How does an anode metal come into solution if it cannot react with an electrolyte? From what I understand, an anode material is chosen so that it does not react with the anode electrolyte solution.

I was just wondering, then, that when you have a galvanic cell which has a Mg metal anode submerged in sodium chloride solution, how the magnesium metal gives two of its electrons and come into solution (as magnesium ion) even though it does not react with (and cannot react with) sodium chloride?

What I mean is, in a typical single displacement reaction, there is a metal that reacts with the solution and comes into solution. For example, with $\ce{Zn}$ and copper sulfate, the zinc replaces the copper and comes into solution as zinc ion. However, magnesium doesn't react with sodium chloride (because the activity series predicts it can't). How then can we say that we have magnesium ion in solution?

Is it that the metal is more stable in its ion state and thus readily gives up its electrons?

Traditionally, the anion is the one responsible for giving electrons: the chloride ion in the sodium chloride solution should come into contact with magnesium anode and give its electrons, which the magnesium metal should then carry. And thus the reaction should be

$$
\ce{ 2Cl- \!(aq) -> Cl2\!(g) + 2e-}
$$

and not

$$
\ce{Mg\!(s) -> Mg^2+\!(aq) + 2 e-}
$$

However, my books don't say this: they say that magnesium gives up two of its electrons and enters the solution.

What am I understanding wrong?

I was just wondering then that when you have a galvanic cell which has a $\ce{Mg}$ metal anode submerged in sodium chloride solution, how the magnesium metal gives two of its electrons and come into solution (as magnesium ion) even though it does not react with (and cannot react with) sodium chloride?

Your misunderstanding stems from assuming that the electrode reaction has to be the same sort of single-displacement redox reaction that would occur with no electrodes present at all.

At a dissolving magnesium anode$^\dagger$, the only reaction that needs to occur is the one you wrote involving magnesium:

$$
\ce{Mg\!(s) -> Mg^2+\!(aq) + 2e-}
$$

The potential applied to the electrode pair is what enables the above reaction to occur without a paired reduction reaction at the anode surface.  Instead of being taken up in another chemical reaction, the electrons are drawn into the anode and on into the connected electrical circuit. Those electrons are then "pushed" into the cathodic reaction, whatever it might be.

  From what I understand, anode material is chosen so that it does not react with the anode electrolyte solution.

This is true, but does not have to strictly hold. MaxW is right: $\ce{Mg}$ does react directly with water. But, the reaction is slow enough that it's perfectly safe, and that one can conduct useful electrochemical operations even while it's going on (e.g., see here and here).

  Is it that the metal is more stable in its ion state and thus readily gives up its electrons?

Despite what one reads from a table of standard reduction potentials or a metal reactivity series, in practical electrochemistry, the above is true for most metals, except for the noblest of the activity series. In my experience, it's usually a lot easier to dissolve metals than to electrodeposit them.  You may have to use additional chemistry to make it happen (e.g., including $\ce{HF}$$^\ddagger$ to strip the passivating oxide from $\ce{Ti}$, $\ce{Nb}$, $\ce{Ta}$, etc.), but once the metals are oxidized/dissolved they tend to stay that way.

 $^\dagger$ Operating at $100\%$ faradaic efficiency. Other reactions that might occur in this system, depending on the applied potential, include the chloride oxidation reaction you noted and water electrolysis to oxygen gas.
 $^\ddagger$ $\ce{HF}$ is very dangerous, do not use at home!


The pentagonal pyramidal structure of XeOF5- Why does the $\ce{XeF6}$ molecule have an octahedral structure while $\ce{XeOF5-}$ has a pentagonal pyramidal structure, despite both molecules having 7 electron pairs? I suppose the extra electron on oxygen would distort the molecule somewhat, but this seems like an extreme outcome.

$\ce{XeF6}$ is not octahedral, but distorted (or capped octahedral), one of the possible arrangement for seven electron pairs.

Wikipedia provides the following, quoted from a paper by Konrad Seppelt in Accounts of Chemical Research[1]:

  ...the structure is best described in terms of a mobile electron pair that moves over the faces and edges of the octahedron and thus distorts it in a dynamic manner...

The molecular geometry of $\ce{XeF6}$ is thus not truly an octahedron, nor is it static. In fact, the referenced article by Seppelt indicates that $\ce{XeF6}$ is only monomeric in the gas phase. In solution and the solid phase it is tetrameric $\ce{Xe4F24}$ clusters. The solid phase appears to be more like clusters of $\ce{[XeF5+ F^{-}]}$ with the fluoride ions bridging two $\ce{XeF5+}$ centers. Thus, in the solid, every $\ce{Xe}$ atom has 7 bonds to fluorine, with two that are more ionic. The structure for this scenario looks like the pentagonal pyramid.

Reference:

  Seppelt, Konrad (June 1979). "Recent Developments in the Chemistry of Some Electronegative Elements". Accounts of Chemical Research 12 (6): 211–216. doi:10.1021/ar50138a004


Relative rate of solvolysis I underestimated this question and just compared the carbocation stability of all the compounds (after removal of chloride via $S_N1$) and got the order D>C>B>A... but my answer is incorrect, what am I missing out?

The rate of solvolysis is highest in (C) as a 6 membered ring is formed which is more stable than 4 & 5 membered rings.

And (B) more readily formed than (D) because the NGP ability of oxygen is more than that of an alkene(due to high charge density of oxygen).

So the correct order is C>B>D>A.


High temperature bath chemicals I have an electric mantle that I want to bathe reaction vessels at temperatures up to $\mathrm{750^{\circ}C}$.  Obviously that rules out oils.  What substances are used as bath media at such temperatures?

I was thinking of using lead or Rose's alloy, but wondering if there are more standard high-temperature bath chemicals?  I suppose an ideal one would be both inert to lab steel, glass, and a standard atmosphere, and liquid down to room temperature.

Different types of solder can give a very high operating temperature, are cheap and offer excellent conduction and heat contact when molten. We have used lead and cadmium free solder to 500 degrees C with no problem. Your issue may be the melting point (ours was 230C), but the right mix may bring this down for you. If you have the time and inclination, you could modify a commercial mix yourself. Otherwise,  sand would get my vote.


Do endothermic reaction have higher energy products than reactants? If a reaction absorbs energy does  that mean it has a net gain of energy?

The simple answer is yes - as energy is required to be absorbed for the reaction, the products have more energy than the reactants.


Iodine medical products After an injury or getting a scare it is typical to use medical products containing iodine. They typically leave a huge red spot on the skin.

Why are creams and solutions containing iodine used in the medicine?

This Wikipedia article gives a good, if brief, discussion of a slow-release form of iodine called "povidone-iodine" in which iodine is the only active ingredient.

  Free iodine, slowly liberated from the povidone-iodine (PVP-I) complex
  in solution, kills eukaryotic or prokaryotic cells through iodination
  of lipids and oxidation of cytoplasmic and membrane compounds. This
  agent exhibits a broad range of microbicidal activity against
  bacteria, fungi, protozoa, and viruses. Slow release of iodine from
  the PVP-I complex in solution minimizes iodine toxicity towards
  mammalian cells.

  PVP-I can be loaded into hydrogels, which can be based on
  carboxymethyl cellulose (CMC), poly(vinyl alcohol) (PVA), and gelatin,
  or on crosslinked polyacrylamide. These hydrogels can be used for
  wound dressing. The rate of release of the iodine in the PVP-I is
  heavily dependent on the hydrogel composition: it increases with more
  CMC/PVA and decreases with more gelatin.


How can I get big bubbles when I electrolyse water with detergent? I've made carbon electrodes by putting 12V through a couple of short 15Ω pencils, but when I try to make hydrogen and oxygen bubbles from soapy water, I get tiny bubbles, smaller than table salt.
If I move the electrodes closer together, it's the same tiny bubbles, but a bit faster.
These don't catch fire very convincingly, and my daughter is unimpressed.
How can I get larger bubbles? Different electrodes? Higher voltage? Just shove the AC power in straight from the wall? :)

This might be tricky. The demos I've seen mostly rely on collecting the hydrogen and oxygen, separately or together, and then bubbling them through a soap solution.
If you've got the soap in the electrolyte, you might have a hard time getting anything more than two piles of foam, one of which is slightly flammable. I doubt the hydrogen foam would even rise in air; the volume percentage of gas is too low.


Why doesn't a catalyst shift the position of equlibrium, when temperature does? Maybe this is a silly question, but it's not one I've been able to answer to my own satisfaction. Increasing the temperature of a reaction favours the endothermic direction, because it makes it relatively more likely for particles to collide with enough energy to make the activation energy. Yes?

But doesn't lowering the activation energy have almost the same effect on the relative likelihood of particles colliding at least that energetically? Someone on a related question about whether catalysts shift the position of equilibrium said:

  The energy difference between the reactants and products is unchanged
  by catalysis

But isn't that equally true for increasing the temperature?

Your presumption in the first paragraph is incorrect. The fact that more particles collide with enough energy to make the activation energy with an increase in temperature means that the rate of reaction increases as temperature goes up, but that doesn't affect the equilibrium of the reaction.

Instead, the effect on temperature has to do with the comparative free energy of the two sides of the equilibrium. The controlling factor in where the equilibrium lies is the difference in free energy between the reactants and products (the ΔG). In fact, the equilibrium constant has a very simple relationship to the free energy: $\ce{\Delta G = -RT ln K }$ ... or $\ce{K = e^{-\frac{\Delta G}{RT} }}$. The ΔG is made up of two components - the enthalpy and the entropy: $\ce{\Delta G = \Delta H -T\Delta S}$, where T is the (absolute) temperature. The side of the reaction which is favored is the one with the lowest (most negative) free energy.

So when we substitute, we get the following relationship for the temperature dependence of the equilibrium constant, in terms of entropy and enthalpy:

$$ K = e^{-\frac{\Delta H}{RT} } \cdot e^{\frac{\Delta S}{R} } $$

The temperature dependence is only in the enthalpy ($\ce{\Delta H}$). Enthalpy, by the way, is what we normally think of as endothermic/exothermic. Roughly, reactions with a positive $\ce{\Delta H}$ absorb heat and are thus endothermic, and those with a negative $\ce{\Delta H}$ give off heat and are thus exothermic.

So imagine we have an endothermic reaction (positive $\ce{\Delta H}$), which means that the exponent for enthalpy in the above equation is negative. When we increase the temperature, that exponent gets smaller in magnitude (closer to zero), which means it becomes less negative. This causes an increase in $e^x$, which means that K gets bigger, which means the equilibrium shifts in the forward (endothermic) reaction. Likewise, if you have the reaction written in the exothermic direction, $\ce{\Delta H}$ is negative, the exponent is positive, an increase in temperature causes the exponent to become less positive, causing a decrease in the $e^x$, which results in a decrease of K -- the equilibrium now shifts in the reverse (endothermic) direction.

So while the statement that increasing temperatures means that the equilibrium shifts in the endothermic direction is true, it has nothing to do with the activation energy of the reaction. The arguments above are from a purely thermodynamic standpoint, and don't involve any consideration about the mechanism of reaction or the activation energy.


