Identifying a meso compound I am still a first year student; I was asked in the exam to identify the meso comound out of some choices. The answer key said that the compound below is meso. But I see no plane of symmetry here. Am I wrong or the question? I hope someone clarifies please. Sorry if the question was super silly.

As the comments imply, having a center of symmetry makes a molecule achiral.  This is because, geometrically, inverting through the center equals a mirror reflection plus a 180° rotation.  If you start with an atom at point $(x,y,z)$ and you reflect it through the $xy$ plane, the atom is now at $(x,y,-z)$.  Now rotate 180° around the $z$ axis and you're at $(-x,-y,-z)$ just like inverting through the center.
Your molecule has no symmetry with respect to just a mirror reflection or just a 180° rotation.  (It might look like you could rotate 180° within the plane of the paper, but the stereochemistry is different.)  But it does have a symmetry with respect to the combination of the two, which is that center of inversion, so you can match the mirror image to the original by allowing the rotation.  That makes the molecule meso.


Synthesis Golf VIII: 4,6-Diacetylhygrophorone A12 As usual we are a bit late on this, but here is this month's molecule:

which is known as (4R,5S,6R)-4,6-diacetylhygrophorone A12, and is an antifungal natural product isolated from mushrooms of the genus Hygrophorus.
As far as I know this compound has not yet been synthesised in the lab, but has been isolated and its relative configuration assigned by NMR measurements in anisotropic media.[1]
The criteria are the same as for the last challenge I posted:

the synthesis must be enantioselective;[2] and
no more than one chiral centre may be purchased in the form of a building block. (Chiral catalysts, ligands, auxiliaries, etc. do not fall under this rule.)

Otherwise, any commercially available starting material (as usual, in the Sigma–Aldrich catalogue) is fair game.

(1) Schmidts, V.; Fredersdorf, M.; Lübken, T.; Porzel, A.; Arnold, N.; Wessjohann, L.; Thiele, C. M. RDC-Based Determination of the Relative Configuration of the Fungicidal Cyclopentenone 4,6-Diacetylhygrophorone A12. J. Nat. Prod. 2013, 76 (5), 839–844. DOI: 10.1021/np300728b.
(2) The absolute stereochemistry was not assigned in ref 1, so you may synthesise either enantiomer. Obviously, though, that doesn't make a difference to the design of the synthetic route, so I have arbitrarily decided to display this enantiomer (following ref 1).

Establishing the correct stereochemistry proved out to be an extremely difficult task, as I have found only one procedure for these systems. This first scheme is a published procedure:

1) Diels-Alder reaction (Paul D. Bartlett, Fred A. Tate
J. Am. Chem. Soc., 1953, 75 (1), pp 91–95);

2) Alpha-allylation and 3) ring closure by anionic addition to ester group to generate two regioisomers which are separated (Siwapinyoyos, Thebtaranonth, J. Org. Chem. 1982,47, 598-599);

4) Epoxidation of the non-conjugated enone and 5) opening of the epoxide by elimination (Chantarasiri, J. Chem. Soc., Chem. Commun., 1990,0, 286-288)

6) Kinetic resolution by enzymatic O-acetylation, followed by 7) hydrolysis of the acetate and vacuum flash pyrolysis at 450 0C and 0.1 mmHg. The acetate is needed in the target molecule, but I guess that it has to be hydrolyzed before the pyrolysis as to avoid side reactions. 8) is epoxidation with dimethyldioxirane and 9) opening of the epoxide with water provides a intermediate resembling the target molecule a lot (Klomkao, Tet. Asymm. 14, 24, 2003, 3885-3889).

From this molecule starts my approach:

10) Acetonide protection of the diol should be selective (syn 1,2 hydroxyl groups react in the presence of anti hydroxyl groups and 1,3 hydroxyl groups). The free hydroxyl group is then acetylated.

11) The acetonide is hydrolyzed. The primary hydroxyl group is oxidized to the aldehyde and the tertiary one is protected as MOM ether.

The following 3 steps are used for protection of ketone in presence of aldehyde:

12) Aldehyde is protected in presence of enone first.

13) Enone is protected by Noyori conditions.

14) Aldehyde is deprotected.

15) Organocopper addition to the aldehyde in presence of the ester to form a mixture of diastereomers (surely it would show diastereoselectivity but this would need to be tried out).

16) If the diastereoselectivity is bad, the mixture of diastereomers is oxidized to the ketone which is then stereoselectively reduced by chelate-control with zinc-borohydride. This could be problematic as the acetonide oxygens could interfere with chelation, but I cannot predict this with confidence. If they don't interfere much the intermediate would look something like in the image attached at the bottom. The hydryde attacks opposite to the acetonide and the alkyl chain. If acetonide oxygens interfere, then a screening of reducing agents is required.

17) Acetylation of the hydroxyl group and MOM and acetal deprotection should provide the target molecule.


Is a non-compound molecule a chemical substance? The Wikipedia definition of chemical substance states:

  A chemical substance, also known as a pure substance, is a form of matter that has constant chemical composition and characteristic properties. It cannot be separated into components by physical separation methods, i.e., without breaking chemical bonds. Chemical substances can be chemical elements, chemical compounds, ions or alloys.

A molecule made up of the same element (e.g. $\ce{H2}$) is not a compound; however, molecules aren't listed in the final sentence. Is a non-compound molecule a chemical substance?

I believe it is and here's my overall understanding of the terms in this space, and any corrections would be appreciated:

Found that the answer is 'Yes' in the IUPAC Gold Book definition of chemical substance:

  chemical substance

  Matter of constant composition best characterized by the entities (molecules, formula units, atoms) it is composed of. Physical properties such as density, refractive index, electric conductivity, melting point etc. characterize the chemical substance.


(P1,T1) -> to (P2,T2) then what will be the change in enthalpy for ideal gas? I need to prove that it is equal to the integral of $C_pdT$ from $T_1$ to $T_2$.
But should not be the case only when the pressure is constant?

The enthalpy of an ideal gas does not depend on pressure.  It is a function only of temperature.  So $C_pdT$ is always equal to dH for an ideal gas of constant composition.  The subscript p on $C_p$ merely indicates that $C_p$ can be measured directly by determining the heat Q absorbed by the gas in a constant pressure experiment.


Aluminium and water safety I have a 8000 gallon water tank that has always had pristine clear water.   I have a brass float valve at the top input to the tank.  Someone put an aluminum ladder in the tank (15 ft deep tank) and after who knows how long, the ladder was encrusted with a white gooey powder and the water turned milky.  what happened?  Main question is IS the water safe to drink or how do I get rid of whatever is in there.

The white stuff is aluminum hydroxide. The water is fine, filter the solids with a coffee filter if it doesn't settle to the bottom. And remove the ladder.


Reactivity of Aluminium Oxide I know that Aluminium Oxide is an amphoteric substance meaning that it reacts with both acids and bases. I recently learned that Aluminium has very good corrosion resistance due to the "Aluminium spontaneously forms a thin but effective oxide layer that prevents further oxidation."

However, in the test for Nitrate Ions, an Aluminium foil is added into the solution, the Nitrate Ion is reduced by the Aluminium, but this Aluminium foil surely has Aluminium Oxide which should prevent the reaction from happening, but it happens anyway. How can this phenomenon be explained?

Source: http://www.aluminiumdesign.net/design-support/aluminium-corrosion-resistance/

Aluminum oxide coating can relatively easily react with either strong base or acid, dissolving in them, exposing naked elemental aluminum to reaction medium. You can read more  here.
To be concise, there are many ways to make aluminum react: you can use a base, you can use an acid, and you can scratch a sample of aluminum and coat it with mercury. Mercury would form an alloy with Al, but that wouldn't affect its reactivity towards the solution of anything, it would just make it harder for oxygen to oxidize the fresh aluminum.
Answering you question: as far as I know, the nitrate test involves strong base, which serves as a reactant for aluminum oxide, dissolving it and letting aluminum to react further.


Which substances have the highest/lowest enthalpy of formation per gram? Which substances have the most extreme enthalpies of formation per gram at standard temperature and pressure?

Beryllium oxide −23.96 kJ/g
Lithium fluoride −23.75 kJ/g
Beryllium fluoride −21.84 kJ/g
Beryllium hydroxide −21.01 kJ/g
Lithium hydroxide −20.24 kJ/g
Lithium oxide −19.93 kJ/g
Boron trioxide −18.29 kJ/g
Magnesium fluoride −18.04 kJ/g
Aluminum fluoride −17.98 kJ/g
Boron trifluoride −16.76 kJ/g
Aluminum oxide −16.43 kJ/g
Water −15.86 kJ/g
Magnesium hydroxide −15.85 kJ/g
Magnesium oxide −14.92 kJ/g
Hydrogen fluoride −13.66 kJ/g
Lithium hydride −11.4 kJ/g
Boron nitride −10.24 kJ/g
Carbon dioxide −8.941 kJ/g
Lithium nitride −4.73 kJ/g
Methane −4.6 kJ/g
Ammonia −2.71 kJ/g
Diborane 1.31 kJ/g
Cyclopropane 1.33 kJ/g
Cubane 5.73 kJ/g
Dicyanoacetylene 6.579 kJ/g
Various unstable nitrogen allotropes >10 kJ/g
Free hydrogen (not bound in diatomic molecules) 216.34 kJ/g


How to determine pH of food in order to avoid unsuitable storage containers In an acidic (lemon-containing) sauce I had placed a metallic spoon. A few days later it had a metallic taste.

Stainless steal might be a significant source of nickel, chromium and iron [1]. Also, using aluminum utensils and packaging when cooking or storing food can contribute to the daily aluminum intake [2].

I want to avoid storing food in metallic containers which could increase my intake of metals.

The pH of some foods can be found here, but in some cases it's not of much use (when the food is a varying mixture of ingredients).

Is there an easy way to determine the pH of food? (whether the food is colored like a tomato sauce or transparent)

You can try wetting the tip of an ordinary pH paper, then put it into a sample of your food with that wet side. The capillary forces will suck up the liquid of your food and you should see a colour change in the formerly wetted area and above.

PH paper is not a very reliable method, but it should give you a general idea and it is easily performed at home.


