Why can hydroxide not be used to synthesise ethanol from ethyl amine? In the following reaction we have to find A and B.

I thought that A could be $\ce{Br_2}$/$\ce{KOH}$ resulting in a Hoffmann bromide reaction.

For B, I thought I would need a base such as $\ce{KOH,NaOH}$. But the answer is $\ce{HNO2}$.
Why?

Hydroxide may be a somewhat good nucleophile but the amide anion is one of the most lousy leaving groups you can imagine. If you want to remove an amino group in an $\mathrm{S_N2}$ reaction, it can only be eliminated as an ammonium cation — but protonating it means we can no longer use hydroxide as a base. Under basic conditions an amino group will never be a leaving group.

Instead, you must turn the amino group into something that is actually a good leaving group. If you include nitrogen and let your thoughts wander, you may arrive at the diazonium group $\ce{-N2+}$. Upon nucleophilic attack, this will be released as nitrogen gas which is great both thermodynamically and kinetically. This leaving group is so good that even relatively bad nucleophiles such as water will react with it.

However, you need to produce it from the amine. This is where nitrous acid comes in. Under acidic conditions, $\ce{HONO}$ will react with primary amines to give diazonium salts according to the following general equation:

$$\ce{R-NH2 + HONO + H+ -> R-N2+ + 2 H2O}\tag{1}$$

Since nitrous acid is produced by acidifying aquaeous solutions of nitrite salts, you have all three components in hand if you ‘just’ add $\ce{HONO}$ or $\ce{HNO2}$.


Gibbs Free Energy and ATP Gibbs free energy describes the maximum of useful work that can be done by a chemical reaction. My question is(Sorry I am not into chemistry and biochemistry) : during a chemical reaction that creates ATP, we say that ATP stores energy between its chemical bonds . For exemple this is the equation of creating ATP: ADP + Pi + energy ----> ATP and of course this reaction has DG. What is the relation of DG with the stored energy? Is it the same?

No, it can't be the same.  If it were exactly the same, then it would mean that ATP was storing every last iota of available free energy.  Another way of saying that is that ATP synthesis would be 100% energy efficient.  That is, the free energy change $\Delta G$ in the reaction $\ce{ADP + Pi + energy -> ATP}$ would be zero.

But perfectly reversible reactions that have 100% efficiency never happen in the real world.  Even theoretically, they can only happen at infintessimally slow rates.

Actual ATP synthesis isn't 100% efficient.  It's efficiency depends on the details of the particular organism and its biochemistry, etc.  Many organisms show efficiencies of 30 - 45% in the synthesis of ATP.


Reasoning for redox reactions to be classified as such Could anyone help me out solving this question regarding redox reactions? I am fairly new to this topic.

Which of the following is a redox reaction?
(1) $\ce{NaCl + KNO3 -> NaNO3 + KCl}$
(2) $\ce{CaCO3 -> CaO + CO2}$
(3) $\ce{Mg(OH)2 + 2NH4Cl -> MgCl2 + 2NH4OH}$
(4) $\ce{Zn + 2AgCN -> 2Ag + Zn(CN)2}$

I know that the answer here is option (4) but could someone help me understand why this is so? I see a change in oxidation in the other three equations but it is not redox.

$\ce{Zn + 2AgCN -> 2Ag + Zn(CN)2}$
Zinc changes its oxidation state from $0$ to $+2$, and silver changes its oxidation state from $+1$ to $0$. So, zinc is reducing agent and silver is oxidising agent.
$\ce{Zn^0 ->[-2e^-] Zn^{+2}}$ (oxidation);
$\ce{2Ag^{+1} ->[+2e^-] 2Ag^0}$ (reduction).
Other reactions are just equations of ion exchange without changing the oxidation states of chemical elements.


Boric acid ant poison I made an ant poison using boric acid and peanut butter for grease eating ants.  They eat the peanut butter (mixed with some olive oil to get it closer to a liquid), but I see a white residue (crystals?) left behind.  How can I get the boric acid dissolved in an oil or grease?

Boric acid is soluble in water, not oil.  Pick an emulsifier and add it to your ant poison.

Boric acid needs to remain acidic to retain its effectiveness as an ant killer.  Ants deaths due to boric acid poisoning is a topic still up for debate; however, it has been suggested boric acid is a neuro-toxin and that the acid damages one of the organs in an ants digestive system.

Peanut butter is usually slightly acidic; however, your particular mixture may be rendering the boric acid inert.  Solutions of sugar and boric acid with a high pH have been known to pass right through ants doing little harm. Perhaps using an acidic additive will increase the bait's effectiveness.

Boric acid in a 1-2% concentration may be ideal if the entire ant colony is targeted. To target the entire colony, enough poison should be used so the ant does not die before it makes it back it its colony.


why are simple organic molecules usually white? I have heard that it is because they absorb electronically in the ultraviolet range, if this is true, why is it true. What is it about the nature of organic molecules that make them absorb in the ultraviolet range.

Most substances are white in powder form if they have no strong absorption of visible light. They whiteness is a consequence of light scattering by small particles and the fact that there is no absorption within the visible range.

The lack of visible absorption is another issue. Light is absorbed by molecules when the energy difference between a filled orbital and an unfilled one in the molecule matches the energy of a photon of light. For light photons the energy is the equivalent of wavelength and therefore colour.

In most small, simple organic molecules, the lowest unfilled orbitals are a long way from the highest filled ones. So they need a lot of energy to cause them to jump to an unfilled orbital. So they absorb only high energy photons (UV light is higher energy than visible light so this implies the absorption bands will be in the UV and we will not notice them as our eyes can't see UV).

The characteristics that create colour in molecules are often long runs of unsaturated bonds where the electrons can spread along the chain of unsaturation. In these delocalised systems the gap between occupied and unoccupied orbitals is smaller than in simpler molecules and often falls into the visible spectrum. These molecules are perceived as coloured.

The molecule below is a simple example of a dye (most are bigger and more complex than this):

The electrons in this molecule can delocalise over all three rings via the NN bridge and this creates orbitals with the right energy gap to absorb blue light.

In simpler molecules with little delocalisation the energy gap is far into the UV. Even in polycyclic aromatic systems the energy gap is often still in the UV.

Of course, this is a simplification, but the essence of what sort of molecules you need to cause visible colour is a relatively large unsaturated system and by definition this doesn't happen in very simple organic molecules. Wikipedia lists many more examples of dyes so you can get a feel for what sorts of structures make strongly coloured compounds.


Is having a lone pair sufficient and necessary for a Lewis base? What about empty orbitals and Lewis acids? Is any molecule with a lone pair and/or a pi bond considered a Lewis base? Also, is any molecule with an incomplete octet a Lewis acid?

No, not every molecule that has a valence lone pair is automatically a Lewis base.

For example, tin forms divalent salts such as $\ce{SnCl2}$ wherein tin is in the $\mathrm{+II}$ oxidation state and thus its electronic configuration is $\ce{[Kr]}\ \mathrm{4d^{10}\,5s^2\,5p^0}$. Even if we say that the d-electrons are not really valence electrons, since they are not really chemically accessable, you cannot deny that tin(II) has a lone pair in its 5s orbital. However, this lone pair does not exhibit Lewis basic properties; instead, tin(II) functions as a Lewis acid. The reason for that is the spherical symmetry of the s-orbital along with its notable dispersion, which means it cannot be reasonably donated in a Lewis-basic matter.

Note that the lone pair is accessable chemically: tin(II) can be oxidised to tin(IV) e.g. by $\ce{HgCl2}$.

A somewhat similar (but much more questionable!) case can be made for electron sextets and Lewis acids. NHCs are a pretty common substance class. The most generic structure is that of imidazole, but with both nitrogens substituted so that one of them bears a positive charge (the one with a double bond to the carbon sandwiched between the two). If a base is added to remove that sandwiched carbon’s proton, a carbene is generated in which said carbon is only bonded to the nitrogen, has an $\mathrm{sp^2}$ configured lone pair and is considered to have an empty $\mathrm{p}$ orbital. However, this carbene is nucleophilic, which is why they are also called nucleophilic heterocyclic carbenes. They react as Lewis bases with metal centres; the second generation of Grubbs’ metathesis catalysts made use of NHCs.

Of course, it is slightly debatable whether the p-orbital on that NHC carbon is truly empty. One could argue that significant electron pair donation from both nitrogens should happen.

(As a side note: radicals, while not having full octets, are typically never considered Lewis acids — but then, their orbital is already occupied by a single electron, so they don’t have an empty orbital that would make them a potential Lewis acid, anyway. Thus, your question should better be worded differently to explicitly exclude radicals.


T∆S implies that heat flows which way? From Schroeder's Introduction to Thermal Physics.

Parts $(a)$ and $(b)$ are quite simple, and we find that $ \Delta H =-2808\ \mathrm{kJ}$ and $\Delta G =-2879\ \mathrm{kJ}$, so by $\Delta G = \Delta H -T\,\Delta S $, we find that $T\,\Delta S = 71\ \mathrm{kJ}$ at room temperature.

What I absolutely cannot understand is how $(c)$ and $(d)$ work. It intuitively makes sense that entropy is increasing. Assuming $1$ mole of glucose, we start the reaction with $7$ moles of reactants and end with $12$ moles of products. Entropy obviously increases, as we can arrange these molecules in many more ways. However, this implies that we are getting usable work from increases in entropy – meaning that the muscles are somehow absorbing heat energy from the environment. But this isn't true in real life – metabolism reactions obviously release heat. What's going on here?

The fact that the book is using $\Delta G$ as the relevant state variable for free energy implies this is being carried out at constant T and p.  So, proceeding:

  "However, this implies that we are getting usable work from increases in entropy...."

Yes, that's correct.  The positive $\Delta S$ lowers $\Delta G$

  "However, this implies that we are getting usable work from increases in entropy – meaning that the muscles are somehow absorbing heat energy from the environment."

No, that's incorrect.  The heat flow here is purely due to $\Delta H$, and it is flowing out of the system.

