at all), and the reaction products have the same color as well. You won't be able to see a thing. Besides, this one is also too sensitive. Besides, it won't oxidize ethanol right away.

So it goes.


Can someone please explain buffers to me? This is what I understand so far:

The concept of a buffer is to minimize any swings in pH. In order to do this you need to create a solution with both acidic and basic compounds.

To create one, you start with an acid:
$\ce{HA + H2O -> A- + H3O+}$

[My first question is why for buffers the acid needs to be weak.]

The buffer is then created by adding in a salt containing the anion of the base which allows for both the acid and its conjugate base to exist in the solution.

Here, why doesn't the conjugate base react with the acid? Second, why does this help minimize pH changes. I see that since that you have a weak acid the conjugate base will be strong and can react with any acid added. However, if you add base what is going to react with it? Isn't the acid too weak to do so? Third, what happens to the $\ce{H3O+}$ that exists in the solution?

Writing out the equilibrium expression for a weak acid in solution:

$$[\ce{H3O+}] = K_\mathrm a\frac{[\ce{HA}]}{[\ce{A-}]}$$

Buffers need to keep the concentration of the weak acid to its conjugate base similar in order for it be effective.

Let’s say you add a small amount of base. Some of the $\ce{HA}$ molecules are converted into $\ce{A-}$ ions. The ratio $\frac{[\ce{HA}]}{[\ce{A-}]}$ changes but not by a large amount.

If we add a small amount of acid, we generate more $\ce{HA}$. Again, the ratio $\frac{[\ce{HA}]}{[\ce{A-}]}$ changes but not by a large amount.

  Why do buffers for the acid needs to be weak?

If you look at the expression, the concentration of hydronium ions is directly related to the ratio of the conjugate acid and its base. If you take a strong acid for instance, hydrochloric, it will almost completely deprotonate in water. Thus the ratio of $\frac{[\ce{HA}]}{[\ce{A-}]}$ is close to 0 and is negligible, and we use the $K_\mathrm a$ to directly find the concentration of hydronium ions. As strong acids have very large $K_\mathrm a$ values, we can directly take the concentration and assume that will be the concentration of our hydronium ions.

  Why doesn’t the conjugate base react with the acid?

It depends on the acid. In buffer solutions, competition arises between acid and the conjugate base. For instance, if the $K_\mathrm a$ is large relative to $10^{-7}$, the acid ionization will proceed and you will have an equilibrium with the conjugate base reacting with its acid resulting in no buffer action.

If however, the tendency of the acid in donating hydrogen ions, say, to water, is outcompeted by the tendency of the base to accept those hydrogen ions from water, this net reaction produces $\ce{OH-}$ ions, the $K_\mathrm a$ of this acid-base pair is less than $10^{-7}$, and we use $K_\mathrm b$ to find equilibrium.

Derivation of the Henderson–Hasselbalch equation:

Let’s use acetic acid and water as an example:

$$\ce{AcOH + H2O <=> AcO- + H3O+} $$

$$ K_\mathrm a = \frac{[\ce{AcO-}][\ce{H3O+}]}{[\ce{AcOH}]}$$

Taking the log of both sides:

$$ \log(K_\mathrm a) = \log\left(\frac{[\ce{AcO-}][\ce{H3O+}]}{[\ce{AcOH}]}\right)$$

Using the law of logs where $\log(xy) = \log(x) + \log(y) $:

\begin{aligned}
 \log\left(K_\mathrm a\right) &= \log\left(\frac{[\ce{AcO-}]}{[\ce{AcOH}]}\right) + \log([\ce{H3O+}])\\
 -\log([\ce{H3O+}]) &= \log\left([\ce{AcO-}][\ce{AcOH}]\right) - \log(K_\mathrm a) \\
 \mathrm{pH} &= \log\left(\frac{[\ce{AcO-}]}{[\ce{AcOH}]}\right) + \mathrm{p}K_\mathrm a \\
 \mathrm{pH} &= \mathrm{p}K_\mathrm a + \log\left(\frac{[\ce{B}]}{[\ce{A}]}\right) \\
\end{aligned}

With buffer systems, you can calculate the $\mathrm{pH}$ of any buffer solution. The Henderson–Hasselbalch equation has several caveats however.

You need have both the acid and its conjugate base in solution.
The extent of ionization must be small enough so that the
concentrations of hydronium and hydroxide ions are small relative
$[\ce{HA}]_0$ and $[\ce{A-}]_0$.


Why is starch insoluble in ethanol? I understand that starch will precipitate from an aqueous solution in the presence of alcohol. I also understand that molar mass effects solubility and starch has a relatively high molar mass. There is also an issue regarding the polarity of the solute and solvent.

Is the insolubility of starch (which has the form $\ce{(C6H10O5)$_n$}$) in ethanol only due to the fact that starch is polar and ethanol is only slightly polar?

Due to the fact that starch (amylum) is a polysaccharide , a long chain compound consisting of glucose monomers it has poor solubilty in most solvents.

Some starches are water soluble because of the hydroxyl groups involved in the polysaccharide chain, some like the branched form amylopectin are more insoluble.

Starches form granules in solution with cold water, the granules break down as the water is heated and can be solubilized.

When ethanol is added, which is less polar than water, it lowers the solubility and potential hydrogen bonds of the starch , because the water is more attracted to the alcohol than the starch, the starch is dehydrated by the alcohol and precipitates.

So while polarity and temperature are variables to starches poor solubility, the critical variable in this case is the starches ability to hydrogen bond with the solvent.


How can a molecular orbital with a nodal plane still be considered "bonding"? Explain how a molecular orbital can have a nodal plane and still be a bonding molecular orbital.

I understand the concept of how nodal planes are formed and how they have a probability of 0 when trying to find an electron in the nodal plane, but I don't understand how it can be considered a bond especially since these nodal planes are usually called "anti-bonding". This question itself seems contradictory and I'm having a hard time trying to understand how this is possible.

The number of nodal planes isn't the determining factor. Yes, having more nodal planes increase the energy of the orbital, but you need to consider the fact that some MOs (specifically the anti bonding ones) are raised in energy. Overall, the total energy of the orbitals is the same as before, so you stabilize some and destabilize others. Take a look at the MOs and you can see that the there's two bonding MOs with one nodal plane perpendicular to the plane of the benzene with $E_{1}$ symmetry. Likewise, there's a pair of MOs with two nodal planes that with $E_{1}$ symmetry that are anti-bonding. These pairs happened to be correlated in that the increase in energy of the anti-bonding orbitals is essentially matched by the stabilization of the bonding orbitals.