Stability between benzyl free radical and tertiary free radical My teacher told me that the hyperconjugation of 9 alpha hydrogen domintes over resonance (for a carbocation )and so the stability of a tertiary carbocation is more than benzyl carbocations due to exceptional hyperconjugation .

I was convinced to the fact and started applying to the questions but then i got stuck and was confused when i saw the stability order of the following groups of free radical .
According to me the order told by my teacher for carbocation should still be valid for free radical .but in the following picture benzyl free radical is ahead of tertiary free radical in terms of stability.

Yes correct....in case of t-butyl cation it is more stable than benzylic cation due to 9 alpha hydrogen hyperconjugation, 3 methyl gropu inductive effect...it is proved by hydride ion affinity that t-butyl cation have ∆H -231 kcal/mol while benzylic cation have ∆H -234 kcal/mol...if talk about any other tertiary cation excpt t-butyl cation then benzylic cation have more stability....and it is the same case in free redical


Thermodynamics: Combining the 1st and 2nd Law I'm confused how combining the first and second laws can derive

$$dS = \left(\frac{C_p}{T}\right)dT+ \left[ \left( \frac{1}{T} \right) \left\{ \left( \frac{\partial H}{\partial P} \right)_T -V \right\} \right] dP$$

The 1st Law states $dU = dq + dw$, and the 2nd Law states $dS = \frac{dq_{rev}} {T}$

I factored out $\frac{1}{T}$ in the given equation to make

$$dS = \frac{1}{T} \left[ \left( C_p\, dT \right) + { \left( \frac{\partial H}{\partial P} \right)_TdP - V  dP } \right]$$

$C_p = \frac{dq}{dT}$

Therefore,
$$dS = \frac{1}{T} \left[  dq + { \left( \frac{\partial H}{\partial P} \right)_TdP - V  dP } \right] $$

Without analyzing $\left\{ \left( \frac{\partial H}{\partial P} \right)_T - V  dP \right\}$, I already have $dS = \frac{dq}{T}$.

Can anyone explain this for me?

Thanks,

Starting with $$dH=TdS+VdP\tag{1}$$H is a function of T and P, so$$dH=\left(\frac{\partial H}{\partial T}\right)_PdT+\left(\frac{\partial H}{\partial P}\right)_TdP\tag{2}$$Substituting Eqn. 2 into Eqn. 1,
$$\left(\frac{\partial H}{\partial T}\right)_PdT+\left(\frac{\partial H}{\partial P}\right)_TdP=TdS+VdP\tag{3}$$But, from the definition of $C_p$, $$C_p\equiv \left(\frac{\partial H}{\partial T}\right)_P \tag{4}$$Substituting Eqn. 4 into Eqn. 3 gives:$$C_pdT+\left(\frac{\partial H}{\partial P}\right)_TdP=TdS+VdP\tag{5}$$Solving for dS then gives the desired relationship.


Understanding Joule-Thomson expansion apparatus I am reading the Joule-Thomson experiment and although I can follow the mathematics I can't understand what is going on. Here is the apparatus

taken from Atkins. So initial the gas is at $P_i$ and because of the opening and the difference in pressure it will eventually reach the right compartment with $P_f$. What I can't understand is what the role of porous plug and how this setup differs from an adiabatic expansion. I mean couldn't we let the the gas expand from $P_i$ to $P_f$ and then record the difference in temperature? Also what is the role of porous plug in this setup if the system doesn't exchange heat and the work is done on the environment in right and left compartment respectively.

The purpose of the porous plug is to cause a viscous frictional pressure drop, resulting from the gas flow through the tiny interstitial pores of the porous medium.  This results in a dissipation of mechanical energy (pressure and kinetic energy) to internal energy, which translates into a "viscous heating" to essentially offset the adiabatic expansion cooling of the gas.  The net result is no change in temperature for an ideal gas, and small change in temperature for a real gas.


Can only gases by diatomic? I know that the elements H, N, F, O, I, Cl, and Br are found naturally as diatomic gases. But can any other elements be found as diatomic gases, or even solids or liquids.

Why or why not?

Any molecule composed of exactly two atoms is diatomic, but I will answer this question assuming that you mean elemental diatomic molecules. Only H, N, O, F and Cl are diatomic elemental gases under STP. Br and I are diatomic liquid and solid respectively, under STP. Oxygen remains diatomic in liquid state, while nitrogen and all halogens(F, Cl, Br, I) are diatomic in all three classical physical states. Gaseous sulfur can be diatomic at or below standard pressure. Structurally similar to diatomic nitrogen, diphosphorus is an unstable allotrope of phosphorus. Diatomic carbon is a gas that exists under very high temperature. There may also be other diatomic allotropes.
Some elements form stable diatomic molecules because of the octet rule (or duplet for hydrogen). Generally a bond can share up to three pairs of electrons, so elements which are three or less electrons short of a stable electron configuration can pair up and share up to three electrons to become stable. Beyond that, elements will find other ways to become stable.
However, chemical stability isn't governed only by the octet rule, there are other factors such as geometry, bond length and nucleophilicity/electrophilicity that affect molecular stability. Therefore, some diatomic allotropes are not the most stable allotropes of the elements. Examples from above are disulfur, diphosphorus and dicarbon.


Do ion-exchange resins soften water through a kinetic equilibrium process? I searched a lot for this and couldn't find any answers.  Is water softening a purely ratio-based equilibrium process, ie, more $\ce{Na+}$ cations on resin than there are $\ce{Ca^2+}$ cations in the water, causing most $\ce{Ca^2+}$ in solution to be displaced? Or do kinetic factors come into play, where at equilibrium $\ce{Ca^2+}$ cations will end up stuck on the resin over $\ce{Na+}$ cations because they are bigger and thus more restricted from leaving the resin due to steric reasons?

(1) Water softeners use an ion exchange material that is reversible, so that after it has been used a while and the ion exchanger becomes loaded with calcium ions, it can be regenerated to the sodium form by flushing with salt (sodium chloride) solution. This would suggest that the affinity of sodium and calcium ions for the resin do not differ that greatly. I would doubt that either kinetic or steric factors count much here either, although a divalent ion like calcium might bind more tightly by being held electrostatically by more than one anionic group on the resin.

(2) I think the primary reason water softeners work is that they typically use the ion exchange resin in the form of a vertical column: you put the hard water in the top and soft water exits at the bottom. When hard water is introduced at the top of the column, a local equilibrium is established between ions in solution and ions bound to the resin, with a somewhat lower concentration of calcium ions and a corresponding higher concentration (at a 2:1 ratio) of sodium ions. As the water moves down the column, this equilibrium is reestablished again, but this time, the concentration of calcium is lower still. This repeats a large number of times, until all the calcium ions are replaced with sodium ions. I have heard this effect referred to as "mass action".


How can the chlorophyll concentration of plants be measured? Is there a way to measure the chlorophyll content of plants (from e.g. leaves) with basic chemistry equipment (at home)?

To measure the total chlorophyll from leaves, probably not, since you need access to methyl isobutyl ether or pyridine.

But you can separate the different pigments in the leaf. From The Naked Scientist: Cut a rectangle 1 cm wide and long enough to stand up in a small jar from a coffee filter. Place your leaf over the top of the strip and roll the coin across the strip about 2 cm from the bottom. Now add a little nail varnish remover to the bottom of your jar, half a cm is plenty. Suspend your strip of paper so the bottom end is sticking a few mm into the acetone. you can hold it up by folding the top end of the strip over forming a hook which will rest over the lip of the jar. Wait a few minutes, with any luck you should see an interesting effect. Try the experiment with a strongly colored leaf like a purple one.


Why is O2 (Oxygen) colorless and odorless while O3 (Ozone) is a pale blue gas with a distinctively pungent smell? Why does adding one more oxygen molecule to the bond make it smelly and have color?!

Our perception of colour and smell involve different chemistry so there is no reason why they should be related. For example, what ever the chemistry of smell is (and its not certain what this is) it clearly does not involving photons and subsequent isomerisation of a rhodopsin.

Ozone is slightly coloured as it has two low lying electronically excited states (Chappius bands 550 to 600 nm) and a second state (at approx 700 to 1000 nm) both of whose energy gaps from the ground state correspond to a wavelength in the visible spectrum. The combined absorption is rather broad, weak and peaks at approx 600 nm which in the red part of the spectrum, thus ozone appears blue. Oxygen also has two low lying electronic states at around 700 nm and 12000 nm but are very weak and the longer wavelength transition is beyond the range of our eyes to detect it (liquid oxygen appears blue as does liquid ozone). Both O$_2$ and O$_3$ have electronic states in the ultra violet at a wavelength beyond the range of our eyes.

( I have added a moderately detailed description as to why high pressure gas and liquid oxygen molecules are coloured. see Does O2 have a color in the gas phase )


Should data tables be made in pen or pencil? For a chemistry lab, I have to make a data table. Should data tables be written in pen or pencil? What is done in the real world of Chemists and why is it done in such a way?

In the real world, chemists have a lab journal to document what they did; Good Scientific Practice guidelines state that results should be written there immediately, readably, non-removably, directly and a few other adverbs. ‘Great’ scientific practice or patent-safe documentation would require a witness.

In universities, I have heard of a trend of more and more lab journals being replaced by computer files. In the labs I have worked in, this was not the case and the paper lab journal remained the way to go. Apparantly, if you want to document something in a patent-safe manner, a paper book is mandatory.

Since the entries into a lab journal must not be removable, the only option you have is ink; usually, ball-point pens (biros) are used.

However, the wording of your question (‘chemistry lab’ and ‘data table’) makes me think you could still be a student. In that case, the only advice you should take is that of your teacher/advisor/…


sp2 hybridisation of alkyl radicals causing formation of racemic mixture What I know about $\ce{^{.}CR3}$ radical is that it has both $\ce{sp^2}$ and  $\ce{sp^3}$ character but the $\ce{sp^2}$ character dominates and so the radical is $\ce{sp^2}$ in nature.
So, suppose we have the above situation. In this, a radical is formed  at  the carbon of the 6 membered ring at which one $\ce{-CH3}$ is attached since the bromination is selective, but will it form enantiomers? That is, can the $\ce{^{.}Br}$ radical attack from both sides?
In this question, the answer given is (c), but I think that enantiomers can not form, as the electron of the radical would be present only on one side of the radical carbon, and so cannot migrate to other side.

In the first step, due to the selectivity of bromine in a free radical reaction, the tertiary carbon which has a $\ce{-CH3}$ group attached to it forms the radical.
Now, since the radical is a $\ce{^{.}CR3}$ radical, the carbon radical in the intermediate becomes $\ce{sp^2}$. This means that $\ce{^{.}Br}$ can attack from both the top and the bottom since this lone electron in the $\ce{^{.}CR3}$ radical is in a p-orbital. This is because the electron is equally likely to be found in both lobes of the orbital.
Now after the attack of the $\ce{^{.}Br}$ radical, you get two products. The first one shows the product formed when $\ce{^{.}Br}$ attacks from the bottom and the second one shows the product formed after the $\ce{^{.}Br}$ attacks the intermediate from the top.
You may have noticed that both products are enantiomers of each other and hence the answer would be (c).
Reference
Chem Libretexts


How can 32 g of Gatorade powder contain 33 g of carbohydrates? A screenshot of the nutrition facts of a can of Gatorade powder was posted on my Facebook feed:

The label reads "Per 32 g (about 3 tbsp)", but if you read through the ingredients, the amount of carbohydrates listed exceed that of the serving size (33 g).