What you're probably thinking is that, at constant T (which we have here), $$\Delta S = \frac{q}{T},$$
and thus, since $\Delta S >0 $,  q must be $> 0$ as well, i.e., that the  positive $\Delta S$ means heat is flowing into the system.

This is wrong.  It is not the case that $\Delta S = \frac{q}{T}$.  Rather:

$$\Delta S = \frac{q_{rev}}{T},$$

and this reaction is not happening reversibly.  Instead,  $\Delta S$  is determined from the difference between $S_{products}$ and  $S_{reactants}$, and each of these individual values are calculated from:

$$S(T') =\int_{0}^{T'} dS= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$$

[For more on this, see my answer here:  Infinite Increase in Entropy when Energy added to Absolute Zero ]

In summary, in a chemical reaction at constant pressure (and with no non-pV work),

$$\text{đ}q_{sys}=dH_{sys}$$

Thus it is the sign of $\Delta H$, not $\Delta S$, that determines the direction of heat flow.

Finally, the answer to (d) is that, as thermal energy is generated by the reaction, it heats up the system, and the 2nd law of thermodynamics says that thermal energy naturally flows from hotter to cooler regions.  So there is heat flow into the surroundings until the reaction mixture returns to the temperature of the surroundings.


Specific and general acid catalysis What is the rationalization for calling reactions with rates independent of pH but dependent on the concentration of a particular acid catalyst under "general" acid catalysis, and pH dependent reactions under "specific" acid catalysis ?

It would seem to me that the pH-independent reactions would be better considered to be under "specific" acid catalysis given as they are catalysed by a specific species.

I have wondered about this too. Although I could be wrong it is reasonable to assume (or at least handy to keep as a reminder) that the term "special" refers to the fact that compound is protonated in a very specific state, namely before the actual reaction occurs. General acid catalysis can happen both before the reaction or at any time during the reaction (i.e. in some transition state), whatever is likely in some range around the pH of the acid. Hence it is more general version of the special case mentioned above.


What significance and applications does nanotechnology have in chemistry? Specifically concerning organic chemistry, but an explanation on its significance and applications in inorganic chemistry would be great as well.

There is branch named Chemical Nanoscience and Nanotechnology Which deals with chemistry asspects.

Talking about organic chemistry, nanotechnology is used for Organic synthesis.

With help of Modern Chemistry today we can synthesis almost all complex molecules, thanks to nanotechnology.

Check out Wikipedia-1,
Wikipedia-2 and nano.gov

For organic chemistry, Polymer nanotechnology is developed notably.  Polymer matrix and nanocomposites have become a prominent area of current research and development

Example:

exfoliation


Validity of Henry's Law for an Oxygen/Bitumen System? Henry's Law is for a gas dissolved in a liquid and states P = kC where P is the partial pressure of the gas above the solution and C is the concentration of the gas dissolved in the liquid. The k is a "constant", that is, it is independent of P, C and time but depends on the type of gas and the type of liquid. According to Pickover, page 184 (reference below), Henry's Law assumes:

A1.  the gas/liquid system has reached equilibrium, and
A2.  the gas does not chemically react with the liquid, and
A3.  the concentration of gas in the liquid is low, and
A4.  the gas pressure is low.

I would be grateful for any help in answering the following:

Q1.  Is the list of assumptions, A1 to A4, correct and complete? I have no reason to doubt it but I would like to confirm it.

Q2.  Some authors (e.g. Dickinson E.J., Road Materials and Pavement Design, 2000, 1(3), 255-280) have applied Henry's Law to the diffusion of oxygen in air at atmospheric pressure into bitumen which is a semi-solid at room temperature. Is there any experimental evidence that Henry's Law can be extended to such a system?

Reference
Pickover, C., "Archimedes to Hawking", Oxford University Press 2008

Regarding Q1, Henry's law states:

  At a constant temperature, the amount of a given gas that dissolves in
  a given type and volume of liquid is directly proportional to the
  partial pressure of that gas in equilibrium with that liquid.

A1 is answered by the definition itself: the system should be at equilibrium.
A2 is straightforward: if the gas reacted, you would not have a constant pressure as a certain (if not all) quantity of it would be removed from the system[1]. This would also keep your system from being at equilibrium, defying the previous assumption.
A3 and A4 both depend on the fact that – as the concentration or the partial pressure rise – the system deviates from an ideal model where the solute is
infinite-diluted and in such system Raoult's Law should be used
instead[2]. So you need a low concentration and partial pressure of your gas.

To answer to Q2, Henry's Law can be applied to any solvent, whether liquid, vapour or solid[3], [4]. It has been used, within certain limits, to study absorption of gas in zeolites[5] or hydrogels, but I couldn't find any precise experimental data on it. Only that in such systems it is only a limited law, adequate at low mole fractions, but useful for practical purposes when high precision is not required.

    1 Henry's Law, UC Davis University ChemWiki
    2 North Carolina State University, CH 431 / Lecture 14
    3 Francis L. Smith and Allan H. Harvey (September 2007). "Avoid Common Pitfalls When Using Henry's Law".
    4 Rosenberg, Robert M., Peticolas, Warner L., Journal of Chemical Education, Volume 81, 11, pp 1647-1652
    5 Eric G. Derouane, Zeolites as solid solvents, Journal of Molecular Catalysis A: Chemical, Volume 134, 1–3, pp 29-45


Why does a silane group react with hydroxyl groups? I read somewhere that a silane group will react with hydroxyl groups.

Does anyone know why this would happen?

It is well known that silicon forms very strong bonds with oxygen (think sand, $\ce{SiO2}$). Hence, any time one can convert a $\ce{Si-H}$ bond into $\ce{Si-OH}$, it will happen.

I believe that silanes are extremely air sensitive due to the moisture content.


Amperage for the electrolysis of water? I am electrolysing water using several different potential catalysts for my eight grade science fair project. The electrolysis will be running at 2 to 2.5
volts over nickel electrodes, as I do not need a long term solution. How much amperage should I put on my DC power supply? Thanks and I would appreciate simplified answer compared to what you would normally give. Yes, I do understand ohms law.

Edit: I did some calculation and if I assume the resistance of the electrolyte is lets say 5 ohms, then we can assume a current of 0.5A at 2.5 volts. Is this correct?

In electrochemistry, the rate of electrolysis depends on rate of charge entering the cell. Second most important point is that one can either control potential or current but not both during electrolysis i.e. you cannot have both values set. If you are fixing potential at 2.5 V, the value of current is not in your hand. I don't think Ohm's law remains valid during electrolysis in general.

So the main question one should ask is: What rate of electrolysis do I need? For example, you may want 200 mL of $\ce{O2}$ in 60 min. From here you calculate the charge Q needed to collect this volume of oxygen. Current is charge per unit time (3600 s) and this is your required current above the decomposition voltage of current. See second figure: http://www1.lsbu.ac.uk/water/electrolysis.html

See the sample calculation here: https://www.quora.com/What-amount-of-current-is-required-to-produce-0-224-cc-STP-of-O_2-at-anode-during-the-electrolysis-of-water


Using the right equipment for successful detection of Novichok Apparently, Navalny’s biomaterials were flown to Moscow's Nii Sklif lab which didn't find any traces of the poison. But they used an American Agilent Technologies GC-MS [1], while Bundeswehr toxicology lab reportedly used an order of magnitude more sensitive, and more expensive, Bruker's apex-Qe FTMS (no official source for this as "Bundeswehr didn’t disclose the details of their work") which successfully detected novichok.
How did German scientists outplay their Russian counterparts in this case and could it be attributed to simply using a more sensitive equipment with perhaps a better MS method?

Lacking additional information, answers likely will be speculation-based only.   To mention a few factors to consider:
Stating GC-MS analysis only states, that there is technique to separate compounds by different rate of retention over a stationary phase (gas chromatography, GC) combined (a.k.a. hyphenated) with an other technique to analyze fractions sorting at different times by mass spectroscopy (MS). Mass spectroscopy typically shatters the molecules into charged fragments which, by their mass/charge ratio, behave differently in electromagnetic fields.  However, depending on the setup, the separation of the fractions by GC may be better, or worse.  On the side of mass spectroscopy, there are multiple methods to ionize and fragment molecules and to record these.  It is a problem that the parameters of running a mass spectroscopy may alter the shape of mass spectra quite a lot; signals may vary in their relative intensity, or sometimes be below the detection limit altogether.  It need not be the engineering advantage among corporations, of Agilent's technology over the by  Bruker (if there is?).
In the line of smaller benchtop mass spectrometers typically mounted behind a GC, larger sector field mass spectrometers, and Fourier mass spectrometers, there is a significant increase of resolution to record tiny differences in the masses of the fragments  the experiment of mass spectroscopy generates.  This may help to attribute the signals to the correct analyte, especially if there already is a suspicion to search for a selection of compounds only.
Routine GC-MS used for forensics and IOC doping screenings, for example compare the mass spectra recorded for a sample against previously recorded trustful references.  (Different to IOC doping screening, it is not evident if the Russian government shares / shared such standards here, though.)   Depending on the compound to identify, though, the analysis does not need to stop at the level of a GC-MS.  Both other types of chromatography may have been used now (e.g., HPLC) to separate the compounds; or multiple mass spectrometers coupled into a tandem (MS/MS) to observe not only the initial fragments obtained by mass spectroscopy, but equally their further fragmented «daughter radicals», too which equally rises the chance to identify your compound in question.
Maybe the Germans did not trace the poison itself, but metabolites judged as sufficient evidence for earlier reported, related compounds.  Especially with small amounts of sample material, the success of an analysis equally depends on every step of sample preparation; here likely (incomplete listing): what was the delay between Nawalny's intake of compound X and obtaining a sample by him (metabolism), the nature of the sample (tissue, blood, hair, etc.), delay and conditions of storage of the specimen to get it to the lab (unwanted sample decomposition), concentration and digestion protocols in the lab to eventually submit the sample to the mass spectrometer; the people's experience and availability reference data / techniques.