Finite probability from discrete distribution and zero probability from continuous distribution in Maxwell-Boltzmann statistics I have difficulty understanding a certain concept with the derivation of the Maxwell-Boltzmann distribution $f(v)$ function from Boltzmann statistics. The derivation starts with the Boltzmann statistics formula that gives the probability for an energy $E_x$:

$$p(E_x) = \frac{\exp(-E_x/kT)}{\sum_{i=0}^\infty \exp(-E_i/kT)}$$

From this formula the Maxwell-Boltzmann distribution function is derived:

$$f(v) = \left(\frac{m}{2\pi kT}\right)^{3/2} \cdot \exp\left({\frac{mv^2}{2kT}}\right)$$

I understand that this formula gives the probability density of a particle with velocity $v + \mathrm{d}v$. The reason that the formula gives the probability density for velocities between $v$ and $v + \mathrm{d}v$ is because velocity and thus energy are considered continuous, and therefore just 1 specific value of energy or velocity would have a probability of 0. Therefore, a range of speeds must be taken to give a non-zero probability.

If one specific value of energy or velocity should give a probability of 0, how come the formula for the Boltzmann statistics gives a non zero probability for a specific value of energy $E_x$?
And how can one reason a formula that considers a specific value of energy or velocity to be zero from a formula that doesn't? Is there a way to explain this?

I don't find this to be a duplicate of my other question. My other question states a misconception about translational energy being continuous in general. This question states how the derivation is done between two formulas, one considering energy to be continuous and the other doesn't.

The probability of any one specific velocity is not zero, it's just very, very small.  Infinitismal, or tending towards zero, is how we usually put it.  The reason is because the spacing between allowed translational states is inversely proportional to the size of the container, and when you are talking about a macroscopic container (a bottle of gas, say, or beaker of liquid) then from the atomic point of view this is gargantuan and the spacing between allowed translational energies very, very small.  (If you want to get a feel for this, look up the allowed energies for a particle in a box.  You'll see that the size of the box goes in the  denominator and as the box gets bigger the spacing between energy levels gets smaller and smaller.  You can amuse yourself by calculating the exact difference between the 1st and 2nd energy levels for, say, an H atom in a box 1cm x 1 cm x 1cm.  It's really small!)

So small that to do the sum over states in your top formula is a gigantic pain, quite challenging.  But if you convert it to an integral, the integral can be done pretty easily.  So that's what we do.  We replace a huge sum, over 10^20 terms or more, each of size 10^-20 or less, with an integral.  Think of it as the reverse of when you learned integral calculus, and in order to figure out an integral you replaced it with a large sum -- and let the size of each term go to zero while the number of terms went to infinity.

Here, you are taking a sum which has a very large number of terms, each of which is very small, and replacing it with an integral.  It's not exact, but it's very good, and it becomes perfect in the limit of an infinite-sized container, and anyway the errors for ordinary-sized containers are ridiculously small, unmeasureable.


Mechanism for Synthesis of Tetranitromethane from Acetic Anhydride and Fuming Nitric Acid In my undergraduate Mechanisms-class the formation of tetranitromethane from acetic Anhydride with fuming $\ce{HNO3}$ was listed as optional knowledge in a chapter about the reactions of enols, enamines, etc. I can't find it in my notes or in the recommended literature, neither can I find information on molbase, and the two papers I can find, the original 1910 synthesis and This, are not mentioning any reaction mechanism.

I guess it somehow decomposes at the stage when the acetic anhydride is tri-substituted on one side, and somewhere there is a trinitromethane-anion involved which forms the product with $\ce{NO2+}$?

The Organic Synthesis preparation you cited for the preparation of tetranitromethane has some clues that are useful in providing a viable mechanism. Upon slowly adding acetic anhydride (Ac2O) to anhydrous HNO3 with cooling an exotherm is produced. The reaction mixture is allowed to sit at room temperature for 7 days. Assuming that they investigators did not go on vacation for a week, I presume that the formation of tetranitromethane (TNM) is slow and that the exotherm is due to the formation of the mixed anhydride 2. [The nitronium cation is also a possible electrophile as well as the mixed anhydride 2.] The enol of acetic anhydride can undergo substitution to afford species 3 and then repetitively on to the trinitro mixed anhydride 4. Anhydride exchange with acetic acid (or nitric acid) affords trinitroacetic acid 5 which decarboxylates to enol 6 (aci-form). Tautomerization of 6 can form trinitromethane. Footnote 3 in the Organic Synthesis link mentions that trinitromethane 7 can be converted to tetranitromethane 8 with HNO3. The transformation of 6 to 8 does not preclude nitration via mixed anhydride 2.


Why do some nucleophiles attack at the carbonyl group while others at an alpha-carbon in ketones? I was reading about the iodoform reaction and it stated that $\ce{OH^-}$ attacks at the $\alpha$-carbon and makes a substitution reaction possible. But there is also the possibility of attacking the carbonyl group directly and forming a gem-diol. Why doesn't that happen?

For starters, you shouldn’t call it a nucleophilic attack at the α carbon; it is a deprotonation (something quite different since the latter doesn’t access the carbon atom in any way).

The second thing to note is that both do indeed happen. However, the nucleophilic attack on the ketone is not productive: nothing will happen that creates a new product or a different pathway. The only productive result of hydroxide’s attack onto the carbonyl carbon in the long run is the re-elimintion of either of the geminal hydroxy groups to reform the carbonyl and we’re back at the start.

Please don’t stop thinking like this (‘what happens when $\ce{X}$ attacks $\ce{Y}$ rather than $\ce{Z}$?’) — it’s a great way to spot side reactions, alternative pathways, problems with functional groups etc.


Is this a valid structure for the nitrate ion? Why is the below structure not considered a valid structure for the nitrate ion? Is it because adding a double bond reduces the formal charges present on the nitrogen? Does the above form exist anywhere?

Nitrate actually exists as a superposition of these three resonance forms:

(Source: Wikipedia)

Essentially, a lone pair from an oxygen with a negative formal change forms a double bond to nitrogen. However, as nitrogen is a period II element, and hence must obey the octet rule, it can not have 5 bonds to other atoms. The already-present double bond then turns into a single bond and the pair of electrons moves onto the oxygen. This "movement" can happen with all three of the oxygen atoms and all three of the bonds. This is called "resonance".

Due to resonance, the energy of the system is minimised making it more stable.

Your Lewis structure is not far off. The "true" structure of nitrate is:

(Source: Wikipedia)

Approximately, each oxygen atom contributes one third of a bond with the nitrogen, lowering the formal charge on each atom from what you have written, which does not exist in nature.

The true structure is sort of an average of the properties of each possibility for where the electrons could be located, including the three resonance forms above, your structure, and any other configuration. However, these other forms are of minuscule probability to be formed (perhaps read up on wavefunctions and electron distributions if you're interested), and so the main contributors to the "real" structure are the three resonance forms.

Note that none of these structures ever exist in isolation for more than the most infinitesimal of time periods.


Why does the C–C bond have extremely weak absorptions? My book, Pavia's Introduction to Spectroscopy, tells me this about the infrared spectrum of alkanes:

  $\ce{C-C}$ : Stretch not interpretatively useful; many weak peaks.

This is very interesting. What makes the C–C bond so reluctant to absorb infrared wavelengths?

I have been doing a bit of thinking lately. Bonds absorb infrared radiation that match with the harmonic vibrational frequency of the bond. I believe the extent of absorption is based on the tendency of the bond to get vibrated. For some reason, the $\ce{C-C}$ bond is very reluctant to absorb the radiation, and I believe it's because the bond is too "rigid".

If a bond is too rigid, it must show large change in intermolecular force per change in bond distance. In other words, I believe, the value of

$$\frac{\mathrm dF}{\mathrm dr}$$

is quite high. In such a case, the bond wouldn't have a tendency to compress and expand, since the bond has such a high tendency to oppose such changes. Assuming the bond to act as a spring, I have,

$$F=kr$$

where $k$ is the spring constant of the bond-spring. Differentiating this w.r.t $r$, we get:

$$\frac{\mathrm dF}{\mathrm dr}=k$$

So, this makes me conclude that a higher value of $k$, implies a more rigid bond. I also know that

$$U=\frac{1}{2}kr^2$$

Plugging the values of $U$ ($\pu{=346000J}$) and $r$ ($\pu{=1.54 \times 10^{-10}m}$) for the $\ce{C-C}$ bond, I get,

$$k_{\ce{C-C}}\pu{= 2.92Nm^-1}$$

If I compare this with the value of $k$ for the $\ce{C=O}$ bond (one of the most prominent absorptions), I get,

$$k_{\ce{C=O}}\pu{= 1.11Nm^-1}$$

The $k_{\ce{C=O}}$ is just 3 times smaller than $k_{\ce{C-C}}$, but the existence of $\ce{C-C}$ absorption is rather non-existent.

I can conclude one thing for sure, that all that I have thought before is completely wrong.

So here's my question. Why does the C–C bond absorb extremely weakly? And what's wrong with what I have thought about above?

For vibrational spectra the primary transition under investigation is the $v = 1 \leftarrow 0$ excitation (because $\hbar\omega >> k_\mathrm{B}T$, so excited states have negligible thermal population). The intensity of the transition depends on the transition dipole moment:

$$R_{10} = \langle 1 | \hat{\mu} | 0 \rangle$$

in that the intensity $I \propto |R_{10}|^2$. Taking a Taylor expansion of the operator $\hat{\mu}$ with respect to $q$, the vibrational coordinate, at $q = 0$, we have:

$$\hat{\mu} = \mu_{q = 0} + \hat{q}\left( \frac{\mathrm d \mu}{\mathrm d q} \right)_{q=0} + \cdots$$

and plugging this in we get

$$\begin{align}
R_{10} &= \langle 1 | \mu_{q = 0} | 0\rangle + \left< 1 \middle| \hat{q}\left( \frac{\mathrm d \mu}{\mathrm d q} \right)_{q=0} \middle| 0 \right> + \cdots \\
&= \mu_{q=0}\langle 1 | 0\rangle + \left( \frac{\mathrm d \mu}{\mathrm d q}\right)_{q=0} \langle 1 | \hat{q} | 0 \rangle + \cdots
\end{align}$$

where we took out the values of $\mu$ and $\mathrm d\mu/\mathrm d q$ evaluated at $q = 0$ because those are just ordinary numbers. The first term in the Taylor expansion vanishes because $|1\rangle$ and $|0\rangle$ are orthogonal (a key property of the harmonic oscillator, which is a good model for the low-lying vibrational states at the bottom of the well), and if we neglect the second- and higher-order terms, we have†

$$R_{10} \propto \left( \frac{\mathrm d \mu}{\mathrm d q} \right)_{q=0}$$

The RHS represents how much the dipole moment $\mu$ changes upon vibration of the bond (which is measured by $q$). As such:

for something like a C–C bond, where vibration leads to nearly no change in the dipole moment, this quantity is small, the transition dipole moment is small, and the peak is weak.
On the other hand, the stretching of the C=O unit leads to a large change in dipole moment and the corresponding peak is therefore stronger.
The C=C stretch of alkenes tends to be weak as well, unless it is conjugated to an electron-withdrawing group like a carbonyl group. This can also clearly be linked to the magnitude of $(\mathrm d \mu/\mathrm dq)_{q=0}$.
And incidentally, this is also why molecules like $\ce{H2}$ or $\ce{O2}$ cannot be studied by IR spectroscopy (no matter how much you vibrate, there is no dipole moment), and also why the symmetric stretch of $\ce{CO2}$ is not IR active. In all of these cases $(\mathrm d \mu/\mathrm dq)_{q=0}$ is identically zero.