My question is, how can this possibly be? Is there a chemical process as some people have suggested that is changing the 32 g of powder into 33 g of carbs once water is added?

Re-asking it here for a chemistry perspective as some users have suggested on skeptics.SE.

Is there a chemical process as some people have suggested that is changing the 32g of powder into 33g of carbs once water is added?

No. Even if there was, it's not relevant in this case: the numbers on the label are an assay of the pure contents of the product before it's mixed with anything, or prepared or altered in any way. Nutrition labels refer to the product as-is.

In the United States, the Food and Drug Administration is in charge of labeling the quantities you're interested in. The relevant document is the Food Labeling Guide, one of many documents which are linked from a more comprehensive list called Labeling and Nutrition Guidance Documents and Regulatory Information. As per the comments (below), I'm noting that it appears as if the product pictured in your post is from the Canadian market and thus subject to nutritional labeling dictated by agencies in Canada. That said, the methodologies used by both the US and Canada are similar.

There are at least 2 regulations detailed in the labeling guidelines that we need to examine here to explain what's going on:

(1) The numbers reported on the container have simply been rounded up or down as per the Food and Drug Administration's Appendix H in the Food Labeling Guide. For instance, they show that for Total carbohydrate and Sugars, anything less than 0.5 g is expressed as 0 g, and quantities greater than or equal to 1 g are to be expressed in the nearest 1-gram increment. This is why of the 33 g of total carbohydrate, only 31 g appears to be accounted for as sugars;

and

(2) Total carbohydrate is calculated in a way that propagates rounding errors introduced as detailed in (1):

  Total carbohydrate is calculated by subtracting the weight of crude protein, total fat, moisture, and ash from the total weight of the sample of food.

In summary: The numbers don't add up because quantifying the components of a food to conform to the rules and regulations for reporting them on a Nutrition Facts label allows for the accumulation of rounding and measurement error.


What happens when an alkylborane is treated with acetic acid? In the hydroboration reactions of alkenes I've seen so far, they were always followed by oxidation with $\ce{H2O2/NaOH}$ (to yield alcohols).

Today I found a peculiar reaction in which acetic acid-d1 was added after hydroboration to give a deuterated product:

I don't quite understand what exactly happens after the hydroboration part. Can someone enlighten me here?

Acetic acid effects protonolysis of the borane:

If AcOD is used then RD (instead of RH) is formed.


Are phenyl and allyl groups in prop-2-enylbenzene coplanar? Wouldn't C1 and C2 in prop-2-enylbenzene be on the same plane as that of the phenyl ring, right? If the double bond connected to C2 was between it and C1, they would be, but is it correct that now they're both planar, but on different planes?

Probably not
Part of the problem with the question is geometry. Any two points always form a plane. What I suspect you mean is "is the allyl double bond in the same plane as the benzene ring?" in which case the answer is probably not.
To make the situation clearer here is a different numbering of the atoms in the allyl benzene molecule:

The question I think you intended is whether the 2* and 3* bond is in the same plane as the benzene ring.
There is no chemical or geometric reason why it would be. the 1-1* bond and the 1-2 bonds will freely rotate so there is no bond-related force driving the orientation of the 2-3 double bond. to illustrate here is a 3D view of one possible configuration:

I say one possible configuration as it is likely that the barrier to rotation of those two bonds is small and, at room temperature in the liquid, the molecule will explore all the possible configurations (it might be different in a solid crystal where packing forces between units might lead to one being favoured).
In short: there is no reason to suppose the double bond and the benzene ring will be in the same plane because the two single bonds connecting it to the ring will freely rotate.


Assign stereodiscriptor to 4th carbon of 3-ethyl-2,4-dimethly-5-(propan-2-yl)heptane-1,7-diol Here is the molecule. Clearly the 4th carbon is chiral. I have drawn the diagraph for it, but the exploratory pathways of the left and right chains are the same. So, the problem is that the two chains are actually different, but same in the diagraph. How to assign stereodiscriptors in such cases?

Not the proper, somewhat complicated, IUPAC ‘digraphs’… but might help.

Bold/colored bonds are the left and right branches' resulting paths taken when advancing to next sphere (because the decision could not be yet done) during determining a higher priority branch by finding a first difference of the compared atom set rankings. Sphere numbers a Roman. Grey digits are semi-conveniently arbitrarily chosen numbering just for reference, not nomenclature numbering nor priorities(!).

Following list shows chosen highest priority atoms in comparison in each nth sphere and left/right branch (bold; also comparison with other sibling atoms in the same branch, if they are not just hydrogen), chosen according to the next bonded atoms from the (n+1)th sphere (in 1st braces; atom number in 2nd brace).

sphere № : [left] vs [right]
I: [ C(CCH) (1) ]   =   [ C(CCH) (4) ]
II: [ C(CCH) (2) > C(CHH) (3) ]   =   [ C(CCH) (5) > C(CHH) (6) ]
III: [ C(HHH) (7) = C(HHH) (8) ]   <   [ C(OHH) (9) > C(HHH) (10) ]

Please note that in the II. sphere, left, the carbon (3) leading to terminal -OH group could not be chosen, because it's C(CHH), compared to higher priority, chosen C(CCH) (2).

Also note that in sphere III, left, any of the two methyls' carbons could be chosen (equivalent).

In this III. sphere, the decision finally could be made. The right branch (R1) has higher priority – finally only hydrogens bonded at left: C(HHH) (7), vs oxygen and hydrogens bonded at right: C(OHH) (9).
(Next priority is left branch R2, then methyl, then H.)

So the compound configuration/name is
(4S)-3-ethyl-2,4-dimethyl-5-(propan-2-yl)heptane-1,7-diol.


Discerning Sodium bicarbonate I have 'sodium bicarbonate' powder , which is supposed to be (NaHCO3) , is there a way for me to do a test on it at home , without some special equipment and determine whatever its 100% pure or it has some additives?

Without additives its a bit hard.

You can test easely if it must be it or not, if you put a bit of it on a flame you must see a yellow flame which means there is sodium $\ce{Na}$ in the compound. Like this.

And to be more sure if you pour some acid (like ethanoic acid or vinegar) it must makes gas. Now if you want to verify that this gas is carbon dioxyde you can try to do the reaction on a glass or whatever you want, light a candle and pour the gas (which is inside the glass) on the top of the flame, if the flame disappear it must be carbon dioxyde. Exactly like in this video obviously you don't use backing powder but your product..

You can also use a solution of limewhater but you musn't have this at your home...but if you have you must obtain this: (but you need to pour the gas only in your solution!) so its not really easy at home.

So if you do all this experiments you can say its $\ce{NaHCO3}$ but its impossible to be sure at 100% whitout doing an Infra-Red exams for example. It can be mass spectroscopy or RMN.. but obviously not at home, and you can detect all additives if they are.

Hope it will helps you a bit,
Friendly.

NB : If you are not sure don't use it for your food.. :)


Can somebody please explain this statement (about the greenhouse effect)? "Methane is a potent gas and over a period of 20 years has a 72 times greater effect on warming the atmosphere than carbon dioxide.
Due to my limited english, i am failing to understand how timeline has anything to do with the effect of a gas on the greenhouse effect. I think it has to do with the half life of the gas in the atmoshpere. Maybe methane is broken down more readily than CO2 and the whole thing means that if they remained in the atmosphere for the same amount of time, methane will do a greater damage. Am i right ?

"Methane has a large effect (24 times as strong as carbon dioxide per unit mole) for a brief period (having an estimated lifetime of 8.9±0.6 years in the atmosphere),[12] whereas carbon dioxide has a small effect for a long period (over 100 years). Because of this difference in effect and time period, the global warming potential of methane over a 20-year time period is 72."[ Wikipedia, quoting Jain, Briegleb et al ].

Simplified, that means:

$\ce{CH4}$ has a shorter half-life in the atmosphere than $\ce{CO2}$, reducing its effect as a green-house gas over time, but
$\ce{CH4}$ is so much more potent in retaining heat than $\ce{CO2}$ that this more than compensates for its short lifetime, making $\ce{CH4}$ a stronger green-house gas over the long term.


What reagents can be used to distinguish pentose and hexose sugars? I have been asked how I can differentiate D-glucose and D-xylose using the following reagents / reactions:

  Orcinol
  Benedict's test
  Ammoniacal silver nitrate
  $\ce{FeCl3}$
  $\ce{C6H5SO2Cl}$, excess $\ce{NaOH}$
  $\ce{NaOH}$, $\ce{H2O}$, heat
  t-Butyl chloride, $\ce{AlCl3}$
  $\ce{Br2}$ in water, dark
  $\ce{NH2OH.HCl}$, $\ce{KOH}$, $\ce{FeCl3}$
  Resorcinol
  2,4-DNPH
  Iodoform test

Usually, chemical tests identify certain functional groups in molecules which react in a particular way. However, here the two compounds only differ in the number of carbons and don't have any real difference in their functional groups, so what reaction can be used to differentiate them?

I'm not going to address all of the reagents, but I want to highlight a specific approach in this partial answer.

Occasionally, knowing some history of chemistry is quite useful when thinking about how to solve problems. In this specific case, I direct you to the work of Emil Fisher (chemistry Nobel 1902), who determined the structure of all hexose sugars. He did this by building up the sugars from smaller subunits while keeping track of symmetries.

This immediately suggests a solution where one performs reduction (or oxidation) of the aldehydes. In other words, make the both ends of each molecule look the same. This is a valid strategy because the underlying symmetries of glucose and xylose are different.

The penta-ol is meso and therefore achiral, while the hexa-ol is chiral and optically active since you started with one enantiomer.

Thus, any reagent that can produce the bis-acid or bis-aldehyde without epimerizing any stereocenters along the way or any reagent that can reduce the aldehyde to an alcohol will provide a suitable method of differentiation in this specific example.


How can the trityl tetrafluoroborate function as a Lewis-acid catalyst Trityl tetrafluoroborate is a reagent sometimes used in synthesis as a very mild Lewis acid catalyst, and recently I've been (unsuccessfully) using it in some protecting group chemistry.  Several other uses are mentioned in eROS:

  ...for various transformations, e.g. the Mukaiyama-type aldol reaction using a dithioacetal and silyl enol ether. It has also been used as the catalyst for the formation of glycosides from alcohols and sugar dimethylthiophosphinates and for the formation of disaccharides from a protected α-cyanoacetal of glucose and a 6-O-trityl hexose. Michael additions of various silyl nucleophiles to conjugated dithiolenium cations also proceed well. Finally, the [4 + 2] cycloaddition of cyclic dienes and oxygenated allyl cations has been effected with trityl fluoroborate.