Determining Kc for the equilibrium involving iron thiocyanate using spectrophotometric data $$\ce{Fe^3+(aq) + SCN-(aq) <=> [Fe(SCN)]^2+(aq)}$$
$\pu{12.0 mL}$ of $\pu{0.00110 mol L-1}$ $\ce{Fe^3+}$ was added to $\pu{6.00 mL}$ of $\pu{0.00140 mol L-1}$ $\ce{SCN-}$. The absorbance of the resulting solution was $0.625$. If the constant $(\varepsilon b) = \pu{3800 (mol L-1)-1}$, determine the value of $K_c$ that should be reported for the equilibrium.
a. $\pu{1.64e-4}$
b. $\pu{3.03e-4}$
c. $\pu{4.67e-4}$
d. $\pu{5.69e-4}$
e. $957$

Answer given: e.
Given what I know about $K_c$ it should be:
$$K_c = \frac{\left[\ce{[Fe(SCN)]^2+}\right]}{[\ce{Fe^3+}][\ce{SCN-}]}$$
which gives me the answer a. I tried calculating $C$ by manipulating Beer's law, but I still don't end up with $957$ no matter what I try.
I would appreciate it if someone could walk me through the steps of solving this problem.

The problem is a bit strange since the details of the absorbance measurement aren't detailed very well. I'll assume the book defines the relationship as:
$$A = \epsilon bc$$

where:

A=absorbance
$\epsilon$ = molar attenuation coefficient
$b$ = path length
$c$ = concentration

So the concentration, c, of the $\ce{[Fe(SCN)]^2+}$ is:

$$ c = \dfrac{A}{\epsilon b} = \dfrac{0.625}{3800} = 1.64\times10^{-4}$$

Considering the dilutions...

$\ce{[Fe^{+3}] + [Fe(SCN)]^2+} = 0.00110 \times \dfrac{12}{18} = 7.33\times10^{-4}$

$\ce{[SCN^-] + [Fe(SCN)]^2+} = 0.00140 \times \dfrac{6}{18} = 4.67\times10^{-4}$

But there is $1.64\times10^{-4}$ of $\ce{[Fe(SCN)]^2+}$ so:

$\ce{[Fe^{+3}]} = 7.33\times10^{-4} - 1.64\times10^{-4} = 5.69\times10^{-4}$

$\ce{[SCN^-]} = 4.67\times10^{-4} - 1.64\times10^{-4} = 3.03\times10^{-4}$

Therefore:

$K_c = \dfrac{1.64\times10^{-4}}{(5.69\times10^{-4})(3.03\times10^{-4})} = 951$

I'd assume the small difference is just due to a difference in rounding somewhere...


Converting mol files to InChiKey in large batches I want to convert mol files to InChiKey. I am using OpenBabel for the conversion but I have over 500 mol files and converting them one by one is too time consuming. Is there some way to convert them all at once?

Edit 1 This is a sample mol file

0187.cdx
  Sample

  8  8  0  0  0  0  0  0  0  0999 V2000
   -0.4125    0.7145    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -0.8250    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -0.4125   -0.7145    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.4125   -0.7145    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0
    0.8250    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.4125    0.7145    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0
   -1.6500    0.0000    0.0000 Cl  0  0  0  0  0  0  0  0  0  0  0  0
    1.6500    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  2  0
  2  3  1  0
  3  4  2  0
  4  5  1  0
  5  6  2  0
  6  1  1  0
  2  7  1  0
  5  8  1  0
M  END

Also I want to know is there some other format which can combine mol files into one.I think I had heard of another format like mol which can combine many mol files,but I cannot remember the name of that format

Edit 2-  What is SDF format? Can it combine many mol files into one?

This is exactly why I wrote Open Babel.

obabel *.mol -oinchikey -m

As long as you're not exceeding the number of command line arguments, you're done.


Recommendations for electrostatic potential surfaces Are there standards for density cutoffs, colours and colour mappings for electrostatic potential surfaces?

I observe that people use many different values.
For instance, this article (in Fig. 1a) uses 0.001 au density cutoffs with colours ranging from blue (< 0 kcal/mol) to red (> 20 kcal/mol).

I understand that not every colour/energy range will work every time, but other publications/websites even switch the colour range.
This website, for instance, uses red for negative energies and blue for most positive ones.
Furthermore, the same website suggests a .02 au cutoff.

Are there recommendations on this?
Which values are more commonly used by textbooks?

You have probably found out from playing around that there are many options. For coloring there is no standard, blue and red are common because they are the easiest to see since they contrast well. As far as the density cutoff, there are some papers that recommend values but the main idea is to choose a value that gets close to the Van Der Waals radius for the atoms. This is after all what is typically is of interest in a chemical sense. The electron density at the Van Der Walls surface is what others atoms/molecules "see". Typically I will just plot both the Van Der Waals and electrostatic surface and modify the electrostatic cutoff to make them close. Then you can set the limits for positive and negative colors.


Why are compounds of V5+ and Cr6+ colored? As far as my knowledge goes, transition metal complex ions show a color due to a transfer of electrons between the levels of split $\mathrm d$-orbitals. In the case of $\ce{V^5+}$ and $\ce{Cr^6+}$ however, there are no electrons in the $\mathrm d$-orbitals, and the electronic configuration of both is similar to $\ce{Ar}$, yet $\ce{V^5+}$ and $\ce{Cr^6+}$ show color(yellow/brown and orange respectively).

Firstly, a note about terminology. As Mithoron noted, the naked ions $\ce{V^5+}$ and $\ce{Cr^6+}$ do not exist because they require a prohibitively large amount of energy to make (just take the sum of the first 5 or 6 ionisation energies of V/Cr to see why). As such your question should really be asking about the vanadium(V) and chromium(VI) oxidation states, which can exist in acidic solution as the oxoanions $\ce{VO2+}$ and $\ce{Cr2O7^2-}$ respectively. The same idea can be extended to the manganese(VII) oxidation state, which exists as $\ce{MnO4^-}$.

Now, on to the question itself. As you rightly note, the central atom in these oxidation states have the same electronic configuration as argon; we usually call it a $\mathrm{d^0}$ configuration to indicate that there are no d electrons. The colours of these species therefore cannot arise from d-d transitions, which are the transfer of electrons from one metal orbital to another metal orbital. They are actually associated with electron transfer from the ligand (in this case, $\ce{O^2-}$) to the central metal atom. The name for this is ligand-metal charge transfer (LMCT).


Find the number of water molecules "embedded" in sucrose Recently I got a question as follows:

  How many molecules of water are joined with 12 atoms of carbon in a molecule of sucrose ?

  (A) 12

  (B) 11

  (C) 6

  (D) 10

I am unable to understand the question.
What actually is the question ?

Sucrose is a carbohydrate.  This means that stoichiometrically, it can be regarded as adding water ($\ce{H2O}$) to elemental carbon $\ce{C}$.
The formula of sucrose is $\ce{C12H22O11}$.
Another way to write this formula is $\ce{C12(H2O)11}$.
Thus, sucrose can be regarded (stoichiometrically, but not structurally) as eleven waters plus twelve elemental carbons.  Therefore the correct answer is (b).
The question is not well-worded, and would be better stated as "How many water molecules are required to join with the 12 carbon atoms in sucrose to give the correct empirical formula for sucrose?".
In the comments Ivan Neretin offers an even better variant, which I will re-word as "How many water molecules could extremely powerful dehydrating agents such as $\ce{H2SO4}$ or $\ce{SO3}$ extract from sucrose, if we assume the only non-water product is pure carbon?".


Cell voltage with different n factors I’m a bit confused regarding cell potentials. If I have two half  reactions each involving a different number of electrons, there are two ways I can see to calculate the net emf of their cell:

Simply add the potentials across the two theoretical half cells, that is, treat each half reaction as a voltage source in series
Calculate the value for the net cell reaction as a whole, that is, make consideration for the energetics of the reaction

As a more specific example, say I have a metal $\text M$, and $$\ce{E^0_{M^{3+}/M}=2V, E^0_{M^{2+}/M}=1V}$$ and would like to figure the standard cell potential, say $x$ of $\ce{E^0_{M^{3+}/M^{2+}}}$.
With Method 1 we have $$x= \ce{{E^0_{M^{3+}/M}}-{E^0_{M^{2+}/M}}}=2-1=\mathbf{1V}$$
With Method 2 we have $$\ce{\Delta G^0_{M^{3+}/M}=-3*F*2=-6F, \Delta G^0_{M^{2+}/M}=-2*F*1=-2F}$$
$$\implies \ce{\Delta G^0_{net} = -6F - (-2F) = -4F}$$
$$\implies x=\frac{-\ce{4F}}{\ce{-1*F}}=\mathbf{4V}$$
How do I rectify these inconsistent values? Does it make any difference here whether I say I am looking for $\ce{E^0_{M^{3+}/M^{2+}}}$ or $\ce{M|M^{2+}(1 molar)||M^{3+}(1 molar)|M}$? What if I use an inert electrode like $\ce{Pt}$ in the latter representation?
Further, if I have another metal $\ce Y$ for which $\ce{E^0_{Y^+/Y}}=\ce{3V}$, which is a stronger oxidizing agent? $\ce{M^{3+}}$ or $\ce Y^+$?
In various sources it seems some use method 1 while others use 2. In this answer on CSE it seems 1 is the right one. But I maybe missing some conceptual context.
Thanks!

If I've understood your question properly, you are asking - in the condition $E^\circ_{\ce{M^3+/M^2+}} =\pu{4 V}$, $E^\circ_{\ce{M^3+/M}} =\pu{1 V}$, $E^\circ_{\ce{Y+/Y}} =\pu{3 V}$ - whether $\ce{M^3+}$ is a stronger oxidising agent than $\ce{Y+}$ irrespective of it going to $\ce{M \text{or} M^2+}$.
This would not be true since $\ce{Y+}$ has a higher reducing potential than $\ce{M^3+/M}$ but not $\ce{M^3+/M^2+}$, and so $\ce{M^3+}$ would be a stronger oxidising agent than $\ce{Y+}$ where it converts into $\ce{M^2+}$ but not when $\ce{M^3+}$ converts into $\ce{M}$.