Why do atoms have electron Affinity? Atoms are stable so how can they pull electrons i.e how do they have any electron affinity at all?

Actually, not all atoms are stable, at least not in the gas phase, which is where we measure the electron affinity. You should first read Why do atoms "want" to have a full outer shell? for detailed elaboration.

Now, electron affinity involves the gain of an electron to an isolated gaseous atom, but not all atoms actually want to gain an electron, i.e., they are already pretty stable. This is precisely the reason why elements like the noble gases or nitrogen have negative electron affinity i.e. you need to supply energy for them to accept an electron.

Also read: Are there any major exceptions when comparing electron affinity?

So, in short: "Atoms are stable so..." not always, in fact, such atoms are in the minority. "...how can they pull electrons..." read the first link. "how do they have any electron affinity at all?" we can always supply excess energy to make an atom accept an electron, forcibly. So, the only difference for stable vs non-stable atoms would be the sign of electron affinity.


How to calculate the number of atoms in a gas molecule? $\pu{20 mL}$ of sulfur vapour at $\pu{1000 °C}$ reacts with $\pu{40 mL}$ of oxygen gas to form $\pu{40 mL}$ of sulfur dioxide gas. Determine the number of atoms in the molecule of sulfur in the vapour state at this temperature.

My working out is as follows:

$$\ce{\underset{\pu{20 mL}}{S(g)} + \underset{\pu{40 mL}}{O2(g)} -> \underset{\pu{40 mL}}{SO2(g)}}$$

I then used the formula $n = m/M$ to find the amount of substance.

$$n = m/M = 20/32.07 = x$$

Then, to find the number of atoms, I used the formula $n = N/N_\mathrm{A}$.

$$N = nN_\mathrm{A} = x\cdot\pu{6.022e23} = \pu{3.80e23}~\text{(atoms)}$$

However, I am uncertain as to whether this is correct as I did not utilize any of the gas laws. For this question, which of the gas laws should I have used?

Any answers would be appreciated.

Reason for rewrite: refer to edit summary.

From Wikipedia on Disulfur ($\ce{S2}$)

  This violet gas is commonly generated by heating sulfur above 720 °C

Since the sulfur is well above $720^\circ\pu{C}$, it would most likely be in the diatomic state, more like an analogous structure to dioxygen ($\ce{O2}$).

We can also judge this from the molar ratios too, as rose22 has pointed out. Since it is given that $40~\pu{mL}$ of $\ce{SO2}$ is formed from $20~\pu{mL}$ of $\ce{S_x}$ (assuming the atomicity of $\ce{S}$ to be $x$ at this temperature) and $20~\pu{mL}$ of $\ce{O2}$, they must come to terms with a simple whole number ratio.

$$\ce{\underset{20~\pu{mL}}{S_x}+\underset{40~\pu{mL}}{xO2}->\underset{40~\pu{mL}}{xSO2}}$$

Statement 1: Since the ratio of volumes of gases at a particular temperature and pressure is equal to the ratio of amount of substance in moles, it is clear that:

$$1:x:x=20:40:40$$

That is to say, $x=2$, so the atomicity of sulfur here is $2$.

In the question, the OP said:

  However, I am uncertain as to whether this is correct as I did not utilize any of the gas laws. For this question, which of the gas laws should I have used?

Statement 1 is a  result from the universal gas law.

$$PV=n\text{R}T$$

When $P$ and $T$ are constant,

$$\frac{V_1}{V_2}=\frac{n_1}{n_2}$$

Some may argue that this relationship is called Avogadro law, but again, this law too is a gas law, evidently.


Sugar dissolution test Is there a way to detect whether sugar is completely dissolved in a cup of water other than looking at the bottom of the cup?

I was wondering if there is a test or something that can be carried out to determine whether sugar is fully dissolved.

The sugar will keep dissolving until it reaches it's saturation concentration of around $\pu{900 g/L}$, which is of course horribly sweet, thick and undrinkable as a beverage.  Assuming you don't completely saturate the water with sugar in this way, the undissolved sugar will continue to dissolve so long as there is still any remaining in the glass.

A common way of measuring the dissolved sugar concentration is by refractometry, as described below (note that home refractometers can be purchased for as little as $30 US):

  A refractometer is a laboratory or field device for the measurement of
  an index of refraction (refractometry). The index of refraction is
  calculated from Snell's law while for mixtures, the index of
  refraction can be calculated from the composition of the material
  using several mixing rules such as the Gladstone–Dale relation and
  Lorentz–Lorenz equation.
  ...
  In homebrewing, a refractometer is
  used to measure the specific gravity before fermentation to determine
  the amount of fermentable sugars which will potentially be converted
  to alcohol.

So, in your case a "homebrewing" type of refractometer is probably your best bet. As long as sugar remains in solution, it will continue to dissolve until it reaches saturation.  This also means it's refractive index and specific gravity will continue to increase.  Once this increase is no longer observed, then the sugar is no longer dissolving, meaning you've either reached saturation or all of the sugar has dissolved.  Again, saturation would mean a thick, honey-like solution. Thorough stirring throughout the measurement process is a must.


Additional Questions Regarding the Auto-Ignition Temperature I had some followup questions regarding a previous post I made here regarding the auto-ignition temperature and ASTM E659

For fuel temperature below AIT, we should still have finite reactants above activation energy, reacting, and heating the remaining mixture so additional reactants are above the activation energy. Theoretically as $t \rightarrow \infty$, could the fuel burn itself out like this, regardless of what temperature it is at (obviously the rate will be very different)? If so, is there an implicit rate requirement in defining the AIT that the above chain reaction process has to happen within a short duration?
Given that ignition is defined as when a flash and temperature rise is seen, does this mean that the actual quantity of fuel burnt is unimportant and it is assumed that a significant enough amount is used?
Is the reason for using an open flask in ASTM E659 likely just for simplicity, as opposed to a piston-cylinder arrangement that offers more control over air-fuel mixture?