  Ref: Encyclopaedia of Reagents for Organic Synthesis

Clearly, the trityl cation could act as a Lewis acid catalyst (or at least its reactivity suggests that it does), and indeed it has a vacant p-orbital allowing it to accept an electron pair (the IUPAC gold book definition being a molecular entity that is an electron-pair acceptor).

What makes less sense to me is the 'mechanism' this catalysis occurs by. The cation itself is pseudo-stable (relative to a standard planar carbocation intermediate such as that observed in an SN2 reaction)  since the aromatic rings are able to stabilise the positive charge, but even if it did react, surely one would expect this to be irreversible, such that a nucleophile would attack the vacant p-orbital to form a stable compound, killing off the reaction.

To give an example of this, trityl tetrafluoroborate may be used to protect an alcohol with a DMB protecting group. What isn't observed is the alcohol attacking the cation to form an O-trityl bond (the trityl group itself is sometimes seen as a protecting group).

One point is that you're assuming that the "p" orbital in the center is the Lewis acidic site. This may not be the case, you can draw a resonance structure that breaks aromaticity but puts the cation at the para position of the phenyl ring. For example, the ortho and para positions are known hydride acceptors (link).

Another issues to consider is that the phenyls are bulky enough that they can't even remain coplanar.

So we can return to your issue of reversibility by looking at trityl via these two properties:

If the Lewis acidic site is via one of the phenyl rings, then the resultant adduct is destabilized by breaking aromaticity. It may be kinetically accessible and allow the trityl ion to serve as a Lewis acid, but the adduct is not stable.
If you do create an adduct with the central carbon, this adduct is greatly destabilized by steric repulsion. Enough so that fragmentation becomes likely.

In both cases, there is a kinetically accessible Lewis acid/Lewis base adduct, but this adduct has some kind of feature that makes it thermodynamically less stable. Therefore, the trityl ion can serve a catalytic role as a Lewis acid.

The exact mechanism of action (1) or (2) is likely dependent on the exact reaction you're looking at.


Cracking of dicyclopentadiene - was my simple Liebig condenser distillation setup adequate? I recently did a distillation of dicyclopentadiene (bp 166 °C) to get cyclopentadiene (bp 40 °C). The setup as instructed by my supervisor consisted of a simple water-cooled Liebig condenser. Oil bath temp. at 180 °C, receiving flask held at -78 °C by a dry ice-acetone bath:

Everything seemed to work fine, and I got plenty of cyclopentadiene after distilling for about 4 hours.

Now (writing the report) I went hunting for some procedure references, and the first hit was this one (http://www.orgsyn.org/demo.aspx?prep=CV4P0238):

My interpretation of these instructions look like this:

My questions (too many I know, sorry):

Is my interpretation of that setup correct?
What is the Friedrichs condenser for?
Why should the water temp. be 50 °C? (and how important is it to get this temp. correct?)
Where would I attach the "calcium chloride drying tube" and what is it anyway, what's it for and how important is it to have one?
Other procedures I found attached Nitrogen to the setup. Why would I do that and how important is it?
How would you interpret "efficient water-cooled condenser"?
Finally, can you put me at ease and tell me that my simple distillation setup was just fine?

The Friedrichs condenser is for refluxing dicyclopentadiene (bp ~170 C). It's kept at 50 C to let the cyclopentadiene (bp ~40 C) pass without condensing.


Conversion of Benzene to 9-Bromo 9,10 Dihydroanthracene I tried to proceed through following steps:
Step-1:
Step 2:
Step-3:
Step-4:
I agree with the fact that yield might not be good. But is this whole reaction possible? Am I going correct? A better method and the corrections to this method are appreciated!

What you are looking for is known as the Wurtz-Fittig reaction - wikipedia article , JACS paper
In this proposed reaction I think you will get a complex mixture (also known as a mess), but you may be able to find some of your desired product in it. Good luck with trying to isolate it cleanly!
If you want a straightforward preparative route then look at the F-C reaction of phthalic anhydride with benzene, the reduction of the resulting 2-benzoyl benzoic acid to 2-benzyl benzoic acid and cyclisation with polyphosphoric acid. This product can be reduced to the dihydroanthracene by Zn/HCl


Reactant energy distribution and spontaneity of reactions I wonder about the implications of substances having an energy distribution around a mean, as opposed to them all having the same energy level, on the spontaneity of reactions.

2 somewhat separate questions:

Can a reaction with a high activation energy, but which in the end has a negative Gibbs free energy change at certain temperature, occur spontaneously even though at this temperature the energy level of the reactants is lower than the activation energy? If I understand correctly, the answer would be yes because the energy of the reactants is just an average of a broader distribution, and so some small fraction of the reactants would attain the $E_a$ even if on average $E_a > E_{\text{reactants}}$?
Wouldn't it also be possible that some non-spontaneous reactions (with a positive change in Gibbs free energy) occur spontaneously? I mean we know that if we invest $\Delta G$ energy, we will force the non-spontaneous reaction to occur. But what if some small fraction of the reactants get to the energy level they would have had if $\Delta G$ energy was invested in them (without it being actually invested). Wouldn't that mean that for that fraction of the reactants, in the high end of the energy distribution around some average, the reaction would occur (while not occurring for the rest and vast majority of the reactants who fail by themselves to attain the same energy level they would have had if ΔG energy were invested in them)? Is there a flaw in this understanding?

I guess the question here is whether the idea that $\Delta G > 0$ reactions are non-spontaneous is a statement of probability that says that such reactions are just very unlikely or is it an hard and fast rule like the conservation of energy?

Yes; you are on the right path in your second paragraph, but distinguish between activation energy and free energy as they are not connected. I'm sorry but I think that your third paragraph is not all correct as it stands. If the free energy change is positive the reaction is not spontaneous by definition. Perhaps you mean, if the overall enthalpy change is positive can the free energy now be negative if the overall entropy change is also positive? The answer is yes.

The thermodynamic quantities only give us information about initial and final states at equilibrium. The activation energy is obtained from kinetic, that is rate constant measurements usually vs, temperature and not from thermodynamic quantities. In the vast majority of reactions the activation energy is greater than thermal energy (approx kBT, where kB is Boltzmann's constant). These reactions are generally slow because, by the decreasing exponential nature of the Boltzmann distribution with energy, not many reactants have enough energy to react in any given time period, say 1 sec. (This is lucky otherwise we would all have reacted eons ago!) In some special types of unimolecular reactions, e.g. electron transfer, bond dissociation & isomerization, with zero or very small activation energy then clearly these will be very fast possibly with rate constants $ >10^{12}\ /sec $.
In the case where the products are of higher energy than reactants, reaction still occurs but the equilibrium lies on the side of the reactants. Again by the Boltzmann distribution argument products react faster to reactants than the other way round.


How does shock trigger nucleation of gases dissolved in a liquid? When you drop a bottle of soda, the dissolved $\ce{CO_2}$ starts nucleating. Opening it before it's 'settled' will trigger more nucleation and usually makes a mess.

Why does the shock trigger nucleation?

Secondly, is it the same mechanism in play when you shake a bottle? And what about the physical state changes as it 'settles'? i.e., why does the $\ce{CO_2}$ rapidly nucleate if you open the bottle right after you drop it, but it doesn't if you wait a bit?

How is nucleation triggered?
There are, as I see it, three major pathways for nucleation of supercritical carbon dioxide in water.
Bottles: Cavitation
As @Michiel points out correctly in the comments, tapping a glass bottle may cause cavitation, which is then the initial nucleation site for the (runaway) bubble formation. However, I think it is noteworthy that not every type of tapping leads to this; It is when you tap on top of an open bottle with a hard object (such as another bottle) that the gas formation is strongest. This is because the shock waves propagating through the glass walls of the bottle meet up at the bottom of the bottle for a huge impulse into the liquid, forming a larger cavity than other methods (don't do this at parties, people will hate you and their beer will be stale). The relative difference of impulse can be immediately grasped from the fact that the tapping bottle will not start foaming, while the tapped bottle most definitely will.
Cans: Deformation
There are however alternative means of containing beverages, for example aluminium cans. Due to the form of the cans, cavitation formation is highly unlikely. More likely, deformation of the can upon dropping will provide the nucleation sites for the carbon dioxide to form bubbles; think jagged edges on the inside, reactive defect sites in the lattice, and a possibly broken oxide layer of the aluminium, exposing highly reactive elemental aluminium for a short time.
Shaking: Air Dissolution
As @Nikolau correctly pointed out in the comments, shaking a bottle will lead to small bubbles of air getting trapped in the liquid and providing nucleation sites for the gas. Nothing magical about that, but I gave it a heading anyway.

Onwards to the second question you had:
Waiting: Why no foam?
The reason that waiting leads to less foam and spewing around of your favourite carbonated beverage is twofold.

As explained in the linked question on Physics.SE, the bubbles will rise to the top of the liquid level. As such, releasing the pressure will cause further nucleation primarily at the surface and not within the liquid, leading to less liquid propagation by gas evolution. In plain English: This means you don't get soaked by the beverage as it is propelled out of the opening, because the gas evolution happens at the surface and "forces" the liquid to stay in place.

The pressure inside the container rises as more gas is formed. If you wait, a new equilibrium state will be reached by redissolving some of the carbon dioxide that nucleated out of the solution, subsequently lowering the pressure again. (I don't think that the pressure goes back to the initial level within a finite amount of time, but I may be wrong here. If someone could clarify this I would be grateful.)
As there are now less nucleation sites, combined with the fact that they rise to the top (see point 1) foam production is severely hindered.


Is there any difference between "autoignition temperature" and "ignition temperature"? Is there any difference between the "autoignition temperature" and the "ignition temperature" of a substance?

The autoignition temperature is the lowest temperature at which a substance can spontaneously ignite in a normal atmosphere without an external source (spark,flame etc).

"Ignition temperature" is what people call autoignition temperature who don't care about details. As Ivan said, there is no such thing as ignition temperature.


Singlet oxygen production Why is in the reaction of hydrogen peroxide and sodium hypochlorite, singlet oxygen obtained? Why isn't triplet oxygen produced instead, as it is thermodynamically more stable?

You are starting with spin-paired starting materials (an even number of electrons, with half in the up state and half in the down state), and they react (exothermically) to create an oxygen molecule. Initially, spin is conserved so that the oxygen molecule also has to be spin paired (singlet), despite the fact that the triplet state of oxygen is thermodynamically more stable. It will then react with a substrate or eventually decay over minutes to hours (most likely thermally) to the triplet state.


Confusion with E/Z configuration Could someone please illustrate how to determine the E/Z configuration of this molecule?

What I understand so far is that if the two highest priority substituent on each side are on the same side, then it's Z and vice versa.

Now, how can I handle such a structure?
I've considered it and I thought it might be $\ce{CH-CH3}$ on one side (up), and $\ce{CH2CH2}$ on the other. However, I don't think it's right.

Bottom line is that I don't understand how to fractionate the ring into two substituents.