The method $1$ that you suggest is a very specific case that can only be used for a general redox reaction. By this, I mean a reaction as follows for a cell having it cell notation denoted as $[\ce{N(s)|N^{y+}(aq)|| M^{x+}(aq)|M(s)}]$
\begin{array} {rlllc} \require{cancel}
\ce{N^y+ + ye- &-> N } &E^\circ = E_n^\circ &\quad|\times(-x)\\
\ce{ M^x+ +xe- &-> M}  &E^\circ = E_m^\circ &\quad|\times(y)  \\  \hline
\ce{yM^x+ + xN  +\cancel{(xye-)}  &-> yM + xN^y+ + \cancel{(xye-)}} &E^\circ = E_{\text{cell}}^\circ  \\
\end{array}
So here, if you apply method $1$, you get $E^\circ_\text{cell} = E^\circ_m -E^\circ_n$
Let's see whether this works using method $2$ (the free energy varient). Now, we know that $\Delta G$ is additive, $\Delta G^\circ =nFE^\circ$ and that when we multiply the total reaction by a value, we also multiply the $\Delta G$ by the same number. Using these properties we find the $\Delta G$ values for the two reactions.
For just one mole of M being reduced , we get
$$\Delta G_{\text{m}}=xFE^\circ_m \tag{1}$$
For just one mole of N, we get
$$\Delta G_{\text{n}}=yFE^\circ_n \tag{2}$$
Now, according to the above cell reaction, the $\Delta G_{\mathrm {cell}} = y(\Delta G_m) + (-x)\Delta G_n$ which would be equal to
\begin{align}
\Delta G_{\text{cell}} &= (-x)yFE^\circ_n + (y)xFE^\circ_m \\
&= xyF(E^\circ_m -E^\circ_n) \tag{3}
\end{align}
Now, the value of $\Delta G_{\text{cell}}$ in terms of $E^\circ_\text{cell}$, we get:
$$\Delta G_{\text{cell}}=xyFE^\circ_\text{cell} \tag{4}$$
Now substituting this value of $\Delta G_\text{cell}$ into equation (3), we get
\begin{align}
\cancel{(xyF)}E^\circ_\text{cell} &= \cancel{(xyF)}(E^\circ_m -E^\circ_n) \\
\implies E^\circ_\text{cell} &= E^\circ_m -E^\circ_n
\end{align}
For a case where you have the same compound disproportionate, this question and its answers should be enough.


Is Al2O3 (aluminum oxide) diamagnetic or paramagnetic? The title about covers it, I am an amateur and I have found that checking 'molecule orbitals' can reveal the answer here but I cannot locate such an orbital. The reason for this question is I am trying to sort aluminum oxide beads from aluminum oxide beads coated in palladium with a magnet via the Lenz Effect.
I have researched magnetic properties. It appears that I can determine them by seeing how many unpaired valence electrons there are. I have no idea how to figure out what the valence orbital of a molecule is, even though such information is widely available for individual elements. For molecules, this information is referred to as molecule orbitals. I expect that it is diamagnetic because both Oxygen and Aluminum are not noble, but this sentence should reveal my level of chemistry knowledge.

Actually the Lenz effect doesn't rely on paramagnetism to sort metallic objects from nonmetallic. See for example this video, where the moving magnets actually impel the nonferrous metals, permitting separation from nonmetallic components. If you take a strong magnet, like the NdFeB magnets you can get at the hardware store or even the Dollar Store, and move it rapidly over the bottom surface of an aluminum frying pan without it touching the surface you can actually feel the force between the two. This dynamic force (dynamic in that you can't feel the force between pan and magnet if they aren't moving relative to each other unless the pan is actually iron) is sufficient to impart more momentum to the metal waste in the video above and so achieve separation.
Looking at this table we can see that $\text{Al}_2\text{O}_3$ is in fact diamagnetic with $\chi_{\text{m}}=-37\times10^{-6}\text{cm}^{3}/\text{mol}$ and $\text{Pd}$ is paramagnetic with $\chi_{\text{m}}=+540\times10^{-6}\text{cm}^{3}/\text{mol}$. I have a strong pyramid magnet that can pick up some of the granules from crushed Cheerios and has a visible pull on some of the lettering on U.S. currency. I suspect it could separate out Pd-coated from noncoated beads statically.


When titrating to find the concentration of a dissolved acid, does the volume of solvent matter? I am conducting a titration to find the mass of aspirin present in tablets. I will be dissolving the aspirin tablet into a solution of ethanol and distilled water to dissolve it. But, does the volume of ethanol or distilled water added actually matter? Or does it just need to be enough to completely dissolve the aspirin?

At very great dilutions, impurities could have some effect on the end point. For example, even previously-distilled water will absorb some small amount of $\ce{CO2}$ from air, making it slightly more acid. However, if the aspirin is present in reasonable amount, that would have negligible effect. So, as stated by @MaxW, consistency, with a moderate theoretical excess of solvent, is important, but don't use greatly excessive amounts of solvent.


Why is the boiling point of water and ammonia so different? Boiling point of water is 100 degrees Celsius, while boiling point of ammonia is minus 33 degrees Celsius, which makes 133 degrees difference. Now when we discuss value of boiling point, we also say that it depends on inter-molecular forces, and in case of both water and ammonia I can see two such forces: London dispersion forces and hydrogen bonds. Relative molecular mass of water is 18 amu, and relative molecular mass of ammonia is 17 amu, so unlikely we can explain the difference in boiling points by difference in London dispersion forces: as they depend on mass, the LDF should be about the same for two compounds. Speaking about hydrogen bonds, it looks as each molecule (water or ammonia) is able to form 4 hydrogen bonds. And in this way amount of hydrogen bonds again should be about the same. So what is an explanation for the difference in boiling point between water and ammonia?

UPDATE: I found some sort of explanation, which partially seems OK, but I see at least one problem with it though:

  Notice that each water molecule can potentially form four hydrogen bonds with surrounding water molecules. There are exactly the right numbers of δ+ hydrogens and lone pairs so that every one of them can be involved in hydrogen bonding. This is why the boiling point of water is higher than that of ammonia or hydrogen fluoride. In the case of ammonia, the amount of hydrogen bonding is limited by the fact that each nitrogen only has one lone pair. In a group of ammonia molecules, there aren't enough lone pairs to go around to satisfy all the hydrogens. That means that on average each ammonia molecule can form one hydrogen bond using its lone pair and one involving one of its δ+ hydrogens. The other hydrogens are wasted. In hydrogen fluoride, the problem is a shortage of hydrogens. On average, then, each molecule can only form one hydrogen bond using its δ+ hydrogen and one involving one of its lone pairs. The other lone pairs are essentially wasted. In water, there are exactly the right number of each. Water could be considered as the "perfect" hydrogen bonded system.

Source: http://www.chemguide.co.uk/atoms/bonding/hbond.html

OK, so each ammonia molecule forms 2 hydrogen bonds and so does each hydrogen fluoride molecule. Their masses are 17 amu and 20 amu respectively (difference is very small). Boiling point of fluoride is 19.5 degrees Celsius, while boiling point of ammonia is minus 33 degrees Celsius, which makes 53.5 degrees difference. So what is an explanation for the difference in boiling point between hydrogen fluoride and ammonia?

If I interpret your edited question correctly, it now boils down to why is the difference between the boiling points of $\ce{HF}$ and $\ce{NH3}$ so large, even though they have almost the same molecular masses ($M_\mathrm{r}(\ce{NH3}) = 17, M_\mathrm{r}(\ce{HF}) = 20$).

Ammonia can self-ionise much like water can according to the following equation $(1)$. And much like water, the process is important to determine a $\mathrm{pH}$ value, but rather neglegible when it comes to determining attractive forces between ammonia molecules where ‘traditional’ hydrogen bonding predominates.

$$\ce{2 NH3 <<=> NH4+ + NH2-}\tag{1}$$

On the other hand, self-ionisation in hydrogen fluoride, especially in the liquid and solid states, is much more important and leads to the two fragments shown in equation $(2)$.

$$\ce{3 HF <=> H-F-H+ + F-H-F-}\tag{2}$$

These two linear fragments both feature de facto four-electron-three-centre bonds and are thus much more prevalent than the corresponding ammonia-derived ones. (Indeed, something analogous to $\ce{FHF-}$ doesn’t exist for ammonia.) Since these also introduce ionic interactions into the equation, the intermolecular forces between $\ce{HF}$ molecules are much stronger than those between ammonia molecules contributing to $\ce{HF}$’s higher boiling point.


Why are there so many DFT Methods? There are many different DFT method available, is this simply because individuals find a better DFT method to suit specific types of molecules?

The issue is that unlike for the methods based on Hartree-Fock, there isn't a systematic path to improving DFT. Unlike for HF, where we can treat the correlation as a pertubation (MP-n) or include excited Slater determinants (CI and CC) to generally improve our description of a molecule, we still don't have a way of determining the true form of the density functional (specifically the part governing exchange correlation) in DFT, so for now we use educated guesses for the general form (Local Density Approximation, Generalized Gradient Approximation, etc) and optimize them for specific systems.

In principle, if the exact form of the functional were known, DFT would be exact. This is why some authors like to use a separate term DFA (Density Functional Approximation) to describe the instances of DFT in use today.


Oligomer Terminology As we know, dimers, trimers etc. can be produced from a certain amount monomers. In the case of a dimer being produced, two monomers dimerise.

Let's take an example of a dimer (glycine anhydride):

Two glycine monomers were dimerised for this - it's a dimer since it has 2 (poly)amide bonds.

Now, let's take an example of an ester (ethyl ethanoate):

This is where my question comes in: would it not be appropriate to call ethyl ethanoate a monomer? As we saw with glycine anhydride, it has two (poly)amide bonds, hence it's a dimer. Ethyl ethanoate has a single ester bond, so would this not render it as a monomer? Or is this incorrect logic, since ethyl ethanoate isn't an oligomer/polymer per se?

P.S. I sometimes put the word 'poly' in parenthesis since I'm not sure whether its usage is necessary in description.