To understand the answer to (1), you have to think about what combustion actually is. It isn't a single reaction, with a single set of reactants, and a single set of products. It's actually a whole bunch of tiny steps (thousands) that all occur together. Those reactions happen among a whole bunch of unstable radical species (hundreds). When you put a match to a pool of gasoline, all that energy from the match starts tearing apart bonds in the fuel, leading to the formation of radicals. That's what triggers the initial release of energy, and what creates the cascade of ignition. So theoretically, no, you could never reach combustion products, through the pathway of combustion, because you'd never have that activation energy, the big kick of energy needed to create radicals and keep them alive long enough for them to create a chain reaction.

The answer to (2) is, no, it doesn't matter how much fuel. Remember you're measuring a mass-independent property. The autoginition temperature isn't "5 degrees per gram". It's a single temperature. What matters is the ratio of oxygen to fuel.

As to (3), it's a lot easier to operate an open flask than it is to operate a piston-cylinder, and it doesn't have a substantial impact on the air-fuel ratio if you do it correctly.


Gibbs free energy change at equilibrium I know that Gibbs free energy change represents the amount of the non-expansionary work that a reaction is capable of doing but what happens to this energy at equilibrium? Why is the system unable to do work? Why is the Gibbs free energy change equal to zero at equilibrium?

Edit: Please don't simply write the derivation that we use to conclude that ∆$G$ must be zero at equilibrium. Please try to explain the physical significance behind this fact.

Please also note that I have already searched this website for an answer to my question, found only the following question but the answers only explained the Math:

Gibbs free energy-zero or minimum

Thanks ever so much in advance :) Regards.

I'm really not getting why that post didn't answer OP's query.

Nevertheless, I'm posting the following argument as an answer.

  Why is the Gibbs free energy change equal to zero at equilibrium?

Well, it's not 'Gibbs free energy change' $\Delta G$ but rather Reaction Gibbs
Energy $\Delta _r G$ which is zero at equilibrium, if written accurately.

Reaction Gibbs Energy is defined as the rate of change of Gibbs Free Energy per unit change in composition i.e.,

$$\Delta_r G =\frac{\Delta G}{\Delta n} \tag 1$$

where $\Delta n$ is the change in composition of the system.

Now, OP must be aware of the first derivative test for finding local minima; that means for reaching the equilibrium position, the rate of change of Gibbs Free Energy per unit composition must be zero and the rate must change from negative in the neighbourhood of the extrema point of Gibbs Free Energy vs. Composition graph viz.

$$\Delta_r G= 0\;.$$

Now, question might arise as that of the physical implication of association of equilibrium position with $\Delta_r G= 0\;.$

The definition of equilibrium position sheds light on this:

A reaction reaches equilibrium position when it has no further tendency to change; that is, the reaction does remain 'spontaneous in neither direction' .

This happens when the reaction Gibbs energy becomes zero viz. $\Delta_rG=0.$


How does one synthesise 2-Monoglycerides from Triglycerides? How does one synthesise 2-monoglycerides from triglycerides with additional ingredients if necessary?

My first instinct would be to add a catalyst for the reaction of $triglyceride \rightarrow 3 FA + glycerol$ (FA=fatty acid; for instance a catalyst could be sodium hydroxide) and then add additional glycerol so as to cause the reaction: $FA + glycerol \rightarrow 2$-$monoglyceride $ to occur but I just wanted to ask if this is true because to me this seems too simple to be true.

One way I have learnt is by taking a page out of the human body's book and using pancreatic lipase, found in pancreatic juice, to catalyse the reaction:

$triglyceride+ 2H_2 O \rightleftharpoons   2$-$monoglyceride+2 FAA$

Where FAA stands for, 'Fatty acid anion.'

If someone has a better answer I'll readily accept it over this one, this is just one method.


What is melting? Which bonds do we break to melt something? To melt diamond, we have to break the covalent bonds, which we can consider 'intermolecular' because it is one giant molecule.

To melt Methane, we have to break the van der Waals (intermolecular) forces.

For $\ce{NaCl}$, ionic bonds, which are intermolecular as well in a sense.

But I've read that melting graphite also requires breaking the covalent $\ce{C-C}$ bonds rather than just breaking the van der Waals forces between the layers.

And for cross-linked polymers, I've read that melting involves breaking only the bonds linking the two polymer chains.

So I guess what I want is a precise definition of melting. Does it involve breaking the weakest bonds? The intermolecular forces? The intramolecular bonds?

When we melt a molecule of, say, $\ce{H2O}$, we imagine molecules of $\ce{H2O}$ roaming around with little intermolecular forces, as most hydrogen bonds have been broken.

What do we 'imagine' when we melt diamond, or a cross-linked polymer, or graphite, or ionic compounds like $\ce{NaCl}$?

This is a good question. Unfortunately there are several criteria by which chemists indentify whether a process is melting or not. One of them is called the Lindemann Criteria which says:

"Crystals are considered to melt, when the vibrational amplitude becomes half of the interatomic spacing in the crystal lattice."

What does it mean? Normally, at temperatures greater than 0K, atoms possess kinetic energy. Infact, temperature itself is a measure of the kinetic energy of the constituent atoms. The kinetic energy of an atom is related to the temperature as:

$$\text{Kinetic Energy}=\frac{3}{2}k\text{T}$$

where $k$ is a constant called Boltzmann Constant, with a value of $1.38×10^{-23}\text{J/mol}$. Atoms in a solid are characterized by their fixed mean positions. They only vibrate within a certain boundary around them. Since they have kinetic energy (and velocity as well), they tend to leave their current location, but repulsive forces from the other atoms push it back to its original position. In this way, you could consider atomic bonds like "little springs".

The extent to which the atom displaces itself from the mean location is called it's vibrational amplitude. As we increase temperature, the atoms will have more velocity, and can consequently displace itself further from the mean position. Lindemann defines the melting temperature as the temperature at which the amplitude becomes half of the space between two adjacent atoms of the crystal.

Yet another criteria for defining the melting point is the Born criteria that says:

"Crystals are considered to melt, when the shear modulus approaches zero"

You might wonder, what is this shear modulus? It's infact, a measure of how much tangential stress a solid object can handle, and the tangential strain caused by it. Tangential forces on an object are forces that act parallel to the surface of the object, hence the name. See the picture below that shows a tangential force applied on an originally cuboidal object:

Tangential (or shearing) stress is defined as:

$$\sigma _{shear}=\frac{F_{shear}}{A}$$

where $F_{shear}$ is the tangential (shearing) force, and $A$ denotes area of surface in question.