Unfortunately most of what you've written is kind of off the mark.

It is indeed "reluctant" to absorb the photon, but that is not because of the stiffness of the bond. If the bond is stiffer ($k$ is larger), then it simply needs a photon of shorter wavelength (as the vibrational frequency $\omega = \sqrt{k/\mu}$ will be larger). Instead it is related to the poor interaction of the C–C vibrational mode with the oscillating electric field of the photon. This is going into stuff that I don't know about, so can't explain much more (search "Fermi golden rule" in Atkins MQM or some other quantum book).
$D_\mathrm{e} = kr^2/2$ is not a valid equality here as it simply makes no physical sense. If the energy of the oscillator is at $D_\mathrm{e}$, then physically speaking, the corresponding bond length $r$ should be infinity, as the bond is dissociated. Furthermore, the equilibrium bond length which you used in the calculation ($r_\mathrm{e} = \pu{154 pm}$) refers to the middle of the potential well. If you really wanted to model the C–C bond as a classical oscillator, then your equation should read
$$U = \frac{1}{2}k(r - r_\mathrm{e})^2$$
because the thing that gets squared is the displacement from equilibrium.
In that calculation you used the dissociation energy for one mole of carbon–carbon bonds, and that doesn't make sense if you're trying to calculate the force constant of one carbon–carbon bond.
Even changing no. 3 wouldn't make any difference, because the harmonic oscillator model simply does not work near dissociation (in fact if a bond were to be a harmonic oscillator, it would never dissociate). Therefore bond dissociation energies cannot be treated within a harmonic model; you have to use something else such as a Morse potential.
A typical value of a C–C force constant is on the order of $\pu{400 N m-1}$. A better way of calculating the force constant is with
$$\bar{\nu} = \frac{\omega}{2\pi c} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}} \implies k = \mu(2\pi c\bar{\nu})^2,$$
where $\bar{\nu}$ is the IR wavenumber and $\mu$ is the reduced mass of the oscillator. A very rough calculation is as follows. Plugging in $\mu = \pu{10^-26 kg}$ (for a $\ce{C-C}$ bond), $\bar{\nu} = \pu{1000 cm^-1} = \pu{10^5 m^-1}$, and $c = \pu{3 x 10^8 m s^-1}$, we have $k = \pu{355.3 N m^-1}$. Correct order of magnitude, at least.
It's also nearly impossible that $k_\ce{C=O} < k_\ce{C-C}$, as force constants tend to be correlated with bond order. A quick Google search shows that $k_\ce{C=O}$ is on the order of $\pu{1000-2000 N m^-1}$. (If you go searching for this, you may wish to know that $\pu{1 mdyne/Å} = \pu{100 N m^-1}$, as CGS units are widely used.)

† Of course, there's an additional point that must be made, which is that $\langle 1 | \hat{q} | 0 \rangle \neq 0$, but you'll have to take my word for it as I'm not keen on proving it here. It is most readily shown using ladder operators, and it should be covered in any QM text under the harmonic oscillator.


What is the senior parent chain in the following compound? Will the IUPAC name be (2-hydroxymethyl) but-3-ynoic acid (or) 2-ethynyl-3-hydroxy propanoic acid?

The most important simplified criteria for the choice of a principal chain are:

greater number of suffixes
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

  (j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);

  (…)

  P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

  (…)

You have correctly identified the suffix (“oic acid”) for the principal characteristic group. The next criterion for the principal chain is the greater number of skeletal atoms (i.e. the longest chain). Thus, the principal chain corresponds to the but-3-ynoic acid part and not to the 3-hydroxypropanoic acid part since but-3-ynoic acid has a longer chain than propanoic acid. Therefore, the correct name is 2-(hydroxymethyl)but-3-ynoic acid.


Reason of dissociation of water molecule in electrolysis $\ce{H2O}$ molecule is dissociated into $\ce{OH-}$ and $\ce{H+}$ ions on reduction at the cathode in electrolysis. But what happens at the atomic level to $\ce{H2O}$ that proton and hydroxide ions get separated after gaining electrons from cathode? What really happens when water molecule gains electrons from cathode?

$$\ce{2H2O(l) + 2 e- ->  H2(g) + 2 OH-(aq)}$$

The dissociation of water to $\ce{OH-}$ and $\ce{H+}$ happens first before any interaction with the cathode. $\ce{H+}$ ions then gain an electron at the cathode to produce hydrogen. At the anode, water is oxidized (loses electrons) to produce oxygen and more $\ce{H+}$ ions.

In pure water there is a very small concentration of $\ce{H+}$ and $\ce{OH-}$ which is why the reaction is very slow and not thermodynamically favorable. It is also the reason why pure water is such a good insulator. But when you add an electrolyte the process becomes much more favourable kinetically and thermodynamically. Further proof that $\ce{H+}$ and $\ce{OH-}$ are generated before any interaction with the cathode.


Reaction mechanism of 2-amino-4,6-dimethylpyrimidine as a nucleophile My understanding tells me that 2-amino-4,6-dimethylpyrimidine can be deprotonated quite easily because the conjugated base is stabilized be resonance and act as nucleophile attacking aldehyde carbon. But I found no resource in literature about this reaction:

Your mechanism is almost correct; this is essentially an aldol condensation. The only issue is with the protonation states, which (as you've drawn) aren't entirely feasible.

Let's start by establishing how acidic that methyl group actually is. You're right that it is more acidic than a typical methyl group, because the charge can be delocalised onto those two nitrogens. But note that this delocalisation is pretty similar to that in a typical ketone enolate:

so to a very, very rough first approximation, that proton you're removing has a $\mathrm{p}K_\mathrm{a}$ of $\sim 20$, roughly that of a ketone. Yes, there are two nitrogens in the ring which improves the delocalisation, but there is also an amino group sticking off the ring, which is also trying to delocalise electron density onto the ring nitrogens. The value of $4.8$ cited in the comments is far more likely to be the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid (where the amino group is protonated).

Secondly, if anything were to be deprotonated, the $\ce{NH}$ in the amino group is going to be more acidic than the $\ce{CH}$ in the methyl groups. This is exactly analogous to how ammonia is more acidic than methane, and how amide $\ce{NH}$'s are more acidic than ketone $\ce{CH}$'s.

And lastly, note that your reaction is being carried out using acetic acid as a solvent. The chances that a carbanion (and a strongly basic one, as previously established) can exist for any reasonable amount of time under these conditions are... low, shall we say!

Now, you've probably drawn mechanisms for aldol condensations in acid. In this case, you don't deprotonate to form an enolate; you just use the enol tautomer, which is still nucleophilic on carbon. So the strategy here is not to deprotonate the methyl group outright, but to use the tautomer. Basically, use the same mechanism you've drawn, but add a proton to everything:*

That includes the final elimination, too. Note that if you protonate the $\ce{OH}$ group and let it fall off, you can get a benzylic cation that is stabilised by the 4-hydroxy substituent.

* You could delocalise all the way onto the other ring nitrogen if you wanted.


Dipole moment of (1s,2s,3s)-1,2,3-trichlorocyclopropane Which of the following compounds has a zero dipole moment?

Only one of the options is correct
According to me, it looks like both options (c) and (d) are right. Since the dipole moments of option (c) molecule seem to negate each other (shown in the image below), so is option (d). However, only option (d) is right. Why?

The issue stems from the fact that you may not have understood what each bond notation means.
Here, the solid wedge indicates that the bond is toward us and so a 3-D view of the same molecule would look as follows:

So as you can see the dipole moments – instead of cancelling, add up. So simply by elimination option (d) is the right answer
For trans-1,3-dichlorocyclobutane, the reason for the zero dipole moment is explained in ron's answer to the question "Does trans 1,3-dichlorocyclobutane have zero dipole moment?".

[...] Conformers A and B both have $C_$ symmetry (the only symmetry element is a plane that bisects the ring and contains the two cyclobutane carbons bearing the substituents) and therefore do have a dipole moment. However, the dipole moments of conformers A and B are equal and opposite, so when flipping is rapid the dipole moment averages out to zero.

Therefore (d) is the correct answer and not (c).


A unit cell for graphene In a study I have run into some questions asking for "How many atoms are contained in the unit cell" for graphene. I have some trouble understanding the unit cell of graphene which is always depicted as a mono-layer of atoms. What is a unit cell in a mono-layer?

The closest I could find is something like this picture:

The upper right image shows something soon-to-be a hexagonal structure. But when is such structure actually appearing?

The unit cell for graphene is a two-dimensional rhombus according to the figure shown on page 31 of this paper.$^1$ (also here.) The result is that two atoms are contained per unit cell. The upper right structure actually appearing in graphite, stacked layers of graphene.

$^1$Zhou, J; Huang, R. Internal lattice relaxation of single-layer graphene under in-plane deformation. Journal of the Mechanics and Physics of Solids, 2008, 56(4), 1609-1623. https://doi.org/10.1016/j.jmps.2007.07.013


What is the derivation of the formula for finding number of pure and hybrid orbital? I am studying Organic Chemistry. Recently I came up with the formula for finding number of pure and hybrid orbital that can be present in a organic compound. That is:

Number of Hybrid  Orbital = number of "Carbons" $\times$  (number denoting the hybridisation of carbon) and
Number of Pure Orbital = number of Hydrogens present $+ 2 \times$  (no of $\pi$-orbitals present)

Example Benzene:

number of hybrid orbital $= 6\times (3) = 18$ (Hybridisation is $\mathrm{sp^2} = \mathrm{1s} + \mathrm{2p} = 3$)
number of pure orbital $= 6 + 2 \times 3 = 12$

which tells that there are 18 hybrid orbitals and 12 pure orbitals present in benzene.
My question is, What is the derivation of this formula?