You can call it a monomer, although your audience will expect that you are about to present its oligomerization or polymerization. And according to IUPAC you should put parentheses if in the brackets you have the repeat unit of the polymer: e.g. poly(styrene).


Why does a low concentration of soap in water behave as a strong electrolytes? According to my text, "a colloid in which the dispersed phase consists of micelles is called an association colloid." An example would be soap in water. They use the term 'associated colloid' and 'association colloid' interchangeably.

They state that, "associated colloids at low concentrations, behave as normal, strong electrolytes."

I can't provide further context as they haven't provided further information.

Could you define associated colloids as it appears in your text? This is not a modern colloid chemistry term. A quick search of Google Books shows that this terminology is popular in Indian test preparation books. Micelle is a modern term. IUPAC defines micelle as "Particle of colloidal dimensions that exists in equilibrium with the molecules or ions in solution from which it is formed" Above a critical concentration, called the critical micellar concentration (CMC), the conductivity of solution drops very quickly. I will let you read more about micelle and conductivity of micellar solutions from the web.

You can watch this video for starting purposes: https://www.youtube.com/watch?v=M6LS4XZswNg. Basically micelle is formed by molecules which have a polar (ionizable end) and a long alkyl chain. Just like any electrolyte they conduct electricity in water. However above CMC, these molecules rather suddenly form a structure called a micelle as shown below (from Google Images). For these reasons, there is sudden change in lot of properties such as surface tension, conductivity and so on. It is not only conductivity.


Increase in Yield of products in an exothermic reaction Given an exothermic reaction $\ce{N2 + 3H2 -> 2NH3}$ (which was initially in equilibrium). The temperature is then increasing with time. I am supposed to predict the yield of ammonia with time.

According to an exothermic reaction, the equilibrium constant K is supposed to decrease with increase in temperature. But it is found that the yield increases initially before decreasing continuously as expected. What causes the initial increase in the yield?

I suppose that the heat overcoming the activation energy could provide an explanation, but is there any other concept which is not fully reliant on the kinetics of the reaction?

Look, you say: "which was initially in equilibrium". Then the temperature are stable. In the reaction you have four mols in the reactants and 2 mols in the products. When you increase the temperature the system pressure increase too.

So to increase the stability the system "walk" for the reactants, because this decrease the pressure (2 mols in products and 4 mols in reactants).

If you remember the gas ideal law, the pressure is directly proportional to the mol numbers. So this is the reason because the temperature favor the products.


What's the difference between enthalpy of combustion and heat of combustion? Wikipedia has two different pages for enthalpy of combustion and heat of combustion. Their definition is the same and they use the same symbol $\Delta H_c^\circ $

It says that enthalpies of combustion are always negative as these reactions are exothermic. The heat of combustion tables list positive values for all substances.
Why do these values differ in sign?

Looking at Wikipedia for the definitions:

The standard enthalpy of combustion is the enthalpy change when one mole of a reactant completely burns in excess oxygen under standard thermodynamic conditions (although experimental values are usually obtained under different conditions and subsequently adjusted).

The heat of combustion $(\Delta H_c^\circ)$ is the energy released as heat when a compound undergoes complete combustion with oxygen under standard conditions. The chemical reaction is typically a hydrocarbon reacting with oxygen to form carbon dioxide, water and heat.

On first inspection, the two seem the same, however, there is a very slight different use in terminology and wording.

1.) What are standard conditions?

Referring back to Wikipedia:

Standard conditions for temperature and pressure are standard sets of conditions for experimental measurements established to allow comparisons to be made between different sets of data.

In chemistry, IUPAC established standard temperature and pressure (informally abbreviated as STP) as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of 100 kPa (14.504 psi, 0.987 atm, 1 bar)

As you know, the standard enthalpy of combustion can be referred to as as $\Delta H ^{\circ} _{\mathrm{total}}$ whose units are in kJ. The molar enthalpy of combustion is how much energy is released per kJ/mol. How is the standard enthalpy of combustion found?

Using the individual enthalpies of formation, and Hess's law, we can calculate the standard enthalpy of combustion, or $\Delta H ^{\circ} _{\mathrm{total}}$.

$\Delta H_{reaction}^\ominus = \sum \Delta H_{\mathrm f \,(products)}^{\ominus} - \sum \Delta H_{\mathrm f \,(reactants)}^{\ominus}$

The standard enthalpies of formation, or the negative values are located here https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table) and as well, generally in your textbook in an appendix along with the standard entropy of formation, and Gibbs free energy of formation, as all three are state functions.

2.) Why do the signs differ in the values from your textbook compared to the table in Wikipedia?

For the heat of combustion, it is generally expressed in units of higher heating value, lower heating value, and gross heating value. As the article states in Wikipedia, the heating value (or energy value or calorific value) of a substance, usually a fuel or food (see food energy), is the amount of heat released during the combustion of a specified amount of it.

In summary, the sign difference that I see is due to different terminology, they are specifically referring to units of HHV, LHV, and GHV and expressing those in the table below. Although, heat of combustion may also be calculated as the difference between the heat of formation $\big( \Delta H_f^\circ \big)$ of the products and reactants just as the standard enthalpy of combustion, it seems to me that the heat of combustion is specialized in a way to be applied to fuels, and as such, different from the standard enthalpy of combustion that we are accustomed to.

Perhaps someone could clarify a little more on this?


Sodium and alcohol reaction I wonder, what happens when sodium reacts with alcohol? What is the reaction mechanism?
I think that sodium is an electrophile that attacks the nucleophile oxygen atom in the -OH-group of the alcohol. Sodium will take one electron and bond with the oxygen molecule, while the bond between the oxygen and hydrogen will break. Am I correct?

Sodium will not take one electron, as you propose. No, it is a redox reaction : the sodium atom will donate one electron to the Oxygen atom of the $\ce{ROH}$ alcohol, so as to repel one $\ce{H}$ atom and make an alkoxide ion. At the end, a $\ce{Na^+}$ cation and a $\ce{RO^-}$ alkoxide anion are produced in the solution. Of course the $\ce{H}$ atom will join the next $\ce{H}$ atom to produce $\ce{H2}$ bubbles that gets out of the solution. $$\ce{Na + ROH -> Na^+ + RO^- + \frac{1}{2}H_2}$$ The reaction is not dangerous. It will not catch fire if done in a sufficiently large amount of alcohol. I have done it plenty of times in the lab. No problem ! It is even a good way for getting rid of old residues of partially oxidized pieces of metallic sodium, in case you don't know what to do with them.


Isotopes and electronegativity Is there any difference between electronegativities of isotopes of the same element. For example consider the above compound. Will all the resonance structures be equivalent?

Also if I place the above compound in water (which will act as a Bronsted acid), which carbon will get the H atom?

(Star represents $\ce{^14C}$ atom)
This figure represents three out of the five possible resonance structures. Of these five structures, four have a negative charge on a $\ce{^12C}$ atom whereas one has negative charge on $\ce{^14C}$ atom.
As the first premise, let's consider stability of the two kinds of resonating structures. The more stable the resonating compound, the more likely it is to be protonated.

We can compare stability using the premise of electronegativity (or rather charge density). According to this property, the more heavy an atom, the larger is its volume. If volume increases, the charge density decreases making it less likely to be able to stabilize the negative charge (becomes less electronegative). Due to this we can say that the resonance structure with $\ce{^14C}$ having the negative charge is less stable and so is less likely to be protonated.
Now, moving onto the second premise, the number of resonating structures containing the negative charge on $\ce{^14C}$ and $\ce{^12C}$ respectively.
If you notice only one resonating structure has a $\ce{^14C}$ with the negative charge, whereas there are $4$ resonating structures containing $\ce{^12C}$ atom with the negative charge. This would mean that probabilistically (considering stability to be equal) The resonance structure having a $\ce{^12C}$ with the negative charge would be more likely to be attacked.
Now, if we combine the two premises, we see that both stability and frequency of the resonance structure containing $\ce{^12C}$ with negative charge is greater its $\ce{^14C}$ counterpart. Concluding, we can say that the resonating structure with $\ce{^12C}$ stabilizing the negative charge would be more likely to be protonated.
Further Reading on how stability of isotopic compounds can be determined :  Inductive effect of hydrogen isotopes


Does aniline react with diazonium ions at C or N? In the azo coupling of aniline with benzenediazonium cation, I thought of two possible products 1 and 2 that could be formed. However, I can't work out which will be preferred. What would be the major product and why?

This is a really interesting question and the answer is that the reaction of benzenediazonium chloride with aniline is a bit different to most of the reactions of benzenediazonium salts in that the initial product is compound 1, diazoaminobenzene. It is possible to run the reaction to isolate diazoaminobenzene prep here.

These diazoaminobenzene compounds are unstable with respect to reversion to a diazonium salt + nucleophile, and so many references suggest that compound 2, 4-aminoazobenzene, is directly produced. The procedure for the transformation of diazoaminobenzene 1 to 4-aminoazobenzene 2 is by heating to 50 °C with aniline (prep here).


Why do liquids freeze into a polycrystal solid when they freeze fast? I don't see any possible explanation. If there's a crystallite at one orientation, why would a crystallite at another orientation start growing instead of more molecules just building onto the surface that's already there in the same orientation? Does the size of the crystallites vary as the reciprocal of the speed of movement of the boundary between the solid state and the liquid state?

Embryo crystals spontaneously and independently form in a supercooled liquid. A crystal successfully growing in one location doesn't prevent others from growing elsewhere. As the temperature decreases, the rate of embryo formation and chance of successful growth increase, so at a low temperature you're likely to get multiple successful crystals.


Electrophilic aromatic substitution with fluorine Why don't we usually use diatomic fluorine in electrophilic aromatic substitution?

Is it because fluorine is a potent oxidizer and will react with the carbons?

What might happen if we reacted benzene with diatomic fluorine in the presence of some Lewis acid such as boron trifluoride? Would we get the expected product: monosubstituted benzene?