Considering the picture above, shearing strain would be defined as:

$$\epsilon _{shear}=\frac{\Delta x}{h}$$

There exists a unique proportionality between the stress applied and strain produced, and is given by:

$$\frac{\sigma _{shear}}{\epsilon _{shear}}=S$$

where $S$ is called the shear modulus. Its a constant for a given material and a given temperature.

In a physical sense, liquids are considered as substances that cannot withstand a tangential stress. When a tangential stress is applied, liquids simply keep on increasing the strain, even under small stress. This itself is a useful way to consider when a solid melts, when its shear modulus becomes zero!

Now, let's consider how we consider melting for the different examples that you have asked for:

Diamond: Diamond has a structure of covalent bonds arranged tetrahedrally to each carbon. Each bond has it's characteristic bond length. Applying Lindemann condition over here would be considered advisable , so that melting point is where vibrations of carbon atoms is half of the $\ce{C-C}$ bond distance. Ultimately some of the $\ce{C-C}$ would break and the molten mixture would largely consist of radicals of varying sizes, that permit free movement between themselves.

Graphite: While each layer of graphite is only weakly bonded to the other layer by van der Waals' forces, each layer (called graphene) itself is a quite a large molecule. You can consider a liquid as a large number of small molecules, so clearly graphene doesn't look like this. To melt it, you would have to break some of the $\ce{C-C}$ bonds so that we can produce small molecules that can move freely. We apply Lindemann condition in this case. The molten mixture would be similar to that of diamond.

Branched polymers: Remember how I told to consider liquids as small molecules that can move freely sliding among themselves? We consider the same thing over here. The branch bonds are comparatively weaker than the rest of the bonds in the polymer, and these are the ones that will break when heated. As far as bond-breakings are concerned, we use the Lindemann condition. While the unbranched polymers are big molecules, they are still small enough to exhibit liquid character at the high temperature required to break the branched bonds. The molten mixture would consist of the straight-chain polymers in radical forms.


How to determine the empirical formula given mass percentages of the elements and the molecular weight? A compound has $67.3 \% \rm~C,4.62 \%~N,6.93 \%~ H$ and O. Determine the molecular formula of the compound knowing its molar mass is $283\rm~\frac{g}{mol}$.

My solution : The percentage for the oxygen is $$100-( 67.3+4.62+6.93)=21.15$$ percent oxygen. Lets determine the number of moles for every substance.

For the carbon we have $$n_C = \frac{67.3 \rm~g}{12\rm~\frac{g}{mol}} =5.6~\rm mol$$

We do the same for N, H and O. We have 0.33 atoms of N,6.93 atoms of H and 1.32 atoms of O.

So the ratio is $5.6 : 0.33 : 6.93 : 1.32$.

We divide by $0.33$ and we have $17 : 1 : 21 : 4$. The empirical formula is $\ce{C17H21NO4}$.

Its molar mass is 303 g/mol. Now we find the quotient:

$$K= \frac{\text{Molar mass of the real formula}}{\text{Molar mass of the empirical formula}} =\frac{283}{303}\approx 1$$

So the molecular formula is $\ce{C17H21NO4}$. Is this correct?

Rather than starting out with 100 g of unknown, I would start out with 1 mol of unknown, i.e. 283 g. That way, you can directly calculate how many moles of each element are in one mole of compound:

$$ n = 67.3 \%  \cdot \frac{\pu{283 g}}{\pu{12 g mol-1}} = \pu{15.86 mol}$$

Or to get the stoichiometric coefficient $\nu_C$ directly, take the percentage and multiply it by the ratio of compound molar mass to atomic mass:

$$ \nu_C =  67.3\% \cdot \frac{\pu{283 g mol-1}}{\pu{12 g mol-1}} = \pu{15.86}$$

If you do that for C, N, O and H, you get the formula:

$$\ce{C_{15.86}H_{19.4}N_{0.93}O_{3.80}}$$

You can try rounding this to the nearest integers:

$$\ce{C16H19NO4}$$

This compound has a molar mass of 293 g/mol. Playing around with the formula, you can find coefficients that match the molar mass exactly, e.g. $\ce{C16H13NO4}$ or $\ce{C17H17NO3}$. In any case, neither one of these nor the formula suggested by the OP matches both the molar mass and the mass percentages.

  Is this correct?

$\ce{C17H21NO4}$ is a good answer, but the data are not consistent, so it is not the complete answer. I would say depending on whether the errors on the molar mass determination or the elemental composition are larger, you would have a different top candidate, and should report the numerically exact formula with non-integer coefficients. That way, the reader can critically evaluate the meaning of the data themselves.


Which has more inductive effect: an aldehyde or a carboxylic acid? A chemistry reference book that I have been reading gives the order of inductive effect ($-I$) as follows:

While the Wikipedia page on inductive effect gives the following order:

$\ce{-NH3+} > \ce{-NO2} > \ce{-SO2R} > \ce{-CN} > \ce{-SO3H} > \ce{-CHO} > \ce{-CO} > \ce{-COOH} > \ce{-COCl}> \ce{-CONH2} >
\ce{-F} > \ce{-Cl} > \ce{-Br} > \ce{-I} > \ce{-OR} > \ce{-OH} > \ce{-NH2} > \ce{-C6H5} >\ce{-CH=CH2} >\ce{-H}$

The reference book has $\ce{COOH}$ ahead of $\ce{CHO}$ but Wikipedia has it the other way around. Which source is correct?

I believe the Wiki page is more accurate, because the Hammett Equation substituent constants (para) indicate -CHO should be more "electron withdrawing" than -COOH, but there is little reason to use such lists as "absolute" - I think they are simply good guidelines for reactivity.

I looked at the data in this link: https://www.bluffton.edu/homepages/facstaff/bergerd/classes/CEM311/handouts/sigmas.pdf

However, substituent constants are not absolute proof of "inductive effect" and it is difficult to conclude from such data that a particular substituent is more "inductively" electron withdrawing than another. To do that, we would need a model (or a set of compounds, equivalently) where we can achieve separability of sigma and pi electrons.


Can Carbon Form bonds without Hybridization? Carbon has two electrons in its p orbital which should be able to form bonds, are there any examples in which this occurs instead of carbon hybridizing before bonding?