All we are doing is counting orbitals, and these formulas are generalizations of counting the same way that multiplication is generalization of repetitive addition. The derivation is exactly as you present it. For the first formula, we are multiplying the number of hybrid orbitals per carbon atom by the number of carbon atoms. For the second formula, we are multiplying the number of non-hybrid orbitals per carbon atom by the number of carbon atoms and adding the number of hydrogen atoms (since each hydrogen atom only has one orbital that cannot hybridize**).
If you took benzene $\ce{C6H6}$ and counted all of the hybrid orbitals, you would count 18. Each carbon atom is $sp^2$-hybridized, so each has three hybrid orbitals. We could count each individually:
$$1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 18$$
Or you could group by each of the six carbon atoms, which eventually simplifies to $3\times 6$
$$(1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) = 18\\
3 + 3 + 3 + 3 + 3 + 3 = 18\\
3 \times 6 = 18$$
Your formula has some limitations. First, it only works for hydrocarbons (compounds with only carbon and hydrogen atoms), and it assumes all carbon atoms have the same hybridization. Here is a hydrocarbon where the carbon atoms do not all have the same hybridization. This is hex-1-en-5-yne and it has two $sp^3$-hybridized carbon atoms, two $sp^2$-hybridized carbon atoms, and two $sp$-hybridized carbon atoms.

Your formula could be extended to:
$$N_{\mathrm{HO}} = 2n_{sp} + 3n_{sp^2} + 4n_{sp^3}$$
Where $N_\mathrm{HO}$ is the number of hybrid orbitals, $n_{sp}$ is the number of $sp$-hybridized carbon atoms, $n_{sp^2}$ is the number of $sp^2$-hybridized carbon atoms, and $n_{sp^3}$ is the number of $sp^3$-hybridized carbon atoms. For hex-1-en-5-yne, we could count:
$$N_\mathrm{HO} = 2\times 2 + 3\times 2 + 4\times 2 = 18$$
There is a more general formula for hydrocarbons. Let's look at it this way. Each carbon atom has four orbitals (either hybridized or non-hybridized). If there are no $\pi$-bonds, then the number of hybrid orbitals is four times the number of carbon atoms. Each $\pi$-bond requires to non-hybridized $p$-obitals, so each $\pi$-bond reduces the number of hybrid orbitals by 2. Here is a general formula for hydrocarbons:
$$N_\mathrm{HO} = 4n_\mathrm{C} - 2n_\pi$$
Here $n_\mathrm{C}$ is the number of carbon atoms and $n_\pi$ is the number of $\pi-bonds$. For benzene, $n_\ce{C} = 6$ and $n_\pi = 3$, so
$$N_\mathrm{HO} = 4\times 6 - 2 \times 3 = 24 - 6 = 18$$
For hex-1-en-5-yne, we have the same number of carbon atoms and $\pi$-bonds, so we get the same result. For the number of non-hybridized orbitals ($N_\mathrm{NHO}$), we have a general formula as well. Each hydrogen atom contributes one non-hybridized orbital ($n_\ce{H}$), and each $\pi$-bond contributes two.
$$ N_\mathrm{NHO} = n_\ce{H} + 2n_\pi$$
For benzene and hex-1-en-5-yene, $n_\ce{H} = 6$ and $n_\pi = 3$, so
$$ N_\mathrm{NHO} = 6 + 2\times 3 = 12$$
This is the same as your second formula.
** I expect one of my physical/theoretical chemist colleagues to come along and inform us that under certain circumstances, hydrogen atoms may in fact act as if they have hybrid orbitals, but for most organic compounds we can ignore this behavior.


How can I remove acids from cola? I would like to taste Pepsi-cola without acid. I like it's flavor but don't like acidic taste. It's acidity comes from carbon dioxide, phosphoric and citric acids (see ingredients list in wiki and answers to this question). Carbon dioxide can be evaporated easily. How can I get rid of phosphoric and citric acids? Preferably without expensive lab equipment and hard-to-get reagents.

Edit: Found one recipe by Josh Velson on quora site. He advises adding limewater until the pH reaches about 12.32. Are there any other methods?

You can't remove the acids, because they're dissolved in the liquid. You can, however, neutralise it.

Acid is basically $\ce{H+}$ ions dissolved in the water. By reacting it with a base, you do the following reaction:

$\ce{H+ + OH- -> H2O}$

So the acid reacts to form harmless water.

The problem is that you can't just add $\ce{OH-}$, you have to add it as some salt. Also, the citric and phosphoric acid you have in the cola were added are also dissociated from a salt or molecular form:

$\ce{H3PO4 -> H+ + H2PO4−}$ for phosphoric acid, for instance.

So whatever you put in there that reacts with the $\ce{H+}$ should also react with the $\ce{H2PO4−}$ ion to make something that's not soluble.

Usually, neutralising acids is done with carbonates such as sodium carbonate or calcium carbonate (also known as soda and lime, respectively). Both of which are safe materials to handle. The problem is that by doing that, you're introducing carbonate ($\ce{CO_3^2−}$) which might alter the taste.

A better solution is to add in hydroxides, such as sodium or calcium hydroxide ($\ce{NaOH}$ or $\ce{Ca(OH)2}$). The problem is that those are much more hazardous things which can easily hurt you if you don't use them properly. I would recommend using calcium hydroxide because then the calcium will react with the phosphate anion in the cola to make insoluble calcium phosphates and with citrate to make (somewhat) insoluble calcium citrate. Then you can just filter the stuff out and have your acid-free cola.

Basically, by adding in $\ce{Ca(OH)2}$ to your cola, you react the "acid" bit to make water, and you react the "phosphoric" and "citric" bit to make a white solid precipitate in the bottom of your glass.

DISCLAIMER

This is all in theory. I highly recommend against adding chemicals to your food and drink. If you don't like the acidic taste of cola, don't drink it. Find another brand or another drink that you do like. Tea? Anyway, do not mess with your food unless you know exactly what you're doing and you know that your chemicals are food-grade and safe. Don't go and do stuff with your cola just because someone on the internet (me) told you it will work.


How many atoms of carbon are needed to produce 0.45 mol aluminium? How many atoms of carbon are needed to produce $\pu{0.45 mol} ~\ce{Al}$?
  $$\ce{3C + 2Al2O3 -> 4Al + 3CO2}$$

I am not sure where to start, whether or not I convert moles straight to atoms or convert moles to grams of $\ce{Al}$, then to moles of $\ce{C}$ and then to atoms? I tried the latter of my suggestions and ended up with $\pu{9.8E24}$, but  I don't know if this is correct.

The key to this is by examining the mole ratio. Your balanced equation shows that 4 mols aluminium: 3 mols carbon

If you are familiar with ratios from maths, you can therefore find the specific number of moles of carbon in this reaction.

$4\ce{Al}: 3\ce{C}\implies0.45\ce{Al} :~?$

I think you should be able to get the answer as 0.3375 moles of carbon (again, it's just ratios)! From here, you use moles = mass / mr to get your answer.


Why is oxygen an oxidizing agent? Why are diatomic oxygen molecules STILL reactive especially with metallic elements like sodium and copper even at room temperature?

You would think that since the two oxygen atoms already have the much needed 8 valence electrons when they bonded with each other that they wouldn't need to react with anything else but that doesn't seem to be the case.

To estimate reactivity solely based on the the octet rule is a crude, (semi-)empirical approach in the first place. Atoms and molecules don't know anything about their number of electrons in shells, but the formation of molecules are based on the minimization of (free-)energy of the system.
The octet rule gives a hint, for what kind of systems the potential energy may be small, but does not say anything about the relative energies of two systems, both fulfilling the octet rule.

To "exactly" predict the reactivity, you would have to solve the Schrödinger equation for both states of your systems, find the eigenvalue of the energies, consider entropic contributions. Furthermore you would need to "sample" all possible pathways of your reaction, to check if the energy barriers are small enough to be overcome at a given temperature.
(To be really exact, you still would need to estimate entropic contributions)

Yet still a chemists' intuition often can predict reactivity very well. Concerning your specific problem, one possible qualitative explanation would be, that the octet rule is not fulfilled for copper, but only for oxygen, before the reaction happens. Taking a look on the structure of Copper(II) oxide on the other hand, shows that it fulfils the octet rule for both, copper and oxygen.(Ignoring d-electrons for copper)

Furthermore, the reaction enriches the electron density in the strongly electronegative oxygen, while reducing the electron density in the "electropositive" copper, which also stabalizes the structure of copper oxide.


Why does Naphthalene turn into needle-shaped crystals by sublimation? I wonder why crystals of Naphthalene are needle-shaped after sublimation. Googling the fact, I could only confirm that it is true but failed to find the reason. Could someone please shed some light as to why this happens?

Every solid material has a crystal structure.  Sublimation tends to produce good crystals because the solid can grow slowly.  In bulk form naphthalene crystals are broken and compressed and so bulk solid naphthalene looks quite different.

The exact same situation applies to snowflakes.  They are flat hexagonal structures when seen individually.  Crunch them into a snowball and they lose that pretty structure.


Can we determine the amplitude of vibration of O-H bonds in water? This is going to be a long shot, and I might be wrong in any and all of my assumptions, be warned (... and please correct me). I might also be trivially right, I have no idea.

At equilibrium, each degree of freedom has the kinetic energy:

$$E = \frac{1}{2}k_\mathrm{B}T$$

So, I have water at, say, $300\ \mathrm K$. Water has three atoms, so 9 degrees of freedom (3 per atom, because space is three-dimensional). One of them is for a $\ce{O-H}$ stretching. I know that usually, you would need to divide into modes of oscillations, but let's say we combine half the energy of the symmetric and antisymmetric stretchings, giving that one $\ce{O-H}$ stretching has a full $\frac{1}{2}k_\mathrm{B}T$ in it. So we have an average, roll of drums,

$$ E = \frac{1}{2}(1.3806488 \times 10^{-23})(300) =  2.071\times 10^{-21}\ \mathrm J$$

in a $\ce{O-H}$ stretching.

NOW, let's say this stretching behaves like an harmonic oscillator, its' energy is given by the amplitude of oscillation:

$$ E = \frac{1}{2}kA^2 $$

where $k$ is the stiffness constant of the chemical bond.

That's nice, but we have two unknowns, $k$ and $A$. So let's go further. If we know the absorption band's wavenumber, we can say that the amplitude corresponds to this wavelength, $2.734\ \mathrm{\mu m}$ (source = Wikipedia). This would mean that we have a stiffness constant:

$$ k = 5.5412 \times 10^{-10}\ \mathrm{N/m}$$

I know that, although it was fun to blindly math around, this result is wrong. Just the fact that the stiffness constant would depend on temperature doesn't sound right. Neither does the bit about wavelength = amplitude.

My question is: is it possible to know the amplitude of oscillation/stiffness constant of a $\ce{O-H}$ bond?

My question is: is it possible to know the amplitude of
  oscillation/stiffness constant of a O−H bond?

Yes.