The CH3 and C2H5 groups are both smaller than the rest of the cycle. So these two groups can be forgotten and this molecule is Z.


Are alkenes with bulky groups attached (like the one shown in the picture) chiral or achiral? Is the compound in the picture chiral or achiral?

I think it should be achiral for the following reasons:

There is free rotation. I know, but for every configuration produced by free rotation there is an exactly opposite configuration.
Also at certain point of time all the groups can come to the same plane.

This question is from my textbook exercise. And the answer says it is chiral. But I've no idea why. Also is it optically active or inactive?

EDIT:
@K_P gave a very useful article related to these type of compounds:

Is the compound in the picture chiral or achiral?

The answer in practice is: it depends. You say that there would be free rotation along the phenyl–vinyl bond. However, every adjacent carbon atom — the 2- and 6-positions of the phenyl ring and the 1-position of the double bond — are substituted. For free rotation to occur, these substituents must pass each other. Since they are all bulky, though, free rotation is very effectively hindered. A planar rotamer with full π conjugation between the phenyl ring and the double bond is not populated at any extend which can be shown by the NMR shifts of the corresponding carbon atoms (and could also be shown if the methyl group on the double bond were a proton, because it’s shift would be vastly different from one in styryl conjugation).

Under low enough temperatures, the compound you have been given will indeed form to separable enantiomers due to the bromine atom which destroys the molecule’s symmetry. These are examples of axial chirality; if the chlorine atom is above the phenyl ring’s plane, it is (aR), if it is below, the compound is (aS). The two enantiomers can be separated, e.g. by chiral HPLC, and they will show optical activity.

At higher temperatures, thermal energy will be enough to overcome the rotation barrier and an optically active single enantiomer will racemise. This temperature is typically above room temperature for a system such as your’s but well below the temperature required for alkene (E/Z) isomerisation.

This concept is called atropisomerism and is also the reason why BINOL and BINAP are chiral.


Why do Grignard reagents act as sources of nucleophilic R− and not Br−? In reactions with Grignard reagents, I've always seen the alkyl group being the nucleophilic component, with the R–MgBr bond being broken. This is always rationalised by the fact that the alkyl carbon has a partial negative charge, whereas the electrophile (e.g. a carbonyl group) has a partial positive charge.
However, the halogen atom should also have a partial negative charge. Why doesn't it attack the electrophile instead, breaking the RMg–Br bond?

In all likelihood the halide ion does attack the substrate.  But, the product such a reaction mode would form does not accumulate whereas the alkylated product does.
Recall that in a nucleophilic reaction halide ions are good leaving groups, at least if you stay away from fluorides which are rarely used for Grignard reagents.  So while the halide ion may add to the substrate, it may also be displaced if an alkyl-anion moiety from another Grignard molecule comes in, or possibly the halide ion may just launch on its own to attach to a magnesium atom (which, in a Grignard reagent, acts as an electrophilic center, for example by adding to the oxygen end of a carbonyl group).
In contrast, when the alkyl-anion moiety attacks, the carbon-carbon bond it forms is strongly favored thermodynamically and a lot harder to break; here we do not have a good leaving group.  So the alkylated product rather than the halogenated product is what accumulates.


Calcination in metallurgical process? In calcination why is the ore heated in the absence of air in a reverberatory furnace,
does the presence of oxygen affect any thing?

Most of the ore will be oxidized if thermally treated under air. That's why it's usual to heat it under controled atmosphere or even in vacuum.


Colour change when excess ammonia added to [Cu(H2O)6]2+ Since ammonia is a stronger ligand than $\ce{H2O}$ shouldn't the colour of the new solution $\ce{[Cu(NH3)4(H2O)2]}$ be lighter as the energy gap will be higher and so light with higher frequency and energy (between blue and indigo) be absorbed which means that the transmitted light is the one between blue and green which is lighter? Why does the solution turn deep blue?

Like the OP mentioned, the replacememt of a weak field ligand ($\ce{H2O}$) by a stronger one ($\ce{NH3}$) certainly increases the $\Delta_o$ of the $\ce{Cu(II)}$ complex.

A chemical sample appears colored when it absorbs a certain wavelength of light. What we see is the sum total of the remaining colors. The colour visible to the human eye would be the complement of the colour(i.e. wavelength) absorbed.

The increase in $\Delta_o$ leads to the absorption of a lower wavelength (or higher frequency) of light, as evident from:
$$E=h\nu= \frac {hc}\lambda $$

Note that a higher frequency would mean a shift towards the blue end.

Quantitatively speaking, the light blue colour of the initial hexaaqua complex was due to the absorption of the orange spectrum range of light.

Back to the color wheel, a shift towards the blue side (in the green direction, please) gives us a lighter shade of orange.

  Why does the solution turn deep blue?

Observe the complement(the colour diametrically opposite) of orange-yellow is definitely a deeper shade of blue.


Titration of multi-basic acid How would a solution made of $\pu{20 mL}$ of $\ce{NaOH}$, $\pu{10 mL}$ of $\ce{HCl}$, $\pu{10 mL}$ $\ce{NaH2PO4}$ and $\pu{5 mL}$ of $\ce{Na3PO4}$, all of them in a concentration of $\pu{10 mmol/L}$, behave? I already tried every way i can think of, but none give the solution of the $\mathrm{pH}$ of this system.

Ways I tried:

After reacting the $\ce{HCl}$ with $\ce{NaOH}$, I use the $\ce{NaOH}$ that i still got to react with $\ce{NaH2PO4}$, and to calculate the $[\ce{H+}]$ I used $[\ce{Na2HPO4}]$ + $\ce{[Na3PO4]}$, but i didnt got any answers.

I used this equation $\ce{NaH2PO4 + 2NaOH -> Na3PO4 + 2H2O}$ to find the concentration and then applied to the amphoteric equation, using $K_{a_2}$ and $K_{a_3}$.

$
\begin{align}
(n_{\ce{NaOH}})_i= \pu{0.2 mmol}\\
(n_{\ce{HCl}})_i= \pu{0.1 mmol}\\
(n_{\ce{NaH2PO4}})_i= \pu{0.1 mmol}\\
(n_{\ce{Na3PO4}})_i= \pu{0.05 mmol}
\end{align}
$
At first the $\ce{NaOH}$ will react with $\ce{HCl}$ as per the following reaction:
$$
\begin{align}
\begin{array}{ccccc}
\ce{NaOH} & + & \ce{HCl} & \ce{->} & \ce{NaCl} & + & \ce{H2O} \\
\pu{0.2 mmol} & & \pu{0.1 mmol} & & \pu{0 mmol} & & \\
\pu{0.1 mmol} & & \pu{0 mmol} & & \pu{0.1 mmol} & & \\
\end{array}
\end{align}
$$
Now, the remaining $\ce{NaOH}$ will react with $\ce{NaH2PO4}$ as per following:
$$
\begin{align}
\begin{array}{ccccc}
\ce{NaOH} & + & \ce{NaH2PO4} & \ce{->} & \ce{Na2HPO4} & + & \ce{H2O} \\
\pu{0.1 mmol} & & \pu{0.1 mmol} & & \pu{0 mmol} & & \\
\pu{0 mmol} & & \pu{0 mmol} & & \pu{0.1 mmol} & & \\
\end{array}
\end{align}
$$
Now the resulting solution is  a buffer of $\ce{HPO4^2-}$ and $\ce{PO4^3-}$. So the Henderson equation can be used.
$
[\ce{PO4^3-}]=\frac{0.05}{45} \pu{M}\\
[\ce{HPO4^2-}]=\frac{0.1}{45} \pu{M}\\
$
$$
\begin{align}
\begin{array}{ccc}
\mathrm{pH} & = & \mathrm{pK_{a_3}} + \log(\frac{[\ce{PO4^3-}]}{[\ce{HPO4^2-}]}) \\
\mathrm{pH} & = & 12.35 + \log(\frac{0.05}{0.1})\\
\mathrm{pH} & = & 12.35 + \log(\frac{1}{2})\\
\mathrm{pH} & = & 12.04 \\
\end{array}
\end{align}
$$


The order of aromaticity of benzene, thiophene, pyrrole, furan, and pyridine I have been taught that the order of aromaticity is: benzene > thiophene > pyrrole > furan. However, I have been unable to deduce any logical explanation for that order. Also, I wish to compare pyridine's aromaticity with these, but my professor does not know about it.

According to me, all the rings are aromatic and satisfy the Huckel's rule of aromaticity with $6\pi$ resonating electrons and a planar ring. So, that point is useless for doing any comparison of aromaticity.

But, we note that $\ce{2p\pi-2p\pi}$ overlap is most favorable in conjugated systems. Hence, thiophene should be the least aromatic (since it has a $\ce{3p\pi-2p\pi}$ overlap instead).

I agree that benzene should be the most aromatic because both its resonating structures are equivalent and have no charge separation. Using exactly the same reasons, pyridine should be the second best aromatic ring (lesser than benzene because the N atom is an electron withdrawing group and is reducing the electron density in the ring)

That said, every resonating structure (except the uncharged one) for pyrrole and furan will create a formal positive charge on the hetero-atom. Since N is less electronegative than O, it will be slightly more stable than O with that positive charge. Hence, pyrrole will be more aromatic than furan.

Therefore, according to me, the aromaticity order should be: benzene > pyridine > pyrrole > furan > thiophene.  But it is obviously wrong.

Hence, my question is: what is the order of aromaticity of benzene, thiophene, pyrrole, furan, and pyridine. And why?

The main criteria of aromaticity is resonant or delocalization energy.
It is difference between energy contained in bonds if they were localized like in Kekule's formula for benzene and actual energy.
For benzene if is about 150 kJ/mol.
The more this energy the more stable molecule is.

The approaches for calculating this energy may differ.
As for β-values mentioned in paper by @orthocresol it looks like it is Dewar's resonance energy per one π-electron calculated based on modified Hückel's MO method as mentioned in GSE article.

There is a table in this article with the same β-values:

In English it is

Benzene                            0.065     Pentalene                     -0.018
Naphthalene                        0.055     Azulene                        0.023
Anthracene                         0.047     Fulvene                       -0.002
Tetracene                          0.042     Pyridine                       0.058
Cyclooctadecanonaene([18]annulene) 0.012     Pyrrole                        0.039
Cyclobutadiene                    -0.268     Furan                          0.007
Azete (Azacyclobutadiene)         -0.160     Thiophene                      0.032


The most variant groups of periodic table Is it okay to call group 11 (formerly 1B) a "group"? I meant that according to my knowledge "A group is a horizontal column consisting of elements with similar valencies and a continuous gradation in physical and chemical properties" but there is no gradual trend or we don't obtain similar reactions for same circumstances. Members do not show similar oxidation states and do not form analogous compounds. Electronegativity values and other bonding characteristics also deviate badly from one another. Are they in a group just because their atomic numbers allot their positions in same vertical column? Shouldn't 1B and other d-block "groups" (which show such variant properties) treated as "ungrouped region" or something like that?

Sure Group 11 is a group. And if you look in the right areas you see similarities between copper, silver and gold:

1)  Among the best electrical conductors even compared with other metals (http://eddy-current.com/conductivity-of-metals-sorted-by-resistivity/).