Hybridisation is a mathematical concept. It is one way of deriving chemical bonds from elements but not the only way. A different one (that computational chemistry typically uses) would be to just plant atoms at certain positions in space, and then mix their unhybridised atomic orbitals with each other to see what comes out. One can then back-calculate this to localise the molecule-spanning orbitals to single bonds and thereby deduce a hybridisation that would have worked to build up the molecule. Take the following example of methane (taken from Professor Klüfers’ internet scriptum for basic and inorganic chemistry at the university of Munich):

On the left you have four hydrogen s-orbitals, on the right you have the orbitals of carbon. As you see, there is no hybridisation assumed at all and the ground state is taken to be a fully populated $\mathrm{2s}$ orbital and three $\mathrm{2p}$ orbitals populated with two electrons amoung them for a total of $\mathrm{2s^2~2p^2}$. The hybrid orbitals are gained from symmetry considerations (in a tetrahedron, the p-orbitals transform as $\mathrm{t_1}$) and hydrogen orbitals were selected as was seen fit. The electrons are added a posteriori. Spin conservation would have produced a twice-excited state with very high total energy, so immediate spin pairing is assumed.

To localise the bonds, linear combine all four orbitals. If you take every orbital with factor $+1$, you arrive at the $\ce{C-H}$ bond to the upper right hydrogen. By shifting around a factor of $-1$ among the three $\mathrm{t_1}$ molecular orbitals during linear combination, you can address every other $\ce{C-H}$ bond. The resulting bonding orbital looks like n $\mathrm{sp^3}$ orbital mixed with a hydrogen $\mathrm{s}$ orbital.


Rubbing alcohol gets warm after diluting with water I've just diluted a 95% rubbing alcohol bottle with water to bring it down to 70%. However, the bottle got noticeably warmer after I did so. I've searched online and I couldn't find any exothermic reaction that happens when alcohol is mixed with water. I did use mineral water by mistake to do so, so my best guess is that alcohol reacted with... one of the salts in the water... maybe?! (I have no idea)

Why did this happen?

Ethanol does the same. It is the consequence of deviation from the Raoult law.

The other consequences are existence of azeotropes ( affects distillation ) and mixing volume deviations. If you mix 1 L of ethanol and 1 L of water, it warms up. When it cools down back, its total volume is not 2 L, but about 1.9 L.

It is closely related to total energy of inter-molecular bonds in pure solvents and in their mixture.


In a heating curve, when it is plateau, where does all the energy go since the temperature does not get higher? So between the change of phases within the matter(solid liquid gas), the temperature reaches a plateau, no matter how much heat you add to it, it just stays the same. Where did all those energy go?

You may have learnt at some point that a change of phase is endothermic (i.e. needs heat to happen). So, when you transition from a phase to another and keep adding heat, the temperature will remain the same, as all the heat you'll add will be used for the change of phase. Hence, the temperature will remain the same until the transition is over (i.e. you gave your system enough heat for the transition to be complete), then will vary again.


Amphoteric amino acid titration Glycine $(\ce{NH2CH2COOH})$ is an amino acid with two $\mathrm{p}K_a$ values $(\mathrm{p}K_{a,1}=2.0, \  \ce{COOH};\  \mathrm{p}K_{a,2}=10.0, \ \ce{NH2})$.
(a) If $\pu{0.01 mol}$ of this amino acid is used to prepare one $\pu{1 L}$ solution. Calculate the $\mathrm{p}\ce{H}$ and the $\ce{NH3+CH2COOH}$ concentration in the solution.
(b) If $\pu{10 mL}$ of the above glycine amino acid solution is mixed with $\pu{90 mL}$ of basic buffer system. If the concentration of the $\ce{NH2CH2COO-}$ was identified to be $\pu{1.05 \cdot 10^{-4} M}$ indicate the $\mathrm{p}\ce{H}$ of the glycine solution.
(c) If $\pu{10 mL}$ of strong base $\ce{NaOH}$ ($\pu{0.01 M}$) was added to solution in (a) what is the final $\mathrm{p}\ce{H}$.

For part (a):
I used $[\ce{H+}]=\sqrt{10^{-10}\cdot0.01} =10^{-6}$.
Then I used $\mathrm{p}\ce{H} =14-\log(10^{-6})$ to get a $\mathrm{p}\ce{H}$ of $8$.
While my $\ce{NH3+CH2COOH}$ is $\pu{10^{-6} L}$.

For part (b):
I used the Henderson-Hasselbalch equation for a basic buffer $$\mathrm{p}\ce{H} =\mathrm{p}K_a     +\log\left(\frac{\ce{[NH2CH2COO- ]}}{\ce{[NH2CH2COOH]}}\right)$$ I got a $\mathrm{p}\ce{H}$ of $10.97$. But I'm   unsure of my answer.

For part (c):
I suspect that it is a basic buffer but I'm not sure which value to use. could anyone help me out with this.

As in aqueous solution:
$$\ce{NH3+CH2COOH_{[b]}<=> NH3+CH2COO-_{[z]} + H+_{[x]}\\NH3+CH2COO-_{[z]} <=> NH2CH2COO-_{[a]} + H+_{[x]}}$$
So,
$$10^{-2}=\frac{[z][x]}{[b]}\\10^{-10}=\frac{[a][x]}{[z]}\\\implies 10^{-12}=[x]^2$$
Notice that a=b as the solution is always neutral; also z is a neutral compund
As,
$$[H^+]=10^{-6}\implies p^H=6$$
$$\implies[b]=100[z][x]=10^{2-2-6}=10^{-6}$$

  You should have done $p^H=-log[H^+]$, your [b] is correct.

For part (b), yes, using Henderson-Hasslebalch equation:
$$p^H=10+\log\left(\frac{1.05\times10^{-4}}{10^{-6}}\right)\approx 10.97$$

In basic solution, the second equilibrium shifts well(as a result the first one too) to the right.As NaOH is a strong acid, we can neglect any other bases.In presence of NaOH,
$$\ce{NH3+CH2COOH_{[b]} + 2NaOH_{[k]}<=> 2NH2CH2COONa_{[a]} + 2H2O_{[y]}}$$
In millimoles:
$$\begin{array}{c|c|c|c|c}
&entity&b&k&a&y\\\hline
&initial&10&.1&0&0\\\hline
&final&9.95&0&.1&.1+110(original)\\\hline
\end{array}$$
Now you can similiarly apply agin the Henderson-Hasslebalch equation:
$$p_H=12+log(.1/9.95)\approx 10.01$$

Note that I am treating the compund as a monoprotic acid of $pK_a =10(-NH_3^+)+2(-COOH)=12$


Understanding Standard Hydrogen Electrode I came across a question today.