Fluorine reacts so rapidly with benzene that aromatic fluorination requires special conditions
and special types of apparatus. Even then, it is difficult to limit the reaction to monofluorination


Weak buffers and accuracy when mixing acid and base What is the weakest buffer possible - that is, what has the lowest buffer capacity possible?

I would just assume that the lower the concentrations of $[\ce{A-}]$ and $[\ce{HA}]$, the lower the capacity is, but it seems like the question is looking for a specific solution?

Also, how precise do you have to be when mixing a strong acid and base to get a pH value of 7 and not 6? since $[\ce{H+}] = 10^{-7}$ and  $[\ce{OH-}] = 10^{-7}$ at $\mathrm{pH} = 7$, while $[\ce{H+}] = 10^{-6}$ and $[\ce{OH-}] = 10^{-8}$ at $\mathrm{pH} = 6$, does this imply you can at most be off by a factor of $10^{-2}$ mol/L?

What is the weakest buffer possible - that is, what has the lowest buffer capacity possible?

You've conflated two different aspects of a buffer. First there is the $K_\mathrm{a}$, or $\mathrm{p}K_\mathrm{a}$ for the species.

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

$$\mathrm{p}K_\mathrm{a} = -\log{K_\mathrm{a}}$$

So weaker acids would have a high $\mathrm{p}K_\mathrm{a}$, close to 7 or greater.

Then there is the buffer capacity which is a function of the absolute concentrations of the various species ($\ce{[H^+]}$, $\ce{[A^-]}$, and $\ce{[HA]}$). So if there is relatively much more $\ce{[H^+]}$ or $\ce{[A^-]}$ in the solution than acid or base being added, then the pH will change by only a small amounts.

  Also, how precise do you have to be when mixing a strong acid and base to get a pH value of 7 and not 6?

If the strong acid is $10^{-6}$ greater than the strong base then the pH will be 6.

$\mathrm{pH} = -\log\ce{[H^+]}$


Hybridization - Mixing one atomic s orbital with 2p orbitals Explain why three ${sp^2}$ hybrid orbitals result when one s atomic orbital is blended with two p atomic orbitals.

When some number of atomic orbitals are mixed together to produce hybrid atomic  orbitals, the same number of hybrid atomic orbitals must result.  So, when N atomic orbitals are mixed together, N hybrid atomic orbitals are produced.  In the case you describe, mixing one s atomic orbital with two p atomic orbitals means that three hybrid atomic orbitals must result.

These three hybrid orbitals will each contain 33.333% s character and 66.666% p character because that is the ratio of the one s orbital to two p orbitals that we are mixing together.  We can describe such an orbital as $\ce{s^{33.333}p^{66.666}}$, but if we reduce that notation to the lowest common denominator we get $\ce{s^1p^2}$, or the commonly used description $\ce{sp^2}$.  This simply means that this hybrid orbital is composed of one part s character to two parts p character.

To summarize if we mix 3 atomic orbitals, one s and two p orbitals, then 3 $\ce{sp^2}$ atomic hybrid orbitals will result.


What is the difference between saturated and unsaturated fatty acids? Chemically speaking, what are the biggest differences between saturated fatty acids and unsaturated fatty acids? What are the effects of those differences on the geometry of the molecule? And what is the effect of that on our body?

Saturated fatty acids are fatty acids with no double bonds and unsaturated fatty acids are fatty acid with one or more double bonds. This is the biggest difference between a saturated and unsaturated fatty acid.

There is two types of structural configuration for unsaturated fatty acids due to the presence of double bond; cis (kink/bend in fatty acid struture) and trans(No kink/bend in fatty acid struture).

.

  cis fatty acid (wikipedia)

  A cis configuration means that adjacent hydrogen atoms are on the same side of the double bond. The rigidity of the double bond
  freezes its conformation and, in the case of the cis isomer, causes
  the chain to bend and restricts the conformational freedom of the
  fatty acid. The more double bonds the chain has in the cis
  configuration, the less flexibility it has. When a chain has many cis
  bonds, it becomes quite curved in its most accessible conformations.
  For example, oleic acid, with one double bond, has a "kink" in it,
  whereas linoleic acid, with two double bonds, has a more pronounced
  bend. Alpha-linolenic acid, with three double bonds, favors a hooked
  shape. The effect of this is that, in restricted environments, such as
  when fatty acids are part of a phospholipid in a lipid bilayer, or
  triglycerides in lipid droplets, cis bonds limit the ability of fatty
  acids to be closely packed, and therefore could affect the melting
  temperature of the membrane or of the fat.

Iodine value test is used to determine degree of unsaturation in a unsaturated fatty acids; that is to measure number of double bonds present in a unsaturated fatty acids.

This link will explain the health benifits of saturated and unsaturated fatty acids


In the following reaction is E2 or Sn2 favorable? Question

Doubt

It is quite obvious that carbonyl compound is going to convert into alkane, but what confuses me is whether Sn2 or E2 will occur at bromide position

Although there is heating( which I believe favour E2) and a strong base, the carbon is 1° which would favor SN2 more than E2.
Please help.

This is simply wrong. The person setting this question has assumed that the hydrazine will be completely selective for the ketone over the alkyl bromide and that the strong basic conditions will also cause elimination of the Br- to give the styrene. Selectivity seems most unlikely give the typical forcing conditions used in W-K reductions

Hydrazine is a powerful nucleophile and reacts with alkyl bromides under mild conditions often giving polyalkylation reference here.

I consider it most unlikely there will be any selectivity unless confronted by literature example.


Oxidation of water by permanganate: Does this reaction happen? Since MnO4- is a strong oxidiser, if one increased its concentration  relative to water, could it oxidise the water (Le Chatelier's Principle seems to suggest so)?

I have attempted to write the possible (redox) reaction, but I fear I've gone terribly wrong:

$$\ce{MnO4^{1-} + 2H2O -> 2H2 + 3O2 + Mn^{3+}}$$

Can someone tell me if I've constructed the equation (and balanced it) right? If not, can someone offer me some guidance in this regard?

You have indeed gotten it terribly wrong. The key hint is that you have more than one element that would be reduced: hydrogen also goes from $\mathrm{+I}$ to $\pm0$ in your sample equation. Furthermore, in neutral or weakly basic solutions permanganate gives $\ce{MnO2}$ after oxidation, not manganese(III).

What you do want to do is to reduce permanganate to something (which I have already established would be manganese dioxide) while oxidising something in water to something. Looking at the oxidation states of hydrogen and oxygen in water, the only element that can be oxidised is oxygen, since hydrogen is already at its highest possible oxidation state $\mathrm{+I}$. Oxidising water’s oxygen leads to a few possibilities of which oxygen gas does indeed sound most likely. Therefore, we have the redox pairs:

$$\begin{align}\text{Red: }\ce{\overset{+VII}{Mn}O4-}/\ce{\overset{+IV}{Mn}O2}&&\text{Ox: }\ce{H2\overset{-II}{O}}/\ce{\overset{\pm0}{O2}}\end{align}$$

We can then determine the appropriate half-reactions by the standard procedure. Instead of walking you through it, I will present you with the final result for comparison.

$$\begin{align}\ce{MnO4- + 3e- + 4 H+ &-> MnO2 + 2 H2O}\tag{Red}\\
\ce{2H2O &-> O2 + 4e- + 4 H+}\tag{Ox}\\[1em]\hline
\ce{4 MnO4- + 4 H+ &-> 4 MnO2 + 3O2 + 2 H2O}\tag{Redox}\end{align}$$

It turns out that the final reaction equation formally doesn’t even need water. However, potassium permanganate is a shelf-stable solid under normal laboratory conditions (if light is excluded), so it is reasonable to assume that the reaction will only proceed in aqueous solution.

To determine whether the redox reaction is feasible in aqueous solution, it is easier to go with the two half-reactions we already have; thermodynamically, they must correspond to the same process with the same reaction enthalpy etc. The Nernst equation together with the inspection of the cell potential allows us to estimate whether a process will be spontaneous or not by mere inspection of the standard potentials.

$$\begin{align}E^0_\text{cell} &= E^0_\text{Red} - E^0_\text{Ox}\\
&= \pu{+1.70V} - (\pu{+1.23V})\\
&= \pu{+0.47V}\\
E^0_\text{cell} &> 0 \\&\Longrightarrow \text{spontaneous process}\end{align}$$

Thus, thermodynamics already predict the decomposition of permanganate in aqueous solution to be spontaneous under standard conditions. As mentioned, the Nernst equation will allow us to estimate whether it also is spontaneous given general laboratory conditions; I will assume $\mathrm{pH} = 7$ and $p(\ce{O2}) = \pu{0.2bar}$

$$\begin{align}E &= E^0 - \frac{0.059}{z} \lg \frac{Red}{Ox}\\[0.5em]
E &= \pu{+0.47V} - \frac{\pu{0.059V}}{12}\lg\frac{p(\ce{O2})^3}{[\ce{MnO4-}]^4[\ce{H+}]^4}\\[0.5em]
E &= \pu{+0.47V} - \frac{\pu{0.059V}\times28\times2.1\times(-4)}{12}\lg {[\ce{MnO4-}]}\\[0.3em]
E &= \pu{+0.47V} + \pu{1.16V}\times\lg {[\ce{MnO4-}]}\end{align}$$

We can calculate the minimum concentration from which the process will be spontaneous:

$$\begin{align}\pu{0V} &= \pu{+0.47V} + \pu{1.16V}\times\lg {[\ce{MnO4-}]}\\
\pu{-0.47V} &= \pu{1.16V}\times\lg [\ce{MnO4-}]\\
-0.405 &= \lg [\ce{MnO4-}]\\
10^{-0.405} &= [\ce{MnO4-}]\\
0.393 &= [\ce{MnO4-}]\end{align}$$

Thus, at any concentration higher than $\pu{0.393M}$ permanganate should start formally oxidising water or decomposing.

Whether or not this happens at exactly this concentration depends on kinetic factors, i.e. can oxygen actually be produced under these conditions, but thermodynamically it would be favourable.


Quantum dot band gap energy equation I have been looking into certain approximations for band gap energies in quantum dots and I found a paper [1]. However, I noticed that the equations which they obtain are not useful to the situation which I would like to use them in.