Your approach is along the right lines but -

The Hooke's Law formula you used

$$E = \frac{1}{2}kA^2$$

is for the stretching of a spring.  Hooke's law changes if the spring has mas(ses) attached at the end(s), now it is a harmonic oscillator.  In this case Hooke's Law becomes

$$E = \frac {h}{2 \pi} \sqrt{\frac{k}{\mu}}$$

where k is the force constant you are interested in and $\mathrm{\mu}$ is the reduced mass of the system.

Since

$$E=h\nu$$

The frequency, which can be experimentally measured, is given as

$$\mathrm{frequency} = \frac {1}{2 \pi} \sqrt{\frac{k}{\mu}}$$

The frequency is independent of the amplitude, affected only by the masses and the stiffness of the spring.

If the masses of the atoms at the ends of the spring are given as A and B, then

$$\mu =\frac{A \cdot B}{A+B}$$

So for an $\ce{O-H}$ bond $\mathrm{\mu=\frac{1 \cdot 16}{1+16}}\ \mathrm{u}$. If we measure the frequency of the $\ce{O-H}$ stretch absorption, then we can calculate the force constant.

Here is a link to a one-page example that shows the actual computation and explains the specific units to be used very nicely.

You also ask about the amplitude of oscillation.  If, by this you mean the bond length, then yes this can also be calculated from spectra (relatively easy in simple diatiomic cases, but more difficult in complex polyatomic cases.  The methodology involves assuming a rigid rotor and then using the Schrodinger equation to solve for the energy states and extracting the bond length.  A nice example using $\ce{HCl}$ can be found here, look down the page for the section entitled "Bond Length of HCl".


How to recycle olive oil manufacturing wastes? The fact that only 20% of the olive used to make olive oil becomes olive oil and the other 80% is spread between hard wastes and liquid water wastes  made me want to ask this question. Isn't there a way to recycle the other 80% and use it in an effective way either in other industries (energy, agriculture, or cosmetics)?

I'd like to know if there is a simple and low-cost way people can do so.

The components of the 80% waste material could definitely be recycled using common procedures ( eg distillation, solvent extraction ) and could be used for many purposes like you mentioned or by converting the waste material to a more useful product.

I'm not sure what they make from the waste from olive oil production exactly but a good example of similar biotechnology is the use of waste cooking oils to synthesize biodiesel.


Major form of methyl 3-hydroxy pent-2-enoate? I came across this question in a test:

  The enol form is the ________ form for 3-hydroxy pent-2-enoate?  (major/minor)

My argument is that the enol form must be more stable because of hydrogen bonding in the enol form and subsequent six membered ring formation (chelation). However the keto form has no such stabilization factor. Hence according to me the enol form must be more stable. But sadly, that is the wrong answer. What's going wrong?

Structure for reference:

quick googling shows that for a very similar compound ethyl acetoacetate enol and ketoforms are in equilibrium with 10% of enol form. However, this equilibrium is influenced by solvent significantly. Actually, the percentage shows that the difference in stability between these two forms is veeery little.

The difference may be attributed to the difference in stability of the bonds, i.e.   C=O + C-H is better then C=C + C-O + O-H. However, internal hydrogen bonding DOES exist and participates in stability of enol form. For example, for hexan-1,3-one, the percentage of enol form is much lower.


The definition of peptide, peptide bond and protein According to what I understand the difference between peptide and peptide bond is that peptide is two or more amino acids (up to 19 -inc.) which linked together, while the term "peptide bond" refers to the bond between each two amino acids. Is that true? If it is, the definition of protein is that it contains at least 20 amino acids which are linked together. Isn't it?

Two or more amino acids form a peptide. The peptide has the repeated form $\ce{-CO-NH-C(\alpha)R - }$ where R is an amino acid side-chain. The peptide bond is the amide bond N-C. The $\ce{CONH}$ group is planar. A peptide chain can be artificially constructed in the lab from any of the amino acids, its length and composition is then for you to determine.

A protein is a naturally occurring chain of amino acids folded so as to perform some function. Usually a protein has many amino acids because it has to perform a function in its organism. I'm not sure what the number of amino acids in the smallest protein is, but 20 seems about right. This number is determined by the protein's function, it can be many hundreds. A part of a protein's  backbone is also called a peptide chain.


How does hyperconjugation in alkenes affect bond lengths? I understand how the diagram below results in hyperconjugation stabilising alkenes.

Essentially, the C-H sigma bonding orbital interacts with the pi bonding orbitals creating two molecular orbitals. It also interacts weakly with the pi* orbital which lowers the energy of psi 1 and 2 stabilising the system.

What I do not understand is the effect this has on bond lengths. We have done a computational investigation in the lab and have found that both the C-H and C-C bonds are weaker, while the C-C bond between the double bond and the alkyl substituent is stronger. I don't understand why.

I have received an explanation for this phenomenon using resonance structures, but I am really looking to understand how to reconcile these observations with the MO diagram above.

The MO interpretation is not really all that complicated: you are allowing the C–H σ orbital to overlap with the C=C π* orbital. This leads to a very slight population of the π* orbital, which is antibonding with respect to the doubly bonded carbons. Therefore, that C=C bond length is slightly lengthened. Likewise, the C–H σ orbital is slightly depopulated, which makes the bond length longer.

At the same time, you are forming a slightly stronger C(sp2)–C(H) bond because these new MOs have bonding interactions between the C–H σ orbital and the p orbital on C(sp2). This is not entirely made clear in the diagram above from Fleming's text, which I have addressed separately here. From the diagram, it seems that there are equal amounts of bonding character (from ψ1) and antibonding character (from ψ2) However, the main point which I wrote in that post is that the π* orbital mixes (very slightly) into both ψ1 and ψ2. This increases the C–C bonding character in ψ1 and decreases the antibonding character in ψ2, such that there is a (very slightly) shorter bond.

The actual text in the book makes this clear, but the diagram does not. I suggest reading the text carefully.


Unit conversion of Molar Conductivity I know this might be a silly question but I am confused with different answers of internet. My question is:

Molar conductivity is given as:

Molar conductivity = $\frac{\kappa}{M}$.

S.I. unit of :
$\kappa$ is $\pu{Sm}^{-1}$  &

$M$ is in $\mathrm{mol \, m^{-3}}$.

Thus, Molar conductivity in S.I. is
$\mathrm{S \, m^2 \, mol^{-1}} \tag{1}$

When I substitute
1 m = 100 cm in (1).
I got new unit as
$[\mathrm{S \, cm^2 \, mol^{-1}}] \times10^4$.
But in a book It was given as

$[\mathrm{S \, cm^2 \, mol^{-1}}]  \times 10^3 \tag{2}$

where they have taken S in S/cm and molarity in mol/litre. But where I was wrong.

If you feel the question too confusing just explain me How I can convert that SI unit given in (1) into unit given in (2).

Refer to page 77 of Chemistry NCERT text book, class 12, part 1.

You were actually right about the $\mathrm{10^4\ S\ cm^2/mol =S\ m^2 /mol}$. Check if the $\mathrm{10^3}$ you are referring to has molarity included in its equation because molarity is given as moles per litre.

You can refer to the online NCERT book at its official website. Thank you, I hope this was useful and sorry if I got it wrong.


Aromatic Substitution in acidic medium Find product A.
I tried solving but why isn't the answer this
The answer is given this

Three parts to this answer:

You have two electro-philes, the tertiary carbo-cation formed by protonating the alcohol and breaking off a water molecule and then the bromine.  Looks like you got this part.

Assume the two electro-philes attack consecutively.  The methoxy group is more strongly activating and ortho/para directing than the alk-yl group for the second attack.  So the bromine and the newly formed carbon-carbon bond will be meta to each other.

The sterically favored positions for each are as shown in the given answer.  The carbo-cation favors the position para to the methoxy group.  The bromine picks the position ortho to the methoxy group that's away from the alk-yl function.

Please excuse my use of hyphens.  Device auto-correct was putting in wrong words and I had to defeat it.


What is the criterion for filling up of electrons in molecular orbitals of increasing energy in simple diatomic species? So far I have found different ways in different books.
For e.g.
One says

  "In MO electronic configuration of diatomic species up to 14 electrons, $\sigma2p_z$ orbital (considering $z$ as the internuclear axis) is greater in energy than $\pi2p_y$ and $\pi2p_x$ orbitals. For species with more than 14 electrons, $\sigma2p_z$ orbital is lower in energy than $\pi2p_y$ and $\pi2p_x$ orbitals."

Whereas the other book used 16 as the no. of electrons instead of 14. This leads to different ways of filling electrons in molecules with 15 electrons such as $O_2^{+}, NO$ etc. I want to know which one is correct.

What is known for sure (and is mentioned in all the books I've seen so far) is that for homonuclear diatomics the "$\pi \rightarrow \sigma$ switch" happens when going from $\ce{N2}$ to $\ce{O2}$: for 14-electron case $\pi_{2p_{x,y}}$ orbitals are lower in energy then $\sigma_{2p_{z}}$, for 16-electon one it is other way around. The usual "explanation" is that moving from $\ce{Li2}$ to $\ce{F2}$ the successive decrease of the $s-p$ interaction between the $2s$ orbital of one atom and the $2p_z$ orbital of another atom decreases the energy of the $\sigma_{2p_{z}}$. But you have to understand that this is merely a manifestation of a known fact: we know how orbitals of these molecules are arranged from the molecular orbital theory calculations and use the notion of $s-p$ interaction to rationalize this knowledge.

The problem is that no information about 15-electron case can be inferred from the above mentioned knowledge about molecular orbitals of homonuclear diatomics from $\ce{Li2}$ to $\ce{F2}$. A priori we could not tell will $\sigma_{2p_{z}}$ orbital be lower in energy than $\pi_{2p_{x,y}}$ orbitals or higher then them for a 15-electron case. Besides, things coul be different for different 15-electron cases, such as $\ce{N2-}$, $\ce{O2+}$, $\ce{NO}$, etc.: for some $\sigma_{2p_{z}}$ orbital might be lower than $\pi_{2p_{x,y}}$ orbitals, for others - higher. There is no simple rule to use, one have to resort to molecular orbital theory calculations to tell the truth.


What is the difference between enthalpy (H) and heat (q)? It seems that in some cases it's the same (that is, $\Delta Q_p=\Delta H$)  yet in others they differ.
For example,
\begin{align}
\ce{2H2(g) + O2(g) &->2H2O(l)} & \Delta H &= -572~\mathrm{kJ}
\end{align}

The answer for one mole of $\ce{H2O}$ produced is $\mathrm{-572\times 0.5~kJ}$.
Is this under the assumption of constant pressure? Therefore it's $\Delta Q_p=\Delta H$?