A galvanic cell is composed of two hydrogen electrodes, one of which is a standard one. In which of the following solutions should the other electrode be immersed to get maximum e.m.f. ($K_a=10^{-5}$ for acetic acid and $k_a=10^{-3}$ for phosphoric acid)
(A) $0.1$ $M$ $\ce{HCl}$
(B) $0.1$ $M$ $\ce{CH3COOH}$
(C) $0.1$ $M$ $\ce{H3PO3}$
(D) $0.1$ $M$ $\ce{H2SO4}$

I thought "Well, seems like all I need to do is find which solution will give the most number of hydrogen ions because its hydrogen ions which will suck those electrons from the other half cell which will be the cause of current. And concentration is same for all. So there is no chance for a weak acid. So option (B) cannot be correct. All other seems like a strong acid. So it should be $\ce{H2SO4}$ as it gives $0.2$ $M$ of hydrogen ions for $0.1$ $M$."
But my thought was absolutely wrong. Why?

This is a concentration cell question where 2 electrodes of same stuff are dipped in their own ions and if there is non equal concentrations of anything a "potential" develops called electrode potential (to be understood in qualitative way)(kind of more concentration of reactant forces product formation)
as you can understand this would be a redox reaction where reduction would occur in one and oxidation in another, Now, by standard it means that all the components of the reaction (participants, i.e. reactants and products) are in standard state i.e. $1M, 1$ $bar$ etc, so, we know that concentration of $\ce{H+}$ in standard one is $1M$, we need to create biggest gap between them, as the concentration is really high and we cant compete with it, just reduce our own $\ce{H+}$ to increase their difference and force ours to undergo Oxidation to gas and standard undergoes reduction

Just for extra information,
$E=E_0-{RT\over nF}\ln Q$ and can be derived by $G=G_0+RT\ln Q,\;G=-nFE$ {$n$ is number of electrons being exchanged ex- in $\ce{2H+->H2},\;n=2$; $F$ is the charge on 1 mol electrons} and Standard Hydrogen is the standard bearer so, we assign its standard value as $0$, $E_0=0$.

EDIT : I can't comment due to reputation
, basically, assume complete dissociation and and concentration of $\ce{H+}$ in $\ce{HCl}$ would be $0.1M$ , $\ce{H2SO4}$'s second ionization is not complete like others but can be assumed for a good approximation, so concentration becomes $0.2 M$ , and basic thing to remember is

IF ONE SIDE  UNDERGOES OXIDATION THEN OTHER MUST UNDERGO REDUCTION

and if we strengthen either reaction, potential increases, concentration is only in a solvent, how can a solid would have active $\ce{H+}$ ions (everything is activity not concentration or partial pressure flat out, they are approximated, solids are assumed to have activity as unity) and log is defined as the power to $10$ that would give us that number


Magnetic moment μ approximation I've been reading a bit about the magnetic moment (spin-only) $\mu_{s.o}$ where they give a formula relating this to the number of unpaired electrons

$$\mu_{s.o}=\sqrt{n(n+2)}$$

where $n$ is the number of unpaired electrons.

However in our lecture today we were using the approximation $\mu_{s.o} \approx n+1$. Is this  an acceptable approximation for the magnetic moment or should I stick to using the previous one.

Obviously using $\mu_{s.o} \approx n+1 $ is easier to use for calculations but I would like someone's opinion on this.

I think you can come at this approximation in two ways. Using more advance methods, the approximation is obtained as a truncation of the Laurent series of $\sqrt{x(x+2)}$ about $x=\infty$.

This is possible, but I think needlessly complex in this case. Using just algebra, we can note $$\sqrt{n(n+2)}=\sqrt{n^2+2n}\approx\sqrt{n^2+2n+1}=\sqrt{(n+1)^2}=n+1$$
By looking at a plot, we can see this approximation is very good, giving essentially the exact result at $n=10$.


Mechanism for synthesis of Meldrum's Acid I have a little bit of problem trying to figure out the mechanism for the synthesis of Meldrum Acid. Meldrum acid is synthesis by a condensation reaction of acetone with malonic acid in acetic anhydride.

First, the carboxyl OH will be acetylated by acetic anhydride. But I can't figure out how the two ester bonds are being formed subsequently. I'm thinking that there's an acid-catalysed aldol condensation somewhere that results in the removing of water. Perhaps by attack of the enol (from acetone) to the acetylated carbon. But I'm not really sure about it either.

Any thoughts?

The first step (reaction with acetic anhydride) creates a mixed anhydride, this is a good electrophile with acetate as a leaving group. This is nucleophilically attacked by the oxygen of acetone (drawn as its enol in the scheme, but it is easier to visualise it as acetone) to give a cationic intermediate which is immediately attacked by the -OH of the neighbouring carboxyl group to give the product.


Partial dissolution of non-polar solutes in polar solvents I have learnt that in order for dissolution to occur, the solute-solvent intermolecular forces must be strong enough to overcome the solute-solute and solvent-solvent forces. However if this is the case, why are non-polar substances such as methane able to partially dissolve in polar solvents such as water? Aren't the dispersion forces always too weak to break the pre-existing hydrogen bonds present in water?

In solids, reorientational and especially translational molecular motion are highly hindered. The motion of individual molecules is constrained about mean fixed positions within a regular lattice. Such regularity is absent in liquids, where significant thermal energy is associated with rearrangements in molecular position and orientation.
This is of course one defining property of a liquid (and more generally of fluids). The corollary of this structural irregularity is that it leads to a greater number of potential "defects" in the network of interactions within the substance.

Consider water. The wikipedia explains that in crystalline ice, water forms hydrogen bonds with 4 other water molecules. Meanwhile,

  [from] TIP4P liquid water simulations at 25 °C, it was estimated that each water molecule participates in an average of 3.59 hydrogen bonds. At 100 °C, this number decreases to 3.24 due to the increased molecular motion and decreased density, while at 0 °C, the average number of hydrogen bonds increases to 3.69.[35]

The net result of this structural irregularity is that even molecules that interact weakly through dispersion interactions (or, in principle, not at all) with water have a significant if small chance of persisting within the liquid. Therefore, it is less a matter of dispersion interactions "breaking" hydrogen bonds, and more of apolar solutes occupying "defects" in the liquid.