What I would like to do is gather information from other articles that show the absorption and emission spectra and understand the size of the quantum dots if the sizes are not given.

I have looked at several papers where $\ce{ZnSe}$ QDs are absorbing and emitting around $\pu{\sim 300 .. 450 nm}$ with sizes from $\pu{2 .. 6 nm}$ in diameter.  However, if you use the LDA+C equation for $\ce{ZnSe}$ quantum dots (QDs) ($E_g = 2.54/d^{1.54}$) with diameters of $\pu{\sim 2 nm}$, you get band gaps around $\pu{0.9 eV}$ or $\pu{\sim 1370 nm}$, not $\pu{\sim 400 nm}$ as you would expect.

Is there something I am missing here?  Is this a $E_{11}$ transition band gap energy and I am seeing like a $E_{66}$ absorption to $E_{55}$ fluorescence energy emission?

Also, I noticed that there are exciton binding energies which may take up a large portion of the total exciton energy. Are these binding energies important to this problem?

If so, is there a better way to search for and obtain some literature on a simple correlation for these type of energy transitions?

Li, J.; Wang, L.-W. Phys. Rev. B 2005, 72 (12), 125325. DOI 10.1103/PhysRevB.72.125325.

I believe the problem was that the LDA+C energies quoted in the paper were extra, additive energies which were supposed to be added to the main band gap energy of the bulk material. In this case that was something like $\pu{2.7 eV}$.

This is because the quantum dots will have higher band gaps the smaller they get. As the quantum dots get larger, in a sense, they approach a bulk material size, hence their band gap energy gets smaller and smaller and approaches the bulk material band gap energy limit.

Hope this helps anyone who happens upon a similar problem.


Derivation of mean kinetic energy I read from a book that average kinetic energy is equal to $3kT/2$ where $k$ is Boltzmann's constant and $T$ is the kelvin temperature. I don't know how the formula was derived. Any help to gain insights into the derivation of this formula would be helpful.

A simple validation
The result you quoted is the average translational kinetic energy for an ideal gas.
First, let's sketch out a rough derivation for the average kinetic energy of a particles of an ideal gas using nothing more than high school physics.
The gas molecules just undergo translations, and don't rotate/vibrate. Also, they don't interact with each other.
Imagine a rectangular box which contains an ideal gas; the gas molecules have a speed $v_x$ in the $x$-direction. Since the box has a uniform cross-section of area $A$, a volume element spanned by a gas molecule in the box is $V'= |v_x\Delta t|A$. On average half the molecules move to the right, and half the molecules move to the left, since there is no preferred direction of motion. Thus, the number of molecules colliding with the container walls is
no. of molecules $= \frac{1}{2}.\frac{nN_a}{V}.|v_x\Delta t|A$
The momentum transferred per molecule is $2mv_x$, so the total momentum transferred is $ \Delta P = \frac{1}{2}.\frac{nN_a}{V}.|v_x\Delta t| 2mv_x$
Since force is rate of change of momentum, $F = \lim_{\Delta t \to 0} \frac{\Delta P}{\Delta t} = \frac{nMAv_x^2}{V}$
Pressure is force per unit area, so the average pressure is $P = \frac{nM \langle v_x^2\rangle}{V}$
Now, note that the total velocity is $v^2 = v_x^2+v_y^2+v_z^2$, and since there is no preferred direction of motion, one can assume all three components of velocity to be equal. $v_{rms}^2 = 3 \langle v_x^2\rangle$
Thus, $P = \frac{nMv_{rms}^2}{3V}$ also from the ideal gas equation $P = \frac{NkT}{V}$ we get $\frac{mv_{rms}^2}{3} = kT$
$E_{\text{kinetic}} = \frac{mv_{rms}^2}{2} = \frac{3kT}{2}$ Which is the result you wanted.
Using the Maxwell-Boltzmann Distribution
The same result for an translational kinetic energy can be arrived at using the Maxwell-Boltzmann distribution.
$$f(v) = 4 \pi \left( \frac{m}{2 \pi kT}\right)^{3/2} v^2 e^{\frac{-mv^2}{2kT}}$$
The average kinetic energy is $$\langle E_k \rangle= \langle \frac{mv^2}{2} \rangle = \int_{0}^{\infty}\frac{mv^2}{2} f(v)\mathrm{d}v = \frac{3kT}{2} $$
You can evaluate the integral as an exercise. I have omitted the details for the sake of brevity. It is relatively simple, you can either do it by hand, or look it up in a table or enter it into Mathematica
Now, the general statement of the theorem (using words and not math) us

For a system at equilibrium the average value of each quadratic cobtribution to energy is $\frac{kT}{2}$

An example of a quadratic contribution would be kinetic energy ($\frac{p^2}{2m}$) or the harmonic potential ($\frac{kx^2}{2}$)
A simple statistical mechanics approach
This theorem is strictly valid at "high" temperatures; we require that seperation between energy levels is small, and a lot of states are occupied so that they maybe treated as a continuum. This works well for translational and rotational modes of motion, but is not as reliable  for vibrational/electronic modes. Anyway, note that the analysis given below is under the framework of these assumptions.
note: $\beta = \frac{1}{kT}$
For a one dimensional container the translational partition function derived using particle in the box energy levels is given below
$$q^T = \frac{X}{\Lambda} \qquad \text{where} \ \Lambda = \frac{h \beta^{1/2}}{(2 \pi m)^{1/2}}$$
$$\langle E^T \rangle = -\frac{1}{q^T} \left( \frac{\partial q^T}{\partial \beta}\right)_V$$
again, you can evaluate this derivative as an exercise. (Hint: $\Lambda$ is constant multiplied by $\beta^{1/2}$). The result is given below
$$\langle E^T \rangle = \frac{kT}{2}$$
This is for translation in one direction (say X). The same process can be repeated for Y and Z, and thus for three dimensions
$$\langle E^T \rangle = \frac{3kT}{2}$$
Now let us examine rotations. Once again, I point out that this analysis is valid in the range $T > \theta^R = \frac{hcB}{k} $
Here, B is the rotational constant.
In this range, an approximate expression for the partition function of a heteronuclear diatmic moeclue is $$q^R = \frac{1}{\beta hcB}$$
$$\langle E^R \rangle = -\frac{1}{q^R} \left( \frac{\partial q^R}{\partial \beta}\right)_V = \frac{1}{\beta} = kT$$
Note: I already pointed out that this result is valid at "high" temperatures, however, I did not give you a sense of scale. For hydrogen molecule $\theta^R = 87.6  \ K$ for chlorine molecule it is $ 0.351 K $. Similarly, the approximations made while deriving the translational functional are valid at room temperatures and for everyday macroscopic containers whose dimensions are much greater than the thermal wavelength ($\Lambda$) of the molecules.
A similar analysis can be performed for vibrational modes, however the temperature range over which it is valid is much greater ($> 805 K$ for chlorine and $> 6332 K $ for hydrogen). These conditions aren't within the realm of everyday experience.


Why does this E2 reaction occur on the less substituted carbon? In compound C it eliminates the beta hydrogen attached to the less substituted carbon, rather than attacking the beta hydrogen attached to the tertiary carbon. Why is this the case?

E2 elimination is stereospecific for anti elimination. In both variants of compound B that you have drawn, the OTs and the proton on the tertiary centre are syn to each other, meaning that the elimination is disfavoured.

For this reason, the NaOMe deprotonates at the secondary centre, generating the products as drawn, despite the fact that this goes against the Zeitsev rule which favours the most substituted alkene.


What pressure does liquid nitrous oxide need at −45 °C? What pressure does liquid nitrous oxide need at −45 °C to remain liquid? How can I calculate its pressure required for other temperatures?

Using NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9, I found the saturation point of nitrous oxide at equilibrium for a temperature of $T=-45.000\ \mathrm{^\circ C}$ at a pressure of $p=783110\ \mathrm{Pa}=7.8311\ \mathrm{bar}$;
for $T=-78.000\ \mathrm{^\circ C}$, it is $p=182290\ \mathrm{Pa}=1.8229\ \mathrm{bar}$.


Why cannot the Schrödinger equation be solved exactly for systems in which more than two particles interact? Is it so because it is impossible to solve mathematically? If so, why? Or is it possible to solve the equation mathematically, but no physical interpretation of the solutions is possible? Or is it because of practical issues, i.e. exact solutions would require more computer power than we have available today?

Several books state that the Schrödinger equation can only be solved exactly for systems of maximum two interacting particles, but fail to explain why. For example, Andrew R. Leach mentions the impossibility, but fails to elaborate on why that is so, in his "Molecular Modelling: Principles and Applications" (2nd ed., page 34).

If the answer could be explained in terms that a new master student in theoretical chemistry would understand, that would be great!

Is it so because it is impossible to solve mathematically?

Yes, in both classical and quantum mechanics there is no known general closed form analytical solution for the three-body problem. To be more precise, for the case of classical mechanics, an analytical solution in the form of a convergent power series exist, but is useless in practice due to very slow convergence.

  If the answer could be explained in terms that a new master student in
  theoretical chemistry would understand, that would be great!

I doubt that is possible. For the case of classical mechanics, Poincaré had a simple argument: one has more degrees of freedom than integrals of motion, and thus, could not solve the equation of motion through the usual integration method. But, as I mentioned, that's not the whole story, and secondly, I'm not sure how to deal with quantum mechanics anyway. So, as Ivan Neretin mentioned in his comment, things are the way they are: for all practical purposes the three-body problems could not be solved analytically.

  Why cannot the Schrödinger equation be solved exactly for systems in
  which more than two particles interact?

And besides, you have to be careful the wording here. I mean, you can easily have a two-particle system for which there is no known analytical solution. This is because the procedure to obtain analytical solution for a two-particle problem relies on the center-of-mass separation trick which works if the potential energy depends only on the relative positions of the particles. More formally, it works only if interactions in the two-particle system are invariant to the translation of the system as a whole. This is the case of two isolated particles that interact just with each other, for instance, an electron and a proton in an isolated hydrogen atom. For this system you can indeed use the center-of-mass separation, reduce the two-body problem to a set of two one-body problems, and get the well-known analytical solutions.