2)  Tend to prefer combining with "soft base" nonmetals:  Cu and Ag found as sulfides, gold is oxidized by $\ce{HNO_3}$ when a chloride source is added.

3)  Have some chemistry in the +1 oxidation state, especially wuth said soft-base nonmetals.


Why do some elements with high electronegativity exhibit metallic properties? Metals typically have low electronegativity, which makes them form ions easily and prefer making metallic bonds to covalent. However, some of them seem to disprove that. Take, for example, gold, lead and phosphorus (for comparison). Their electronegativity values are 2.54, 2.33 and 2.19 respectively. Phosphorus is a typical non-metal, while gold and lead should have higher electron affinity and be even stronger non-metals that phosphorus. Why isn't that true?

Whether an element adopts a metallic or nonmetallic structure does not depend on electronegativity alone.  Aside from electronegativity at least two factors promote formation of a nonmetallic structure:

1)  If the atoms are close to filling their valence shells they are more likely to do so through covalent bonds.  Iodine needs only one more valence electron per atom and can get it by forming covalently bound diatomic molecules; gold needs seven valence electron per atoms to complete the valence shell and goes the metallic bonding route instead.

2)  elements in earlier periods tend to form stronger covalent bonds favoring the nonmetallic structure.  Silicon form the familiar tetrahedral semiconducting structure but lead, below it with similar electronegativity and valence electron configuration, is metallic.

Other factors may enter, too, and Group 14 is an excellent place to look for subtle effects.  Above I said that silicon has the tetrahedral structure we all know and love, but that is under ambient conditions. The tetrahedral structure of silicon is energetically more stable than the metallic one, but the latter has more entropy and thus silicon may be converted to a metal upon heating.  In the case of silicon the temperature for this switchover is high enough so that the metal is molten, so we have simultaneous metallization with melting -- and, therefore, contraction with melting.  Germanium does the same thing, whereas with lead the metallic structure is more stable all the time.  Tin is a transitional case where it's a semiconductor when cold but switches to a metal below the melting point.

So there is indeed more to what makes a metallic element or a nonmetallic one than just electronegativity.


Does resonance affect the polarity of the molecule? Ozone is a resonating structure, is it because of its resonance and bent structure that it is a polar molecule? Or Is it because of the formal charges whereby the formal charges determine the direction of the dipole moment. If yes, how does this work with resonating structures. Initially, I thought that resonating structures automatically are polar, however, the azide ion is a clear counter-example. I am curious, how one deals with such a case, should we take each resonating structure by itself and consider the superposition in the end.

Edit:
Ultimately, my question is how can you explain ozone's polarity without formal charges because on a wikipedia page, they explain it using formal charges. If that's the case, shouldn't resonance be considered. If yes, how?

When talking about resonance it is important to know that this is only a concept to visualise certain bonding features. It is very important to understand, that the actual bonding situation is a mixture of all possible resonance structures. The Molecule itself exists at all times as this mixture, hence neither of these resonance representations can be observed. One can regard it also as a superposition of these structures.

Resonance structures had to be introduced into the Lewis formula context, when it was obvious, that one formula can no longer describe a bonding pattern remotely accurate. While the Lewis formalism is great for many things, as it keeps chemistry visualisable and simple, it unfortunately has its boundaries. Only for molecules with very simple bonding patterns it is somehow possible to guess some physical properties (qualitatively).

Formal charges are also only a concept, or a result from the limited depiction of the Lewis representation - hence the name formal. They do not usually represent any physical charge. However, sometimes, the physical charge will coincide with the formal charge. Therefore it is usually not possible to derive any actual dipole moment from them.

Charge separation within molecules depend of the geometric arrangements of the nuclei and the electrons around them, i.e. the electron density. It is a physical property which can be observed and also predicted by means of quantum chemical considerations. Molecular Orbital Theory is a very fundamental concept which allows to derive these properties (at least on a qualitative level).

The bent arrangement of ozone is a result of its electronic structure. And the dipole moment is a result coming from that.

Upon the extension of the question and some comments:

Like Thomij already stated, the formal charges are only for bookkeeping. They always have an integer value. A partial charge, however, is a physical observable, resulting from the difference from the electron density and the field of the nuclei.

In all cases the overall formal charge has to add up to the overall physical charge, which can also be calculated from all partial charges. These partial charges can also be calculated from quantum chemical calculations, since you are approximately solving for the optimal electron density at a certain nuclear arrangement.

There is absolutely no possibility to predict a dipole moment without knowing the electron density (or wavefunction) at a given nuclear geometry. A Lewis formula can be generated from these facts, showing the simplest but still well matching bond description. This representation has to sometimes include formal charges. The linked wikipedia page does not employ this concept to explain the polarity, they rather use partial charges.

The partial charges in ozone result from a superposition of the molecular orbitals (click here to enlarge). These are the valence orbitals obtained from BP86/cc-VPDZ:

Especially the HOMO (very right) and the HOMO-3 (4th from right) and HOMO -6 (4th from left) contribute massively to a slight charge separation in ozone.


Denticity of bridging ligands How would we assign denticity to bridging ligands?
For eg, consider $\ce{NH^-_2}$, it has 2 lone pairs and can hence act as a bridging ligand, making 2 metal-ligand bonds. So I thought it should be bidentate. However, my professor says it is monodentate, why so?

It is important to clearly separate the terms denticity, hapticity and bridging.

An atom or a ligand is bridging is one single atom bonds to two different acceptors.

A ligand (but never a single atom) can be multidentate or multihaptic if two atoms of that ligand can bond to the same acceptor.

A ligand such as the amide anion cannot be multidentate or multihaptic as it only has one coordinating atom: nitrogen. This nitrogen can coordinate to two different metals (in a μ-amido complex) but it still remains a monodentate ligand.
A ligand such as ethylenediamine or en for short ($\ce{H2N-CH2-CH2-NH2}$) can coordinate to the same central atom via each of the two nitrogen atoms. This is a bidentate ligand as it has two ‘teeth’ in the central atom.
A ligand such as ethene ($\ce{C2H4}$) can coordinate to a single central atom in a bihaptic (η2) fashion. Drawing shortest lines from nucleus to nucleus, both carbon atoms would be the same distance from the acceptor atom (in reality, it is better thought of the π system donating). Hapticity is reserved for such cases in which a continuous set of atoms coordinate to a central metal.
A ligand can be both multihaptic and multidentate. An example would be dibenzylideneacetone or dba ($\ce{Ph-CH=CH-C(=O)-CH=CH-Ph}$), in which both of the $\ce{C=C}$ double bonds act as a bihaptic ligand. (Due to geometry constraints, dba usually coordinates to two different acceptors rather than to a single one in a chelate fashion.)
Finally, it is possible for a ligand such as en to be both bidentate and briding, if either of the nitrogen atoms coordinates to two different metals. This requires deprotonation of at least one nitrogen.
The bottom line is: briding ligands can be monodentate or multidentate; they can be monohaptic or multihaptic. Only bridging ligands based on a single donor atom must be monodentate.


Why is 1H-pyrrol-1-ium non-aromatic? It has 4n electrons and is completely conjugated. It should be anti aromatic.

Why is this compound non-aromatic?

It is non-aromatic even though it has 4$\pi$ electrons because nitrogen cannot form five bonds. Hence delocalisation is not possible and so it is non aromatic.


Hybridization of orbitals and forming of bonds in the nitrogen dioxide molecule I considered the nitrogen atom hybridizing its orbitals and then ending up having 3 $\mathrm{sp^2}$ orbitals and the oxygen with 2 $\mathrm{sp}$ orbitals. The thing which bothers me is when people draw a radical (single electron) over the Nitrogen atom after there is the bond formation between nitrogen and 2 oxygen atoms.
Let us consider the Nitrogen, it has one full $\mathrm{p}$ orbital (AKA lone pair) and 3 single electrons in 3 $\mathrm{sp^2}$ orbitals, so that means that it can form 3 bonds – first it bonds with one oxygen forming a σ bond, leaving the nitrogen with 2 electrons, now the nitrogen forms one π bond with the same oxygen and the remaining single electron forms another σ bond with another oxygen atom.
Where is the single radical on the nitrogen after all the bonding has taken place?

Moreover, as in the above picture, people draw the other oxygen (the one that did not form a π bond with nitrogen) as having 3 lone pairs, whereas, after the hybridization of the oxygen atom, I can only see it having 2 lone pairs in 2 $\mathrm{p}$ orbitals.
What is actually correct? Am I going wrong somewhere?

In your simple model, you start off with:

nitrogen: $\mathrm{3(sp^2)^1 + p^2}$
left oxygen: $\mathrm{(sp)^1 + sp^2 + p^1 + p^2}$
right oxygen: $\mathrm{(sp)^1 + sp^2 + p^1 + p^2}$

The σ bonds are easy. Use $\mathrm{sp(O) + sp^2(N)}$. That leaves you with:

left oxygen: $\mathrm{sp^2 + p^2 + p^1 + (\sigma_\ce{O-N})^2}$
nitrogen: $\mathrm{p^2 + (sp^2)^1 + (\sigma_\ce{N-O(1)})^2 + (\sigma_\ce{N-O(2)})^2}$
right oxygen: $\mathrm{sp^2 + p^2 + p^1 + (\sigma_\ce{N-O})^2}$

Now the question is how to do a π system (obviously there can only be one σ bond between two atoms). The orbitals used for the π system must be orthogonal to the bond axis, since that axis must be a node. Both oxygens have two p-type orbitals which fulfill this condition. Nitrogen also must use its p-type orbital to form the π bond. The $\mathrm{sp^2}$ orbital cannot because it is in the same plane as the other two $\mathrm{sp^2}$ orbitals, angled at $120^\circ$ rather than $90^\circ$. This gives us a π system composed of two oxygen p orbitals (one on each oxygen) and one nitrogen p orbital; choosing our orbitals well will result in this π system having four electrons. Essentially, this is theme and variation of Your Average 4-electron-3-centre bonding; but you can also consider it akin to the allyl anion.
The one remaining unpaired orbital would be in $\mathrm{sp^2}$ on nitrogen, pointing away from both $\ce{N-O}$ bonds but being in the same plane.

The above are the considerations and reasonings that you should use from the starting point you had. It is not the only way. You could assume that nitrogens five electrons are actually distributed as $\mathrm{2(sp^2)^1 + (sp^2)^2 + p^1}$. In this case, the argument outlined above is essentially the same except that the π system now contains three electrons rather than four; it is an allyl radical system, if you wish. Since this ab initio construction will lead to all orbitals with s contribution ($\mathrm{sp^2}$ on nitrogen, $\mathrm{sp}$ on oxygen) being fully occupied, it is more consistent with Bent’s rule and likely to be a better proposition.
If you do it like this, you can – in the Lewis 2-electron-2-centre formalism – choose either oxygen and have the radical form an $\ce{N-O}$ π bond on that side; the other will remain a radical. By resonance, you can draw the mirrored molecule.
$$\ce{O=N-\overset{.}{O} <-> \overset{.}{O}-N=O}$$
In these structures, the radical is never centred on nitrogen although we find that dimerisation at low temperatures leads to $\ce{O2N-NO2}$.