References in wikipedia article

[35] Jorgensen, W. L.; Madura, J. D. (1985). "Temperature and size dependence for Monte Carlo simulations of TIP4P water". Mol. Phys. 56 (6): 1381.


Why are transition metal hydroxides insoluble? From what I understand, adding $\ce{NaOH}$ to a solution containing transition metal ions gives transition metal hydroxide precipitates. Why is $\ce{NaOH}$ soluble but transition metal hydroxides insoluble? I have a feeling it has something to do with ligands and complexes formed, which I am familiar with.

The rule of thumb for hydroxide salts is actually that all hydroxides are insoluble save those of the alkali metals and the heavier alkaline earth metals (namely $\ce{Ca^{2+}}$, $\ce{Sr}^{2+}$, and $\ce{Ba}^{2+}$). There are many resources you can look up to check the solubility rules of certain ions, this one, for example.

Before researching this, I suspected that the solubility of the salts with a certain anion might be determined by the cation's control over its valence electrons, which is measurable through ionization energy. Sure enough, when you compare the first ionization energies of various metals, you find that those with the lowest first ionization energies are those that are soluble hydroxides.

Alkali Metals

The salts of all alkali metals are completely soluble because they have relatively little control over the outermost electron orbiting them, in addition to this being the only electron they must lose in order to attain an octet structure.

Alkaline Earth Metals

For the alkaline earth metals, the complete outermost $\mathrm{s}$ orbital must be lost in order to attain an octet structure. Because the orbital is complete, and among other reasons, the first ionization will require more energy than the alkali metal in the same period. In addition to this, the second electron must then also be lost. This means that overall, it requires much more energy to ionize this group than the alkali metals. This explains the insolubility of $\ce{Be(OH)2}$ and $\ce{Mg(OH)2}$.

The solubility of the hydroxides of the metals lower in the alkaline earth metal group can be explained by the shielding effect, which effectively says that the ability of a positive nucleus to attract electrons decreases as the amount of orbitals surrounding the nucleus decrease.

Transition Metals

The insolubility of the hydroxides of the transition metals can be explained both by the increasingly large ionization energies that result from successively removing electrons (or negatively charged hydroxide molecules in this case) from the metal cation, and the greater effective nuclear charge on the electrons across a period.


Is it possible to determine the composition of a gas mixture from the effusion of a gas through a pinhole? Typically all the problems which I've seen regarding rate of diffusion involve relating the rate of one gas to the other and from there finding either speed of effusion or diffusion or getting the formula weight. However what sort of procedure should be used when it is needed to find the composition of a mixture?

The problem is as follows:

  A certain volume of nitrogen gas $(\ce{N2})$ pass through a pinhole in $\pu{20 s}$. Find the composition of a mixture between $\ce{O2}$ and $\ce{CO2}$ if this mixture is let pass through the same slit and the measured time for this effusion is $\pu{24 s}$ and of Avogadro's conditions.

  45.5% $\ce{O2}$
  69.3% $\ce{CO2}$
  12.4% $\ce{O2}$
  69.3% $\ce{O2}$
  45.5% $\ce{CO2}$

I'm not very sure if it is possible to find what it is being asked here. It's obvious that the approach here is to use Graham's law of gas diffusion. But the problem arises on what volume should be used to establish an equation from where to find the composition of the mixture.

What I thought was:
$$\frac{v_{\ce{N2}}}{v_\text{mixture}} = \sqrt{\frac{FW_\text{mixture}}{FW_{\ce{N2}}}}$$

The thing here is that the average molecular weight can be used to find the composition of the mixture.

$$\frac{v_{\ce{N2}}}{v_\text{mixture}}=\sqrt{\frac{\chi_{\ce{O2}} \cdot FW_{\ce{O2}} + \chi_{\ce{CO2}} \cdot FW_{\ce{CO2}}}{FW_{\ce{N2}}}}$$

But the thing here is I am still stuck on where to go from here. What should be the right way to go to get the composition of the mixture. Could it be that the volume of the nitrogen is needed?. So far I'm going in circles with this. Regarding the source of this problem it doesn't indicate as far I'm aware since it is a collection of problems from different topics in gases.

Go on! You are nearly at the end. If you square your equation, and replace $y$ by $1-x$, you obtain: $$(24/20)^2 = \frac{32·x + 44(1-x)}{28}$$
You finally obtain: $x = 1/3$ and $y = 2/3$.


Liquid-Liquid phase separation Consider a container that contains a mixture of many liquids. Can two of these liquids phase separate without having any significant effect on other liquids inside the container? Is there a feasible scenario where the above can actually happen? I am interested in a scenario where the two liquids only separates themselves from each other and not the rest of the mixture.

More precisely:
Assume the container contains a mixture of liquids A and B with many other liquids (state 1). Can these two liquids separate (triggered by change in the pressure, temperature, ...) so that at the end (state 2) the liquids in the container are divided into two regions with the following properties:

The first region does not contain liquid B. It contains liquid A with the exact same mixture of the other liquid (same as state 1).
The second region does not contain liquid A. It contains liquid B with the same mixture of other liquids (same as state 1).

Indeed we want other liquids in the container to be neutral to the phase separation of the liquids A and B.

Separation can be done in two ways:

Natural:

Natural separation term is not a good one but i'm gonna use it here. If at equilibrium (When the free energy of all the chemical constituents are same), those two specific liquid like to separate themselves out from the rest of the mixture, than you just need to give it enough time to reach equilibrium to separate them out. And this is the only method i can think of which will have no significant effect on the liquid in that container.
Forced:

Forced separation: Now if the above mentioned equilibrium condition needs different temp. and pressure than room temp. and pressure. Then by simply changing temp. and pressure you can achieve the above separation. Or you can utilize the different in various properties to separate them, e.g. Viscosity, boiling point, melting point etc.
Distillation: If they have separate boiling point than you can use distillation column and collecting the liquid at various height will give you separated solvent.
Solvent extraction, Filtration, mixer settlers, centrifugal force, ion exchange, membrane filtration, using bio separation like algae etc you can also achieve liquid liquid extraction.