Enthalpy and heat are entirely different things.

Enthalpy is a function of state. If you know the state of a system, you know its enthalpy. If you know the starting and ending states of a process, you can find the enthalpy change.

Heat, on the other hand, is an inexact differential. Knowing the initial and final states of a process is not enough information to tell you the heat transfer. Instead, the heat transfer depends on the particular path taken between the states.

For a simple example of why this is important, consider a heat engine. The process is a cycle, meaning the initial and final states are the same, so a cycle of a heat engine has zero enthalpy change. However, the entire point is that the heat engine converts heat into work, so the heat exchange during a cycle is not zero.

That is why the heat exchanged is equal to the enthalpy change only under constant pressure. We need to give extra information (such as that the pressure is constant) to know the path taken during the process in order to say anything meaningful about heat exchange.


Merge rigid parts and flexible results after docking (Vina) If I use flexible residues in the binding pocket, they are cut from the file.

The protein is broken and the amino acids are unbound in the room.

How can I get my results and the protein together again?

A Useful Script for AutoDock4 and Vina That Merges Flexible and Rigid PDBQT Output
« Prosciens.
http://prosciens.com/prosciens/a-useful-script-for-autodock4-and-vina-that-
merges-flexible-and-rigid-pdbqt-output/
Link to the script: http://prosciens.com/prosciens/oldproscienssarl/files/flexrigidpdbqt2pdb_template.sh


Is there a certain amount of titrant that is ideal? For an AP chem lab dealing with determining the amount of citric acid in orange juice by titrating with NaOH, one of the discussion questions goes like this:

  Choose an amount of beverage to be titrated that will require at least 10 but less than 20 mL of titrant. Explain why this range of titrant is optimal.

This question makes no sense to me. Why would there be an "optimal range" of titrant? I would think that I should fill the buret to full capacity if enough titrant is available. Would it not be better to have excessive titrant than not enough of it?

We completed the lab, and the titration required 23.05 mL of 0.1 M NaOH solution to titrate 20.00 mL of orange juice. If we had followed the instructions, we would have added less orange juice, and used less NaOH, but what is the benefit in doing so? Budget cuts? If it was about not wasting solution, why would there be a minimal requirement of 10 mL of titrant?

In a perfect world there would be no optimal range. We do however live in a world full of errors and practical considerations.

Why use over a minimum amount?

Let's assume for demonstration, that each drop has a volume of 0.05 mL. Of course this is not an accurate value as various factors would affect this.

The titration has reached its end point when one last drop changes the colour. If your titre value is very small e.g. 1 mL, then each drop is 5% of your titre.

If only part of the drop is needed to reach the end point then all the extra part of the drop is inaccuracy, which in this scenario could be up to 5%! If you have the misfortune of accidentally adding one drop too many you are introducing an even larger error. This will impact concentration calculations based on the experiment.

Why not just use 'a lot' then?

There are a couple of things here:

1) As noted by Mithoron, this would be a waste. The accuracy gained from requiring additional titrant will level off rather rapidly.

2) If you use more than one full burette's worth there is an error penalty.

3) Using extra titrant is a waste of your time. Consider a 50 mL burette, if it needs to be filled up after each titration, that would be inefficient. I'm assuming you're doing repeats so this has real merit.


Why do C-F bond has less dipole moment than C-Cl bond? Larger the difference in electronegativities of bonded atoms, the larger the dipole moment.

(this statement has been extracted from chemwiki-dipole moment)
The electronegativity difference between carbon and fluorine is about $1.43$ (Pauling scale), $1.649$ (Allen scale). The electronegativity difference between carbon and chlorine is about $0.61$ (Pauling scale), $0.325$ (Allen scale). Although, the electronegativity difference of $\ce{C-F}$ bond is greater than electronegativity difference of $\ce{C-Cl}$ bond, dipole moment of $\ce{C-Cl}$ bond ($\pu{1.860 Debye}$) in $\ce{CH_3-Cl}$ is greater than dipole moment of $\ce{C-F}$ bond ($\pu{1.847 Debye}$) in $\ce{CH3-F}$, why?

The bond dipole moment ($\mu$) of a chemical bond is equal to the magnitude of the separation of charge ($\delta$) times the displacement of the charge (or the distance over which the charges are spearated) $d$, or:

$$\vec{\mu}=\delta \vec{d}$$

Thus, there are two factors controlling dipole moment. The charge separation can be estimated from the electronegativity difference, but not perfectly. Disparities in size and polarizability also effect the separation of charge between two atoms. The displacement is the length of the bond. The $\ce{C-Cl}$ bond length (176 pm) is longer than the $\ce{C-F}$ bond length (134 pm) - data from here.

Thus:

For $\ce{C-F}$, the electronegativity difference on the Pauling scale is 1.43, and the bond length is 134 pm: $1.43\times 134 = 191$ (ignoring units at the moment).

For $\ce{C-Cl}$, the electronegativity different on the Pauling scale is 0.61, and the bond length is 174 pm: $0.61\times 174=106$.

So, my quick and dirty method did not work so well. However the disparity between these two bonds by my method $(106/191)=0.55)$ is smaller than by just comparing electronegativities $(0.61/1.43=0.42)$
 and that is because of the other factor mentioned above: size and polarizability. Chlorine is bigger and more polarizable than fluorine (and carbon). The carbon-chlorine bond is polarized toward chlorine not just because of the electronegativity difference. Chlorine's larger size decreases the effectiveness of orbital overlap in bonding, and chlorine's increase polarizability implies that it can support more partial negative charge than suggested by its electronegativity alone.


proving an unknown organic compound using reactions I have been given three spectras (MS, IR, H NMR) to analyse, from them I have found that the compound is chlorobenzene (and I've confirmed online). The task I have gotten asks me to further justify my findings by using reactions and predicting and analysing the spectra of the products. However, I am quite unsure how I'm going to predict and analyse the spectra of the products from the reaction to prove my compound.

The task you had being assigned is called identification by derivatization. It's one of the strongest proof for the confirmation of the structure of a molecule.
If you have an hard-to-identify compound X, you can make it react with Y and then analyze the product of the reaction, X-Y, instead.

I once synthesized a polysubstituted aromatic compound and wasn't sure whether or not a iodine atom was present on the ring. So I performed a Heck reaction on the molecule obtaining the corresponding substituted olefin, which was a known compound, present in the literature and could be identified by its NMR spectra.

Chlorobenzene is quite an anonymous compound and its derivatization does not lead to significative compounds. Just an example to answer your question, you can perform a nitration to get o-chloronitrobenzene which is a solid and can be identified by its melting point.


How was it determined that the electron observed in the cathode ray experiments was the same particle that gave an atom its balancing negative charge? How was it determined that the electron observed in the cathode ray experiments was the same particle that gave an atom its balancing negative charge?

Couldn't there have been an entirely different negatively charged particle? What justified that assumption?

The idea that the atom contains small indivisible quantities of electric charge was developed in the 19th century by various contributors (e.g. Faraday, Stoney, Laming, Weber, and Helmholtz).

When Thomson (Nobel Prize in physics 1906) discovered the electron in 1897 and estimated its mass and charge, he used the free electrons of cathode rays.

The important experiment that showed that atoms contain electrons was the discovery of the Zeeman effect in 1896, i.e. the splitting of spectral lines in the presence of a magnetic field. Lorentz explained the phenomena with his electron theory. In 1902, Lorentz and Zeeman jointly received the Nobel Prize in physics.


How does amine inversion affect basicity? What is amine inversion, and how does it affect basicity?

For instance, why is $\ce{Et3N}$ less basic than (2.2.2.)cyclohexyl-amine despite the +I from ethyl groups? The answer to the question is option 3.

  Compare the basic strengths of the compounds given:

  (i) > (ii) > (iii)
  (ii) > (i) > (iii)
  (ii) > (iii) > (i)
  (iii) > (ii) > (i)

Amine inversion is also known as nitrogen inversion. You can find more here: http://en.wikipedia.org/wiki/Nitrogen_inversion

It has nothing to do with relative basicity of amines, that is defined by 1) energy of particular N-H  bonding (increases with +I effect) and 2) energy of ammonium cation hydration. (decreases with alkyl group number and their size). The relative balance of said factors is fragile and cannot be easily predicted without reference to experimental data except cases when amine has lone pair involved into bonding interactions (aniline, pyrrole, amides) or connected with extremely strong electron acceptors.


Why is graphite brittle? Carbon exists in more than one allotropic forms, and the allotropes have different physical properties due to different arrangement of atoms. I am going to compare their hardness for now.

Below is the structure of graphene:

Source: Wikipedia article on Graphene

From what I read, graphene is hard, but graphite is brittle.

If graphite is made up of many layers of graphene, then why is graphite not stronger than graphene?
I suppose when you say graphene (or diamond) is hard, it means that, for example, when you take a hammer to strike the layer of graphene, the layer is not easily broken, that is, the covalent bonds are hard to be broken by physical means.

By this reasoning, the layers in graphite are easily separated because they are held together by weak van der Waals forces, whereas it should be difficult for us to cut across the layers (in direction perpendicular to the plane of layers, cf: image below)

Does this mean that we can only cut the graphite in one dimension? I thought the graphite being brittle already implies that graphite can be cut in more than one dimension?

Original image source: Tirupati Graphite

Hardness and brittleness are not contradictory properties. There are plenty of substances that are both hard and brittle (cast iron is a good example, glass is another).

And "strength" is different as well. It depends on context. Tensile strength is not necessarily related to ability to survive a blow: some compounds with high tensile strength (which basically means the compounds resist being stretched) are incredibly fragile (or brittle if you prefer that term). Some compounds with low tensile strength are very hard to break (nylon rope, for example).

Part of the problem is that many of the properties of materials are based on the structural properties of the bulk material not the idealised properties of the chemical bonds making up the material. Some metals, for example, are tough (not brittle) and are hard to break but this property arises because defects in the crystal structure can move around absorbing external stress. Brittleness is often caused because no such mechanism exists and, once a surface bond breaks, the stress on the next bond gets worse, releasing energy and propagating a crack, tearing the material apart. Glass often has many small surface cracks which, under tension, concentrate stress making them brittle and subject to falling apart. But very pure, carefully grown glass fibres can be nearly crack-free making them much stronger under the same stress. In short to explain brittleness, toughness or strength, you need to know the macroscopic makeup of the bulk material not just the chemical composition.