Actual measurement shows, that the radical is centred in an orbital that transforms as $\mathrm{a_1}$. As $\mathrm{a_1}$ is symmetric with regards to all symmetry transformations, it cannot be a π type orbital which can transform either as $\mathrm{a_2}$ (node through nitrogen) or $\mathrm{b_1/b_2}$ (no node through nitrogen; depending on your choice of coordinates). You see, it gets more and more complicated the more you rely on experimental data.


What is the chemical reason talc is used in water-based paint and not in solvent based paint? As I understand, paints & coatings use talc predominately in water-based products and not in solvent-based products. Why is this? Is it because the solvent-based products don't need fillers? Or do they already have the characteristics that talc would provide?

Most water based paints are emulsions of plastics particles dispersed in water, it is common to use vinyl acetate to make polyvinyl acetate. To do this an emulsion of vinyl acetate in water is formed, then by adding a free radical initiator.

One purpose which fine clay can furfill in the paint polymer synthesis is to increase the stability of the emulsion of droplets of the liquid monomer. These clay particles will be incorperated into the final paint.

It is also common to add other inorganic substances such as barium sulfate and titanium dioxide. These are used to make the paint more opaque.

One classic organic based paint class is the alkyd paint, these are made by reacting acid anhydrides such as phthalic and maleic anhydride with a polyalcohol such as glycerol or 2,2-dimethylpropane-1,3-diol under acidic conditions. It is normal to distill water out of the reaction, often an unsaturated fatty acid is added in the form of a glycerol triester (vegtable oil).

The idea is that with glycerol the fatty acid ester will react in a transestification reaction to add pedent groups onto the main polymer. Then when the paint is exposed to oxygen a radical cross linking reaction occurs. The other main option I see is to add some maleic acid to the synthesis of the paint binder. The to add some styrene to the paint. This again will cross link with a radical reaction.

For the alkyd paint synthesis there is no need to add talc, these polymerisations are done under homogenous conditions by heating a close to solvent free mixture of monomers. Rather than being a radical polymerisation (example of an addition polymerisation) this reaction is a condensation polymersation where many small chains are converted into a few long chains.


How does adding lemon juice to sugar make better caramel? It's a common cooking advice: if you want a somewhat softer caramel, add some lemon juice to the sugar:

  To help prevent the caramel from crystallizing, you can add an acid to the sugar before you begin: add about half a tablespoon of lemon juice to each cup of sugar and mix it with your hands

But how does it work?

The citric acid in lemon juice hydrolyzes the sucrose molecules into glucose and fructose as you heat the sugar. The water in the lemon juice is also part of this process. This mixture of monosaccharides is much less crystalline in the same way that most mixtures are less crystalline than pure substances. Wikipedia has a good article on this process under the title "Inverted sugar syrup". The syrup is "inverted" in the sense that the sign of its optical rotation has changed because sucrose has a large positive optical rotation while fructose has a large negative optical rotation. Glucose has a positive optical rotation, but of a lower magnitude than sucrose.

The reaction is:
$$\ce{C12H22O11}(sucrose)+\ce{H2O -> C6H12O6}(glucose)+\ce{C6H12O6}(fructose)$$
The reaction can occur to a small degree by simply heating sugar with water in absence of acid catalysis. I assume the acid is necessary to get good conversion before you burn your sugar or vaporize all of the water.

Honey is a naturally occuring inverted sugar syrup of this type. Honey is resistant to crystallization because it is a mixture of gluclose and fructose. Honey is also slightly acidic.


Why do thiols have such a propensity for bonding with mercury? Historically, thiols $\ce{-SH}$ were known as "mercapto-" due to their strong propensity to bond with mercury.  Certain drugs still carry this designation, though mercaptopurine has an $\ce{=S}$ group instead of a thiol.

What makes the thiol group bond so strongly with mercury in particular? Why don't alcohols make such a strong bond with $\ce{Hg}$?

If it's known, was there any historical reason why it was important to be able to "capture" mercury?

Well, sulfur in general, and thiolates in particular, are known for bonding very strongly to gold, lead and mercury. In the particular case of mercury, which was known very early in the history of chemistry (indeed since the times of alchemy), this is the reason thiols are often called "mercaptans", from the Latin mercurium captans (capturing mercury).

In Pearson's HSAB theory (hard and soft acids and bases), the reason the S-Hg bond is be stronger than the O-Hg can be explained because S2- is a soft basis and Hg(II) is a soft acid, making a good fit… while O2- is harder.

Finally, regarding the historical use of thiols to bind mercury (and lead to), let's just say a chelating agent for mercury and other toxic compounds was quite useful as an antidote. You can read a very nice book on the historical use of mercury and lead in John Emsley's The Elements of Murder.


Quick-and-Dirty Molecular Dynamics by Mass-Weighted Atom Translations? I'm thinking about a quick-and-dirty approach to molecular dynamics, mostly for teaching purposes.

At a given temperature, statistical mechanics tells us molecules have translational and rotational and vibrational energy as given by the equipartition theorem. At a basic level, all three motions derive from movement of the individual atoms along Cartesian axes.

So it seems like a very simple quick-and-dirty approximate molecular mechanics could be implemented by "shaking" the individual atoms with mass-weighted random displacements. You'd need to implement some sort of force field to make sure the added displacements would fall into the correct potential energy distribution for a given molecule.

Quick searching doesn't seem to find this as an approach, likely because it won't obey a particular canonical ensemble.

I'm curious... Has anyone implemented such a method for qualitative visualization? What are some of the pitfalls compared to "proper" molecular dynamics? (e.g., presumably ensuring the proper velocities?)

The random displacements in your suggested method are uncorrelated from step to step.  What you are suggesting is essentially heavily damped molecular dynamics at the Brownian limit.  The limitations are obvious, you won't sample correlated, large-scale motions efficiently.  (That might be a student project right there: turn up the Langevin parameter, see when you lose rotations of phenyl rings, and ask yourself what this means.)


If there are multiple bases in solution and a strong acid is added, will both bases react? For example, if there is one mole of carbonate ions ($\ce{CO3 ^2-}$) and sulfate ions ($\ce{SO4 ^2-}$) in a solution, and one mole of hydrochloric acid ($\ce{HCl}$) is added, will the one mole of hydrochloric acid react just with the sulfate ions as that is stronger, forming hydrogen sulfate ions?
Or will the acid react with both the ions forming hydrogen carbonate ions ($\ce{HCO3-}$) and hydrogen sulfate ions ($\ce{HSO4-}$)?

Both based will indeed react with the acid.  But wait, there's more.  When the bases react to form their conjugare acids, the conjugate acid from one base could react with the other base.  When one base is strong and the other weak, this secondary reaction will favor the direction where the strong base gets protonated.  So you see the strong base gets more protons while the weak base tends to be pushed back to the basic form.

What if both bases are similar in strength?  Then we apply equilibrium thermodynamics.  Say hydrochloric acid is added to a solution containing both sodium acetate and sodium nitrite.  The equilibrium between each of the latter two ions is governed by the appropriate acid dissociation constants:

$\ce {HC2H3O2 = H+ + (C2H3O2)-} $  $K_{a1}={\ce {[H+][(C2H3O2)-]}}/{\ce {HC2H3O2}}$

$\ce {HNO2 = H+ + (NO2)-} $  $K_{a2}={\ce {[H+][(NO2)-]}}/{\ce {HNO2}}$

To get the equilibrium that relates the competing weak acids and bases directly, just divide to eliminate the hydrogen ion concentration; in this case

${\ce {[HNO2][(C2H3O2)-]}}/{\ce {[(NO2)-][HC2H3O2]}=K_{a1}/K_{a2}}$

So your acid dissociation constants aren't just about acid dissociation.  They also govern how a limited supply of protons gets distributed between competing bases.


Why/how is blood red? (colours of hemoglobin) Oxyhaemoglobin is red, deoxyhaemoglobin is bluish-purple, and carboxyhaemoglobin is a cherry red colour.

Evidently in the porphyrin around the iron molecule in heme there is a conjugated $\pi$-system which could account for the colour, but how exactly does this work? Also, what is the role of the metal ion (specifically $\ce{Fe}$) in the middle, its oxidation state, or the ligand, in changing/determining the colour in conjunction to that? Do the globin chains play any role in determining the colour (if so, how)?

Chlorophyll, for example, also has a porphyrin motif, but it is green due to having magnesium in the middle. Vitamin B12 is also intensely red in colour (according to Wikipedia).

The $\ce{Fe}$-porphyrin complex is a vastly conjugated $\pi$-system, in which the oxidation state of the $\ce{Fe}$ plays a central role. The wavelength of light absorbed by hemoglobin exactly corresponds to the the difference in energy between the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) of the heme $\pi$-system, and is what gives hemoglobin its color. These values are exact and quantifiable. Changes in this color, therefore, are the direct result of changes in the energies of the molecular orbitals of the hemoglobin $\pi$-system.

To understand the color change, it is important to think about the geometry of the molecule. When oxygen is not bound, the iron atom is in the +2 oxidation state. It's slightly too big to fit into the hole in the center of the plane of the immediately-surrounding "heme," so it actually rests just on top of the heme plane. Here is the molecular orbital of the iron species with no $\ce{O2}$ bound:

When $\ce{O2}$ does bind, the $\ce{Fe}$ and $\ce{O2}$ orbitals bond and their energies change. In particular, the $\ce{O2}$ $\pi$* orbitals interact with the $\ce{Fe}$ $\mathrm{xz}$ and $\mathrm{z^2}$ orbitals. A side effect of the $\pi$* accepting properties of all three of these ligands is that iron formally changes from being in the +2 oxidation state to the +3 oxidation state. You can think of it as having "given up" one of its electrons to the ligand via the ligand's $\pi$* orbitals. As a side effect of losing one of its outer electrons, the iron atom is now small enough to align perfectly with heme plane.

Almost counterintuitively, however, formation of oxyhemoglobin is actually energetically disfavorable, which can be explained by its color. Deoxyhemoglobin appears purplish-blue, meaning that it must be absorbing yellow light (lower energy),and is therefore more stable. Oxyhemoglobin appears red, meaning that it must be absorbing at blue-green light (higher energy) and is therefore less stable. This is actually vital to oxyhemoglobin's functionality. Because it is a higher energy, and thereby less stable structure, it easily releases the bound oxygen in the presence of hypoxic cells.

[Source]


How does 1-bromo-2-fluorobenzene react with lithium amalgam? How does this organometallic reagent react with 1-bromo-2-fluorobenzene? Does it react like Grignard reagent? If the product of this reaction is then reacted with furan, what will be the final product? I can guess there will be a benzyne intermediate, but I am not sure.