How to determine whether a organic compound is diamagnetic or not? I have faced several questions regarding the diamagnetic behaviour of certain organic compounds. I know that one has to check the electronic configuration of the atoms in the compound and see if there is an unpaired electron or not? Of all aromatic compounds (benzene, cyclopropene cation, cyclopentene anion) that I have come across one thing was common they were all diamagnetic. I have the following question in my mind

Are all aromatic compounds necessarily diamagnetic and if it is true what may be the proper reasoning for that?
Moreover, the compounds which are stabilised by resonance (hyper conjugation,+M effect) are necessarily diamagnetic. I have this doubt because by the electronic configuration tertiary butane cation should be paramagnetic ($2p^1$ configuration) but is still diamagnetic.

The above is the structure of tertiary butyl cation. It is clear that if we consider the electronic configuration of the central carbon it
 is $1s^{2}2s^{2}2p^{1}$.So it must be paramagnetic. This is where my confusion is.

If all the electrons are spin paired then the molecule will be in a singlet state$^*$ and will be diamagnetic, if they are not so paired then it will be  will be paramagnetic. The rules are the same whether or not the molecule is 'organic'.

Most 'normal' organic compounds, e.g. methane, benzene, etc. have paired electrons in each of the orbitals from the lowest one to the highest occupied one (HOMO), and thus are diamagnetic. The orbitals including the LUMO and up are unoccupied.

Radicals such as $\ce{CH3\cdot}$ have an unpaired electron (doublet state) and so are paramagnetic.

If a biradical is formed e.g. carbenes (see IUPAC Gold Book 'biradical' for an example structure) then it will depend upon whether the spins are paired in a singlet state (diamagnetic) or parallel as in a triplet state which will be paramagnetic. Often singlet and triplets are close in energy with the triplet being lowest. Similarly, dioxygen has two unpaired electrons and forms a triplet ground state so is paramagnetic. Nitric oxide  has one unpaired electron and is paramagnetic.

If a molecule is put into an electronically excited state by absorbing a photon then both singlet and triplet states can be produced, however, theses are short lived and decay back to the ground state.

$^*$(singlet /doublet/triplet refer to spin multiplicity, spin quantum number  $\pm 1/2$ and total spin e.g. $S=s_1+s_2$ gives multiplicity g  as $g=2S+1$)

example of biradical from IUPAC Gold Book.


Saltwater Desalinization If a bucket of seawater is put into a centrifuge, would the spinning force be enough to move the salt in the solution toward the outer wall of that centrifuge?

This will not work. Centrifugation will only separate heterogeneous mixtures by density. For example, when centrifuging blood, the cells and platelets will separate to the bottom of the tube because they are not in solution. Depending on the lipid concentration in the plasma, a lipid layer may phase separate (not in solution) and move to the top.

The salt in sea water is dissolved in solution. It will not separate. Even if the centrifugation would produce a concentration gradient across the solution, diffusion would counteract that gradient.


Getting the wrong pH value, according to data and online calculator I'm doing a titration problem with a week acid and strong base, I used experiment data and an online calculator to check my answer, but my pH I calculated is a bit off. Can anyone see where I messed up? I can't figure out where...

AcOH
0.0984M
0.02500L
0.00246mol
Ka 1.8x10^-5

NaOH
0.096M
0.02580L
0.00025mol

After rxn:

0mol AcOH left

0.0025mol - 0.00246mol = 4.0x10^-5 mol OH excess

0.00246mol AcO-

4.0x10^-5mol OH / 0.0508 L total soln = 7.87x10^-4 M OH-

-log([OH-]) = pOH = 3.104

pH = 10.90

The pH should be 10.52 according to my data and the online calculator.

As you can see from the point that $n_\ce{OH} > n_\ce{AcOH}$, there will be an excess of hydroxide ions. What you want to calculate is the pH for this amount of hydroxide ions within the total volume at this point.

Thus the calculation looks like:
\begin{align}
[\ce{OH-}] &= \frac{n_\ce{NaOH}-n_\ce{AcOH}}{V_\text{a} + V_\text{b}}\\
\Leftrightarrow \mathrm{pOH} &= -\lg ([\ce{OH-}])\\
\Leftrightarrow \mathrm{pH} &= 14+\lg ([\ce{OH-}])
\end{align}

Your error is based on a rounding error ... please don't round intermediate result as all rounding errors will sum up. You just need to round the final results to a proper amount of decimal places.

The correct calculation:

$$\mathrm{pH} = 14 + \lg \left(\frac{0.0024768 - 0.00246}{0.025 + 0.0258}\right) = 10.5194 \approx 10.52$$

vs. your calculation:

$$\mathrm{pH} = 14 + \lg \left(\frac{0.0025 - 0.00246}{0.025 + 0.0258}\right) = 10.8962 \approx 10.90$$


Solubility Product and Buffers The ionization constant of benzoic acid is $6.46 \times 10^{-5}$ and solubility product for silver benzoate is $2.5 \times 10^{-13}$. How many times silver benzoate is more soluble in a buffer of $\mathrm{pH}$ 3.19 compared to its solubility in pure water?

Equation used to calculate solubility of silver benzoate in water:

$$\ce{C6H5COOAg <=> C6H5COO- + Ag+}$$

I've calculated the solubility of silver benzoate in water , it is $5 \times 10^{-7} \pu{mol/L}$ (since that is all I calculated, this is the only equilibrium equation I know).

But I cannot understand what to do next.


Nomenclature of Organic Compound from Condensed Formula I am just unable to name $\ce{C6H5CH2C6H5CH2C6H5}$ using IUPAC nomenclature.
I tried searching online however I couldn't find a name for it.

Are you looking for this? 1,4-dibenzylbenzene

http://www.chemspider.com/Chemical-Structure.205991.html?rid=80e8a67d-0782-4794-bf27-db26ccd1ce39&page_num=0

What I did was go to their Structure tab and input what I thought to be the compound you typed. If this isn't it, you can probably draw what you think it is.

Hope this helped!


Biphenyl symmetry group dependence on torsional angle What is the symmetry group of the biphenyl when the torsional angle between the phenyls is 20 degrees? What about when it is 90 degrees?

The $20$ degree structure exhibits the $D_2$ point group and the $90$ degree $D_{2d}$.

The figure below shows an $D_{2d}$ structure (a spiropentadiene) but the similarity in symmetry should be clear (I hope ). (From molecule-viewer.com)

The biphenyl at an angle not $90$ degrees is shown below