So it is with graphite. There is no easy mechanism in a single graphite crystal to dissipate stress if a bond breaks, so it will tend to be brittle. But most samples of graphite are not single crystals: in most cases they will consist of a bunch of small units with the layers oriented in different directions. So bulk graphite isn't just brittle but also soft as some orientations will slide easily around. And it is hard to even talk about bulk graphene because by definition it is a single layer or a very thin bundle of layers. When people say graphene is strong they usually mean that it has high tensile strength along the plane of the layers. This doesn't mean it isn't either brittle or hard and even if it were in some orientations, it probably would not be in others.


Mechanics behind auto-ignition temperature I have some questions about the auto-ignition temperature for fuels. Per ASTM E659:

  AIT (auto-ignition temperature) is the lowest temperature at which the substance will produce hot-flame ignition in air at atmospheric pressure without the aid of an external energy source such as spark or flame. It is the lowest temperature to which a combustible mixture must be raised so that the rate of heat evolved by the exothermic oxidation reaction will overbalance the rate at which heat is lost to the surroundings and cause ignition.

The last sentence seems to say that AIT is the temperature where enough of the particles are above the activation energy such that enough heat released goes to perpetuate the remaining fuel to combust (cause ignition). Is this correct?
Per ASTM E659, ignition is subjectively defined as the appearance of a flame with a sharp rise in temperature. Why it necessary that we observe the appearance of a flame (in other words, can we have combustion without an observed flame)?
Is there an assumption of thermal equilibrium between fuel and oxidizer when defining AIT? The testing method above says air at predetermined temperature will be used, but I'm not sure how they plan to control air temperature when heating begins.

To be clear, I'm no expert on the ASTM standards writers' thinking, and all I have to go on is the standard you've posted.

Yes, that's exactly right.  Some of the reaction's heat will be lost to the surroundings, but enough of it will go into the reaction mixture to raise its temperature and promote further, more rapid, combustion.
There are reactions that produce lots of heat but no flame, hydrogen peroxide decompostion, for example.  The standards writers had to come up with a clear definition for ignition, and as you note, they did: "the appearance of a flame accompanied by a sharp rise in the temperature of the gas mixture."  They even say that ignition is a subjective term, so they are careful to define it clearly.  There might be some chemical reaction that produces a large amount of heat in air with no visible flame (even in the dark, as the standard requires) and it might even qualify as an ignition event to you or I or any other sensible person watching, but it would not fit the standard.  They have to set some limit, after all, and that was what they set.
My original thinking on this was flawed; I misunderstood the air temperature reference.  The reason that they mention the air at all is that they are being specific that the reaction mixture should be an air/fuel mixture, rather than something else (oxygen/fuel, for instance.)  The predetermined temperature aspect of that point has to do with the AIT determination procedure itself.  They are setting the temperature of the system to some value T, then adding the fuel, then waiting for ignition.  So yes, the air in the flask is more or less in thermal equilibrium with the substance being tested.  It can, however, leak out and exchange with the air outside the flask. If the substance does not autoignite, the flask is purged and the temperature increased (including the air in the flask.)  I'm not sure whether they want a completely new sample of the substance for each temperature measurement.  It seems likely, since we're dealing with small amounts of generally volatile compounds.


How can inconsistencies in these sodium chloride solutions be explained? I'm entirely chemically-naive and am confounded when trying to convert $\ce{NaCl}$ brine concentrations to moles per litre.

The situation is that I'm performing linear multiple regressions using published data form multiple studies. Because the studies report their brine concentrations using different units I am trying to convert all of the concentrations into molarity (moles per litre). To do that I'm using the conversion tool at EndMemo - here's a screen shot:

But in more than one place in the literature people report different M values ... for example, below is an extract from one of the studies. Their M values don't match what I'm getting using the online converter (for example, 4 g/L $\ce{NaCl}$ is coming out as 0.068 M/L...hopefully I'm missing something very basic here?)

(Broseta D., Tonnet N., and Shah V., Geofluids, 2012, 12 (4), 280-294.)

As an aside, I'm not finding consistency with the unit symbols for Molarity and litre (sometimes they are capitalised, sometimes they are not...are they referring to different things or is this another example of inconsistency)?

example, 4 g/L $\ce{NaCl}$ is coming out as 0.68 M/L...hopefully I'm missing something very basic here

Yes, you are off by a factor of 10.

4 g/L $\ce{NaCl}$ is 0.068M (0.068 moles per liter of solution).

As far as the journal article, I would used the 4 grams per liter value (and other grams per liter values) and convert to moles per liter properly and not rely on the 0.08 (small capital "M" l-1) value.

If the value is critical to your research, email the corresponding author of the paper and and ask about it in a polite way.


Hardened polyurethane remover I've looked at the MSDS of a couple of products for removing hardened polyurethane (PU) foam. Both contained only ethanolamine as a dangerous substance, but only between 1-5%. So I'm guessing this might not be the active substance. The Wikipedia page on polyurethane doesn't indicate any degraders (and the page on polymer degradation doesn't mention polyurethane), so what (active) substances are typically used to degrade hardened polyurethane in commercial cleaners?

I also found that some metal acetate-ethanolamine complexes can be used as catalysts for preparing PU foam. So possibly ethanolamine is the active substance or just a catalyst for removal as well. Can someone shed some light on the purpose of ethanolamine?

Example for one of the products I was looking at "Soudal PU remover SDS"; there's also a consumer leaflet for that one which might have additional hints. It's "very active", "odourless", "free of aromatic hydrocarbons", pH 10.5.

The one solvent I see referenced the most for polyurethane foams is N-methyl-2-pyrrolidone or as the the cool kids call it: NMP. You don't have to just take my word for it a certain supplier that I have no affiliation with sells it for this purpose: Spray Foam Systems NMP.

As far as the purpose of ethanolamine, it must either be reacting to reduce chain length.  If you look at the urethane group below the Carbon bonded to two oxygens and a nitrogen is susceptible to attack by the basic ethanolamine. This reaction would break the molecular bonds making the polymer chains shorter and thus more mobile and more easily dissolved, thus saving time. This explination is a little hand-wavy but I didn't want to go deep into organic reaction mechanisms.


Alloys for inkless metal pen What alloys can be used to make a metal pencil like this one?

"Silverpoint" nibs have a long history in art and significantly predate modern drawing tools such as the lead pencil (made of graphite and clay but possibly called lead as lead metal was one possible material in metal pens). Silverpoint writing and drawing doesn't smudge or fade and many examples have survived almost unchanged over hundreds of years.

There is a useful site with a good history of the original artistic uses of silverpoint pens here. According to this site a wide variety of metals will work. Historically which was used depended on status:

  Wealthy and royal persons who needed to keep records (business inventories and so forth) employed these scribes, and the stylus material served as notice to all of the status of the employer - gold and silver for the upper classes, lead for somewhat lesser beings.

It seems tht many metals will work. The dictionary of Art is quoted as saying:

  Metalpoint. Drawing instrument (the forerunner of the pencil) made from a small, pointed metal tip, usually of lead, silver, copper or gold, encased in a wooden holder.

The current preferred metal for artistic use seems to be fairly pure silver.

  Best results are obtained from a thicker point that is "pure" silver, or ".999 fine." This is much softer than sterling, so it will deposit more metal on the drawing surface

The silverpoint.com web site sells ready made points of both silver and gold.

It isn't obvious what metal or alloy is used in the pens mentioned in the Wired story, but it isn't silver. It apparently contains at least some lead as the pencils come with warnings not to let children chew them!


Why is high molar molar mass a requirement of good primary standards for volumetric analysis? I have been taught that using primary standards with a relatively high molar mass will reduce the error between the equivalence point and the end point during titration. However, I am unsure as to why this is the case.

It would be great if someone could point out to me what the purpose of using primary standards with a high molecular mass is.

This practice is done in analytical chemistry in order to minimize the relative weighing error on the balance. Preferring a larger formula weight for a primary standard has nothing to do with impurity levels. We have to start with the highest purity standard.

For example, you wish to prepare a 0.010 M solution of oxalic acid dihydrate in 1 L flask. Its formula weight is  126.07 g/mol. You need to weigh out 1.2607 grams of the acid on the balance. Imagine the balance has a small constant error, the relative error in weighing is smaller for a larger formula weight compound as compared to a smaller formula weight standard for the same concentration. Imagine if the molecular weight of oxalic acid were 60 g/mol, how much do you need to weigh out?


Baking Soda + Oil at high temps My mom while making Potato chips put some Baking Soda in Oil with it. Just curious to know, it was first time Baking Soda in Oil at that high temperature. Is it safe to eat?

Answer ASAP before my chips get cold.

Baking soda is essentially sodium hydrogen carbonate, which should not be expected to react with oil because of its poor solubility. $\ce{NaHCO3}$ melts at 50 °C and starts to decompose slowly to sodium carbonate, carbon dioxide and water.

$$\ce{2NaHCO3 ->~ Na2CO3 + H2O + CO2\uparrow}$$

By the release of carbon dioxide, $\ce{NaHCO3}$ acts as a raising agent when used in baking. Under the conditions of deep frying (150-200 °C), the decomposition is fast as the reaction is entropically favored (generation of gases). $\ce{CO2}$ and $\ce{H2O}$ escape from the baking mixture, leaving sodium carbonate behind.

The ingestion of small amounts of $\ce{Na2CO3}$, which are generated from typically used amounts of baking soda, should not be a health problem. Acidic ingredients in the baking mixture will convert sodium carbonate into the sodium salts of the respective acids, carbon dioxide and water. Higher amounts of sodium carbonate, stemming from the use of too much baking soda, would cause an unpleasant metallic taste of the baked food. Sodium carbonate is actually used as a food additive (E 500) and in toothpastes as a foaming agent and abrasive.


Gel filtration separation of cellulose, starch, oxytocin and palmitic acid I am studying biochemistry and I have been asked by my teacher to find a procedure for separating cellulose, starch, the peptide oxytocin (9 a.a) and palmitic acid ($\ce{C16}$) using gel filtration chromatography. I have some ideas about using Fehling's test to tell the cellulose from the starch and the peptide since they all react with the $\ce{Cu^{2+}}$ in the solution. And I'm thinking that the palmitic acid will not dissolve in the solution and can therefore be separated from the others.

However, I have a problem with how the gel filtration would work. Obviously cellulose and starch are much bigger than palmitic acid and oxytocin but, how could I possibly separate cellulose from starch using gel filtration??

Gel filtration: https://en.wikipedia.org/wiki/Size-exclusion_chromatography

I think by using a gradient mobile phase that changes by time. At the beginning, the mobile phase would have higher affinity to cellulose and lower to starch and then the gradient changes where it would have higher affinity to starch than cellulose.