But what if you put that hydrogen atom in a inhomogeneous electric field? Center-of-mass separation won't work than, since potential energy terms due to interaction of both the electron and the proton with the external field depend on their actual position with respect to the field, not just on their relative positions. The same thing happens when considering two electrons in a field of a fixed proton, i.e. just the electronic subsystem of a helium atom. The electronic two-body problem is also unsolvable analytically in that case due to the presence of external (for the two-electron system in question) inhomogeneous electric field created by the proton.


What is the nature of the bonding in cyclotriphosphazene? The bonding in $\ce{(NPCl2)3}$ (and other compounds of the form  $\ce{(NPR2)3}$) has historically been described using the ‘Dewar island model’.

By analogy to benzene, $\ce{(NPCl2)3}$ is π-electron precise, containing 6-electrons in the π-system, and as such was thought to exhibit some degree of aromaticity.

The Dewar model accounted for this benzene-like structure by allowing phosphorous to use an occupied d-orbital to engage in a dπ(P)–pπ(N) interaction, forming the delocalised π-system of the phosphazene ring. This rationalised experimental observations that $\ce{(NPCl2)3}$ is (almost) planar and has equal P-N bond lengths which are somewhere between standard P-N (single) and P=N (double) bond lengths. What it fails to explain however is why other members of the series $\ce{(NPR2)n}$ do not show planarity.

Despite these flaws, and despite it being widely acknowledged that the d-orbitals aren’t capable of being involved in bonding to any great extent, the Dewar island model is still commonly found in inorganic chemistry textbooks and taught to undergraduates (possibly because its an easy to understand explanation and draws analogy to the familiar).

  Is there a better model to explain the bonding in $\ce{(NPR2)3}$ that is (1) consistent with the observed properties of the molecule, (2) makes sense in terms of the orbitals/interactions used (i.e. no invoking d-orbitals) and (3) can be backed up by calculation? [*]

[*]: My understanding is that negative hyperconjugation can be invoked, but this doesn't seem to be widely accepted and possibly isn't the most intuitive explanation for bonding by itself.

The bonds between the phosphorus and nitrogen in phosphazenes are best described as having a significant ionic component with each of the N atoms having a negative charge and each of the phosphorus atoms having a positive charge. There is also some negative hyperconjugation resulting from the donation of electrons from the nitrogen atoms' lone pairs to antibonding orbitals  between the P atoms and the R groups. The degree of negative hyperconjugation increases with the electronegativity of the R groups and it imparts a partial double bond character to the bonds between the P and N atoms.
For more information see http://pubs.acs.org/doi/abs/10.1021/ic0511266


Intrinsic Reaction Coordinate - Converting To Reaction Coordinate with Units of Distance I've located a transition state and ran two IRC calculations (one to get to the product and one to get to the reactant). Is there some type of way of approximating the IRC to something that has units of distance (Angstroms, Bohr, etc.)?

Intrinsic reaction coordinate (IRC) is not some internal coordinate of a molecular system (such as, bond length, bond angle, etc.), rather it is a curvilinear coordinate that describes the intrinsic reaction path (IRP), which is the reaction path along the direction of the gradient. Only for some simple reactions one can visually (and approximately) identify IRC as one of the internal coordinates, but in general that is not possible.


what kind of suspension liquid should be used with ferrofluid (so it does not stain the glass) Please take a look at the following video:

I am working on a new project, and I need to find whats the best liquid to hold ferrofluid inside the glass (or maybe even plastic) container, so that ferrofluid (that easily stains everything) does not stain or stick to glass?

Also, is there any special preparation for the glass needed?

I had success with the following:

1) Glass bottle, ideally with flat sides.  If you have a round bottle you may not be able to see the ferrofluid due to diffraction.

2) A solution of ~25% distilled water (a.k.a. deionized water) and ~75% of store bought isopropyl alcohol at 91% concentration.  That means that ~70% of my solution was actually isopropyl alcohol.  I eyeballed the ratio between the water and the isopropyl alcohol, those numbers are likely to not be exactly correct.

Procedure that I followed:

Add some isopropyl alcohol to the glass container.  Then add some (or all) of your ferrofluid to the container.  Then add some water to bring your isopropyl alcohol concentration to about 70%.  Next, use a magnet to settle the ferrofluid to the bottom and then remove the magnet.  If your ferrofluid has little bits that are floating, then you need to add more water.  If your ferrofluid looks too stringy then I would add more alcohol.  When I was done the ferrofluid had zero staining, like you see in the videos online, except for in a few spots.  I think for better results I would need a higher quality glass container.

notes: Don't put too much ferrofluid in your container, otherwise you will have an uninteresting blob.  I recommend using gloves.  To clean the ferrofluid stuff off my hands I used warm water, laundry detergent, and lots of scrubbing with a towel.  If you mess up, you can use a magnet to hold your ferrofluid in place while you pour off some of your suspension fluid.  In addition, you can use a straw and your thumb to pipet some of the ferrofluid, however some will get stuck in the straw, so I wouldn't do that unless you have a bunch of ferrofluid.  Finally, I only finished like 30 minutes ago, so I hope it will all look good tomorrow.  If something bad happens I will make another post.

notes, part 2: If all you want is a cool desk toy, then I recommend just buying one of the pre-made ferrofluid toys online. (I saw some for ~$30).  You will get a lot more ferrofluid for your money if you buy just the ferrofluid, however making the toy is a bit of work.

GOOD LUCK!

edit:  no combinations of the following worked for me: acetic acid, hydrogen peroxide, pep plastic.


Lack of colour in Nitric Oxide Usually compounds' colours are explained by the presence of unpaired electrons which is free to excite and de-excite giving the compound a charectistic colour. NO has atleast one unpaired electron but even then it is colourless. Can anyone tell me what the reason is?

I am not a physics major, this post will be rather free of equations and spherical chickens. This is a chemist's view of the situation ;)

You need more than an unpaired electron to create color in a substance. You need two places for it to be, with a given energy difference in the visible color range. Looking at the molecular orbital diagram of NO we see that no such organisation is possible (within reasons). The only choice is to go from one $\pi^*$ to the other, but those are at the same energy level. There is no way for this diatomic molecule to deform in any meaningful way so as to create a difference in energy levels between them. Metal complexes, especially iron- and cobalt-based ones, can do this, but not with the $\pi^*$ in any case. There is only $\sigma^*_2$ left, but the distance from $\pi^*$ to $\sigma^*_2$ is in huge, larger than visible spectrum (an event that knocks an electron up there is likely to unravel the entire molecule).

All in all, there is no possibility for the electron to move anywhere that requires the emission of a photon in visible range. It cannot produce the color you seek.


What is there in this 3D representation of organic molecules I fail to grasp? I think there's a basic idea I'm missing about 3 Dimensional representation of molecules. I think it may be about converting to Fischer representation. Consider the following two products of bromination of an alkene, the first one being for cis-form and the 2nd one for trans-form.

I know the question may sound stupid, but it's crucial to get the basics. Why is the product II) the meso form and the products I) are the optically active forms. By what I know, the solid wedge goes towards the observer, so in product II), shouldn't the Bromine substituent(or maybe the $CH_3$ substituent) and $H$ atom be on the 'same side', making it the one optically active?

You have drawn the compounds correctly, and yes

a solid wedge means the bond is coming out of the plane of the screen towards you
a dashed wedge means the bond is going behind the plane of the screen away from you
a solid line means the bond lies in the plane of the screen

Here is a Newman projection of your molecules, sometimes Newman projections can make it easier to "see" things.

Build a model (I did) or flip the drawings around, there is no way you can interconvert the isomers labeled "S,S' and "R,R"; they are chiral enantiomers.  The meso compound and its mirror image are easily converted into one another (try rotating the molecule about a $\ce{C_2}$ axis that lies in the plane of the screen, 2) is perpendicular to the mirror plane [the dashed line in my drawing] and 3) bisects the $\ce{C_2-C_3}$ bond in the molecule); they are indeed representations of the same molecule.


How to interpret TLC data regarding the purity of one’s sample The picture provided is a sketch of my TLC data from the lab along with the $R_f$ values for $i$ (initial, un-recrystallized aspirin) $= 0.429$, $a$ (acetylsalicylic acid) $= 0.429$, and $s$ (salicylic acid) $= 0.571$.

One of the questions in my post lab was:

  Based on the TLC data, discuss the purity of the impure aspirin sample, explain your reasoning.

So far, I have said that based on the data, the aspirin sample would be considered pure because there were no multiple spots in any lane. I don't know how else to elaborate on this. What can I say about the $R_f$ values?

TLC is a great and poor method at the same time. It is great because it is quick and accurate, and it is poor because it can only show substances that can be coloured in some way and are sufficiently non-volatile to not evaporate away.

You can be rather sure of the following points:

Your impure sample does not contain any more free acid since a spot of the same $R_f$ value is missing.
Your impure sample contains a probably very high amount of aspirin, since the spot has the same $R_f$ value.

However, you cannot be sure about:

Impurities that evaporated away during spotting, development or colouring of the TLC plate
Impurities that simply don’t show up in your method of colouring (e.g. due to the lack of a benzene ring for UV fluorescence).
Solvent impurities that will never show up due to both points above.

To prove purity you would need to resort to different analytic methods such as melting point (historically), HPLC-MS or GC-MS, NMR, etc.


How do Gilman reagents open epoxides? From what it looks like, Gilman reagents are good nucleophiles, and will attack at the less-substituted side of an asymmetric alkane because Gilman reagents don't necessarily need a substantial amount of partial positive character at the site of attack.

However, don't strong nucleophiles open epoxides in an SN2 like manner? Where's the inversion?

In the first reaction, the reaction doesn't take place at a stereocenter, so there's no way to observe the inversion.

In the second reaction, the stereochemistry in the product is R, R. The Gilman reagent approaches opposite the C-O bond. After bond rotation, we see the product in the indicated conformation.