Quoting from Page 1063 of Advanced Organic Chemistry FIFTH EDITION, Part B -
Reactions and Synthesis, FRANCIS A. CAREY, and RICHARD J. SUNDBERG:

  Benzyne can also be generated from o-dihaloaromatics. Reaction with lithium amalgam or magnesium results in the formation of transient organometallic compounds that decompose with elimination of lithium halide. o-Fluorobromobenzene is the usual starting material in this procedure.

As @orthocresol said in the comments that Benzyne undergoes Diels-Alder reaction with furan.

The first Diels–Alder reaction of benzyne with furan was performed before giving a 76% yield of the adduct, which was converted to 1-naphthol with acid${^1}$ ${^2}$.

From Diels-Alder adduct to 1-naphthol

In the absence of the furan trapping agent, biphenylene and triphenylene were obtained in yields of 24 and 3% respectively${^3}$.

References

${^1}$:https://www.researchgate.net/publication/273556121_The_Benzyne_Story

${^2}$ :Shashidher et al., Pharm Anal Acta 2011, 2:7
DOI: 10.4172/2153-2435.1000137

${^3}$ :https://www.researchgate.net/publication/273556121_The_Benzyne_Story


Does hyperconjugation stabilize a carbocation more than a free radical? Which of the following statements is correct about methyl group?

  It stabilizes both carbocation and free radical equally.
  It stabilizes a free radical more than a carbocation.
  It stabilizes a carbocation more than a free radical.
  None of these.

I thought the answer would be (1). Keeping in mind these structures: The methyl group can stabilize the carbocation and the free radical group attached to it equally through hyperconjugation. However, the answer given is (3). How is this possible? Why is the carbocation more stable than the free radical? Are the structures I drew wrong?

A carbonation is deficient in 2 electrons whereas a free radical is only 1 electron deficient. Thus any type of stabilisation working on a carbonation will be much more effective as it is more electron deficient. Hence, hyperconjugation or inductive electron donation stabilises a carbocation more than it does a free radical.


Which of the following is a secondary alkyl halide a $\ce{(CH3)2CHCH2CCl(CH3)2}$

b $\ce{(CH3)2CHCHClCH3}$

c $\ce{(CH3)2CClCH3}$

d $\ce{(CH3)2CHCH2Cl}$

This is just to check that I'm understanding the material correctly.

I'm thinking it's c because when drawing the line structure the alpha carbon is connected to two other carbons.

Well, no. Here is the quote from Wikipedia:

  From the structural perspective, haloalkanes can be classified
  according to the connectivity of the carbon atom to which the halogen
  is attached. In primary (1°) haloalkanes, the carbon that carries the
  halogen atom is only attached to one other alkyl group. An example is
  chloroethane ($\ce{CH3CH2Cl}$). In secondary (2°) haloalkanes, the carbon
  that carries the halogen atom has two $\ce{C–C}$ bonds. In tertiary (3°)
  haloalkanes, the carbon that carries the halogen atom has three $\ce{C–C}$
  bonds.

So, the presence of $\ce{CH2Cl}$, $\ce{CHCl}$, and $\ce{CCl}$ are the indicators of primary, secondary, and tertiary haloalkane respectively.

Here is b with the $\ce{C}$ atom to which $\ce{Cl}$ is attached as well as its two $\ce{C–C}$ bonds highlighted in red.


What does "oxidized form" and "reduced form" mean? What does the phrase "oxidized and reduced forms of the reactants" mean?

For some context: The book was talking about half-reactions in a half-cell of a galvanic cell, and it said this,

  Sometimes, both oxidized and reduced forms of the reactants in a half-cell are soluble and cannot be used as an electrode.

I am confused as to what the "oxidized and reduced forms of the reactants in a half-cell" might mean because in a half-cell either only oxidation or reduction takes place.

In this case, the adjective "reduced" does not imply something is actually being reduced. It simply refers to the form of the element in question which has the lower oxidation state. Let's say something is getting oxidised in a half-reaction, for example

$$\ce{Fe^2+ -> Fe^3+ + e-}$$

Iron takes two forms here, $\ce{Fe^2+}$ and $\ce{Fe^3+}$. $\ce{Fe^3+}$ is the oxidised form, that much is obvious, right? Then, $\ce{Fe^2+}$ is the reduced form.

No $\ce{Fe^3+}$ is actually being reduced to $\ce{Fe^2+}$. However, it is still correct to refer to $\ce{Fe^2+}$ as the "reduced form", since it is the form of iron with the lower oxidation state.


Why is this PH3 considered to be an unhybridized molecule? Also, don't hybrid orbitals appear in all molecules?

The answer by "ron" says "the molecule can be viewed as being unhybridized":
How does lone pair of a central atom affect the dipole moment?

I'll try to explain this by comparing this with the case of NH3.
N is a better electron withdrawing group than P. So in the case of NH3 the electron cloud is more closer to N. And therefore the repulsion in the bonds causes the 3 H atoms to be farther away from each other resulting in a bond angle of about 109°. But in the case of PH3 the electron cloud is farther away from the central atom(P) than in the case of NH3. So the bond pair - bond pair repulsion is comparatively lesser, causing the 3 H atoms to move closer together to an angle of almost 90°, resembling the px, py, and pz orbitals, as a consequence it's lone pairs are present in almost pure s - orbital. So it is is said that PH3 can be viewed as unhybridized.


Enantioselectivity of Noyori Asymmetric Hydrogenation I'm unable to understand the enantioselectivity of Noyori's hydrogenation reaction, and would appreciate any help with the same:

Here's a mechanism I found on Wikipedia:

but, to my disappointment, the highlighted statement, i.e. The (R)-BINAP-Ru catalyze the synthesis the (R)-Product, and the (S)-BINAP Ru catalyze the synthesis the (S)-product with high ee. contradicts the attached paper about Noyori's hydrogenation.

The last two entries of the table mention formation of S products from R-BINAP.

Also, I feel it's not really possible to determine the absolute configuration beforehand, i.e. without knowledge of surrounding groups.

It'd be great if someone could describe the mechanism based enantioselectivity of this reaction in detail, so that correct predictions about the stereochemistry of the product can be made, given the catalyst (R or S BINAP) and the substrate.

Thanks in advance!

I think this is easy to answer than you think. Let's look at the stereochemistry of compounds in the question, 1g and 1h (I also put the common product, 1a, for convenience):

The stereospecific reduction is on carbon-3 ($\ce{C}$3) in all cases. Accordingly (see diagram above), if you apply Cahn–Ingold–Prelog (CIP) sequence rules for each, you'd find (3R)- for 1a (methyl group is priority 3 based on $\ce{CHH}$ vs $\ce{HHH}$), (3S)- for 1g (isopropyl group is priority 2 based on $\ce{CHH}$ vs $\ce{CCH}$), and  (3S)- for 1h (phenyl group is priority 2 based on $\ce{CHH}$ vs $\ce{CCC}$). All these assignments are $\ce{C}$2 versus $\ce{C}$4 since priory 1 for all three groups on $\ce{C}$3 is $\ce{OH}$.

Thus, we can conclude that the change of stereochemistry is mainly due to the priority rules (not due to the change in stereochemistry on transition states or intermediates). This is also supported by the example given in the same reference (Ref.1) in the box in depicted diagram. Even when bulkiness at $\ce{C}$2 was increased by adding a methyl group, the reaction (with R-BINAP) has undergone same manner to give (R)-configuration to $\ce{C}$3 center.

References:

Ryoji Noyori, Takeshi Ohkuma, Masato Kitamura, Hidemasa Takaya, Noboru Sayo, Hidenori Kumobayashi, Susumu Akutagawa, "Asymmetric hydrogenation of $\beta$-keto carboxylic esters. A practical, purely chemical access to $\beta$-hydroxy esters in high enantiomeric purity," J. Am. Chem. Soc. 1987, 109(19), 5856-5858 (https://doi.org/10.1021/ja00253a051).


Python package for modelling chemical reactions I'm looking for a package that is compatible with Python 3 and can actually do work with reactions along with, for instance, referencing values. Ideally, it should also not be a package that will take 20 days to learn how to access one value.

More specifically: there should be a way to figure out all possible reactions a certain state could lead to, a way to manipulate lists of possible reactions and applicable reactions in certain states, and to apply reactions to states to get a resulting new state (this, of course, may require some coding on my part, but just an idea of what it should lead too - the more the package can do the better).

Thanks; I'd be glad to provide any clarification in the comments.

This is not really my area of expertise, but a quick search for "python chemical reactions" revealed three hits I've never seen before that may be of interest, the first one being closest to what you want.

chempy

from chempy import ReactionSystem  # The rate constants below are arbitrary
rsys = ReactionSystem.from_string("""2 Fe+2 + H2O2 -> 2 Fe+3 + 2 OH-; 42
    2 Fe+3 + H2O2 -> 2 Fe+2 + O2 + 2 H+; 17
    H+ + OH- -> H2O; 1e10
    H2O -> H+ + OH-; 1e-4
    Fe+3 + 2 H2O -> FeOOH(s) + 3 H+; 1
    FeOOH(s) + 3 H+ -> Fe+3 + 2 H2O; 2.5""")  # "[H2O]" = 1.0 (actually 55.4 at RT)
from chempy.kinetics.ode import get_odesys
odesys, extra = get_odesys(rsys)
from collections import defaultdict
import numpy as np
tout = sorted(np.concatenate((np.linspace(0, 23), np.logspace(-8, 1))))
c0 = defaultdict(float, {'Fe+2': 0.05, 'H2O2': 0.1, 'H2O': 1.0, 'H+': 1e-7, 'OH-': 1e-7})
result = odesys.integrate(tout, c0, atol=1e-12, rtol=1e-14)
import matplotlib.pyplot as plt
_ = plt.subplot(1, 2, 1)
_ = result.plot(names=[k for k in rsys.substances if k != 'H2O'])
_ = plt.legend(loc='best', prop={'size': 9}); _ = plt.xlabel('Time'); _ = plt.ylabel('Concentration')
_ = plt.subplot(1, 2, 2)
_ = result.plot(names=[k for k in rsys.substances if k != 'H2O'], xscale='log', yscale='log')
_ = plt.legend(loc='best', prop={'size': 9}); _ = plt.xlabel('Time'); _ = plt.ylabel('Concentration')
_ = plt.tight_layout()
plt.show()  # doctest: +SKIP

OEChem

Cantera

There is RDKit and Pybel as well, but those are more traditional cheminformatics tools.


Are ionic bonds stronger than covalent bonds? A covalent bond involves overlapping of orbitals while an Ionic bond involves charge separation.

Why are bonds formed by the overlapping of orbitals weaker than charge separation; why is an ionic bond stronger than a covalent bond?

In a fully covalent bond, you only have the mixing of the orbitals, as you said. But it's not only that, you also have coulomb interactions between the nuclei, that slightly raise the energy of the molecular orbitals, which results in asymmetric energies.

In a fully ionic bond you also have a coulomb interaction, but this time with the different sign, because you have a positive and a negative charge. These want to be as close together as possible, resulting in a stronger bond.

Generally speaking: Coulomb interactions have a greater impact than molecular orbitals on the energy, and thus the strength, of a bond.

It's almost like saying that electromagnetic forces are stronger than gravitational forces.