Specifically, which mobile phase you would use and what gradient would take some time and research to find the answer but it think you get the idea


Why do people assume 28 oxygens in chlorite mineral formula? I have been trying to calculate chlorite mineral formulas from microprobe data. However, all of the mineral formula spreadsheets on the internet, including those from the Open University assume 28 oxygens (http://www.open.ac.uk/earth-research/tindle/AGT/AGT_Home_2010/Microprobe-2.html)

However, all the formula I have ever seen for chlorite only have 18 oxygens? (see DHZ or http://www.mindat.org/min-1016.html)

By then calculating my formula assuming 28 oxygens, it means I have easily more cations than the mineral structure would have sites for?

Many thanks for any help

It doesn't mean that there are 28 oxygen atoms.  It means a charge equivalent of 28 oxide anions, considering that each oxide anion has a -2 charge and each hydroxide anion has a -1 charge.

This is based on the chlorite having an formula including $\ce{O20(OH)16}$

For example see $\ce{Mg10Al2(Si6Al2)O20(OH)16}$ in The effect of Cr on chlorite stability and the appearance of a new Al-rich silicate

In otherwords, as explained in OCTAHEDRAL OCCUPANCY AND THE CHEMICAL
COMPOSITION OF DIAGENETIC (LOW-TEMPERATURE) CHLORITES

  All chlorite analyses were re-calculated as chlorite structural formulae on the basis of 28 oxygen equivalents (56 negative charges) i.e. assuming an ideal anion framework of $\ce{O20(OH)16}$


Why does not aqua regia burn filter paper? We were doing basic radical analysis in lab and we were asked to add aqua regia on the filtered PPT collected on the filter paper and collect the formed solution , my question is if aqua regia can dissolve even gold and platinum why didn't it burned the filter paper(or made holes in it )

For "burning" of paper by sulphuric acid, the better term would be "aggressive dehydration".

Note that aqua regia is rather a diluted acidic mixture without significant dehydrating or oxygen based oxidation behaviour, but manifests aggressive chloration behaviour, that dissolves precise metals.

Cellulose has quite good endurance against chlorine, which is used in various solutions for cellulose whitening.


Light sensitive AgF decomposition reaction Silver compounds such as silver fluoride are light sensitive (as are all other silver halides) so they will decompose when exposed to light.

I'm stuck with a problem : I want to expose silver(I)fluoride to light to then produce elemental silver and fluorine gas - I'm worried that because of the high reactivity of fluorine silver(I)fluoride will be formed again...

So how could I successfully split silver(I)fluoride by light into elemental silver and fluorine gas without the fluorine reacting again with the silver?

As discussed in the comments, the asker only wanted to create some fluorine in situ for use in another flask, with no desire to store it. Chemistry with fluorine is difficult due to its dangerously high reactivity. Do not attempt anything without the proper safety procedures.

I actually still have doubts whether it is possible to obtain fluorine gas reliably from the exposure of silver fluoride to sufficiently high energy photons, and if there is any simple way to produce photons with the required energy. It is well known that the other silver halides easily suffer photolysis, especially for the heavier halides, but reactions which produce fluorine gas tend to be extremely endergonic, and therefore don't happen except in rather severe situations.

In case it is somehow possible to produce fluorine gas directly by photolysis of silver fluoride, then it is possible to transport the gas to another container where it can react, simply by connecting the flasks with a pipe. Few materials are capable of withstanding exposure to fluorine, however, so the containers must all be made of very well dried gas. Unless the asker has access to a very high power light source, I suspect that the rate at which silver halides decompose is quite slow, which may preclude their efficient usage as halogen generators. This is especially true for silver fluoride, which likely decomposes the slowest, and the fluorine formed in very low concentrations could easily be lost by reaction with impurities or residual water before even reaching the target. I do not know whether there are any catalysts which can be added to increase the photolysis speed of silver fluoride. Presumably some semicondutor with a high bandgap, such as titanium dioxide, could absorb the light and cause photoreduction of the silver/photooxidation of fluoride, but the catalyst would be quickly consumed by the fluorine formed.

If fluorine gas can be formed by silver halide photolysis fast enough, then by simply linking it to another flask with the desired reactants should cause the fluorine to diffuse and react, followed by a continuous equilibrium shift in the production of more fluorine in the flask containing the silver fluoride, and its consumption in the flask with the other reagents.


Assigning labels and answering questions using given DSC graph I'm currently working through past exam papers in preparation for my upcoming Inorganic Chemistry exam and I got to this question. We really haven't looked at DSC graphs, so I'm not completely sure how to answer this. I've shown my attempt, but am looking for someone to confirm if I'm right, if not, where I went wrong.

Question:
The DSC thermogram depicted below was run on a sample of a new compound suspected of possessing mesogenic properties. The thermogram was obtained at a scan rate of 10 °C min−1 and the solid line represents the heating scan, while the dashed line represents the cooling scan.

Interpret the DSC thermogram, naming and explaining the processes occurring at peaks (a) – (e), using the correct terminology.
Does the DSC thermogram confirm that the new compound is a mesogen? If so, what can you say about the liquid crystal phase(s), and what name is given to this class of mesogen? What word describes this particular kind of thermal behaviour?

My answer: e = Crystallization, a = melting point (crystalline -> smectic), b = transition point (smectic -> nematic), d = (re)crystallization (nematic -> crystalline), c = melting point (crystalline -> nematic). Since a nematic phase is a subset of liquid crystal (synonym for mesogens), it confirms that the new compound is a mesogen. Molecules in the nematic phase are (somewhat) orientationally ordered (in contrast to the smectic phase which is (somewhat) orientationally and positionally ordered). I don't have an answer to "What word describes this particular kind of thermal behaviour?".

As example, Skoog / Leary's principles in instrumental analysis contained a section about DSC in the chapter of thermoanalysis.  Your university library likely has some of them or other books about the topic, too.

Then, beside monitoring glas transitions, melting, decomposition (while heating) or solidification / crystallisation (while cooling) I suggest to attribute the other phase transitions with (in the best case: simultaneous) visual observation of the sample (heat stage microscopy, like here) at these critical temperatures.  As these transitions take time, from a practical perspective, a heating rate of $\pu{10 K/min}$ is maybe rather quick.

After reading "liquid crystals" in the title, I thought a cholestic phase would be mentioned, too, but if your data support its absence, ok.  On the other hand, one of PerkinElmer's brochures includes this representation

(source)

which might be useful to complement the structural description of your liquid crystals.


Why does adding salt to water reduce the pH? I recently heard someone claim that adding table salt to vinegar caused HCl to form which helped them clean coins - clearly not the case - however I thought I'd disprove it by doing an experiment.

So I added NaCl to ethanoic acid and measured a 0.5 reduction (3 to 2.5) in pH using a pH probe. This seemed a bit weird so I repeated using tap water and the same happened (7 to 6.5).

The pH of water when NaCl is added should not change from 7!

The probe measures by detecting a change in voltage through the solution so it could be that, but you'd think that salts shouldn't have such an effect or be adjusted for. Can anyone repeat please?

What's going on?

This is probably an effect in which sodium ions from the salt are being weakly detected by the glass electrode, giving the appearance of a lower pH. Try add the same concentration of salt to samples of your reference buffers and recalibrate your pH meter to those before measuring the pH of the salt water. I would expect you would see pH 7. Also, keep in mind that the pH of unbuffered water is very sensitive to low concentrations of certain impurities -- from the air or from the added salt. For example, carbon dioxide in the air will lower the pH over time.


Is methanol optically active? Strictly on theoretical basis, not based on experimental results.

I can very well be wrong, but I think that methanol should be
optically inactive because it has a plane of symmetry passing through it.
My teacher however insists that it is optically active as you can create two non-superimposable mirror images out of it.

Who is correct and why?

Just to sum up what has already been pointed out in the comments: Methanol can have infinite conformations in free space due to free rotation of the C-H and C-OH sigma bonds. Hence, for any particular spatial conformation you can think of for the molecule, it can always develop a perfect mirror image for the same almost instantly(in case you are wondering about some time lag between them) as conformers are in chemical equilibrium.

So what this essentially means is that, suppose some conformation A of methanol has polarized light in some direction (assuming the right-handed coordinate system, let's say the positive X direction) in some fashion(let's say circularly polarized, clockwise). Then, this particular interference pattern of light will be internally compensated by a conformation B of the molecule which will circularly polarize the light in the anticlockwise direction in the negative X direction. So, the emergent light ray will appear to be unpolarized for the observer


Why isn't just a fully occupied valence s orbital stable? Why arent elements like calcium with an electron configuration of $[\text{noble gas}]\ n\mathrm{s}^2$ stable, although all the populated orbitals are fuully filled? Why is it necessary to obtain an octet configuration instead? (Ignoring the trivial cases of hydrogen and helium)

Elements with just 1 or 2 electrons in the most outer $\ce{s}$ orbitals, like $\ce{K}$ or $\ce{Ca}$, have very well screened off the central charge of the kernel. As consequence, the effective charge of the kernel and related energy needed to ionize the atom are relatively low.
OTOH, elements with just few missing electrons in the most outer $\ce{p}$ orbital, like $\ce{O}$ or $\ce{Cl}$, do not have well screened off the central kernel charge, which is relatively high, what leads to considerable electron affinity to some extra electrons.
Combining these 2 kinds of elements leads to energy release, as energy spent by ionization of the former is  compensated by energy release by capturing electrons by the latter and by a covalent contribution to the element bond.
In case of solid ionic compounds, there is additional large energy release du the ionic lattice energy.


Compounds that release carbon dioxide What are some compounds that release carbon dioxide when heated, hydrated, or reacted with sodium, aluminum, silicon, or iron?

A solvent such as water would be required and carbonate or bicarbonate salts will release CO2(g) when reacted with most acids.

Dry ice will turn to CO2(g) when exposed to room temperature.

These are only a couple of examples as others have stated there could be many other methods.


Structure of the compound C9H7O2Cl? A compound having formula $\ce{C9H7O2Cl}$   (O = oxygen) exists predominantly in enol form (A) and also in keto form (B). On oxidation with $\ce{KMnO4}$ it gives m-chlorobenzoic acid as one of the products. Identify (A) and (B).
(subjective question; no options were given)

I tried it and found the double bond equivalent as 6, 4 of which were utilized in benzene ring. Also, the benzylic carbon must have at least one hydrogen. Therefore I determined the following structure as (A):

But my book suggests (A) should be:

Please explain whether both are correct or only one product will exist.
Source: Joint Entrance Exam (JEE) 2003 India

Both the answers are correct but the most appropriate will be B as the chelation occurs because of hydrogen bonding which makes it more stable.
H-bonding is  possible in your structure but the ring is less stable
.