Quantum Mechanics/Wave Mechanics in Chemistry Beginners Thread

Quote: Originally posted by aga

Are the OC answers open to everyone, and should be U2U'd to you Darkstar, same as for the QM answers to blogfast ?

Quote: Originally posted by aga

Might be an idea to go U2U on this with a time limit, same as the QM Q/A thing.

Quote: Originally posted by aga

The Carbon at the second position from the left will be having the electron density pulled away from it towards the more electronegative Oxygen (orbital probability density closer to the O than the C) effectively making it relatively electron-poor. That should make the O of the OH- the nucleophile and the 2nd C of the ethyl acetate the electrophile.

Quote: Originally posted by aga

The O on the ethanol has the extra electron so must be the nucleophile.The right-hand H on the sodium acetate (where's the Na ?) couldn't give a shit, as it has it's orbital full and happy, so the electrophile must be where the electron ends up, which is on the Right-most O of the sodium acetate.

Quote: Originally posted by aga

The homeless electron ends up with the right-hand O for no good reason i can think of, other than that it has a spare p orbital for it to half-occupy.

Quote: Originally posted by aga

After Step #1 there's already an electron excess, so further excess electrons can't have much effect ?

Quote: Originally posted by aga

Oh ! So the whole resonance thing is imagining (even calculating!) where the electron densities are on the molecule, especially the active bits, and the relative electrostatic charges that implies ?

Quote: Originally posted by aga

A Carbon atom with 4 bonds should be happy. What we're discussing is that it's not always the case. An adjacent bonded Oxygen is being greedy with the electrons, and the Carbon wants more, so is receptive to suitors bearing negatively charged Gifts ?

Quote: Originally posted by aga

Aha ! "Thermodynamics makes no pronouncement about Kinetics" leaps to mind. AKA "Just cos it is possible does not make it happen"

QQuiz Questions for next week: U2U me the answers!


Question 7: (for 10 points)

Estimate the Enthalpy of Formation (at 298 K) of fluoromethane (CH₃F), from bond Enthalpies. Show your calculation.


Question 8: (for 20 points)

When 1-chloropropane is refluxed with a small stoichiometric excess of strong NaOH solution, 1-propanol and a solution of NaCl is obtained according to the following overall reaction equation:



Propose a simple reaction mechanism (use curly arrows to show the movements of electrons).


[Edited on 8-11-2015 by blogfast25]

Quote: Originally posted by aga

Oh ! So the whole resonance thing is imagining (even calculating!) where the electron densities are on the molecule, especially the active bits, and the relative electrostatic charges that implies ?

Quote: Originally posted by Darkstar

To help further illustrate the impact that resonance can have on OC reactions, in my next post, I'm going to show you a rather interesting reaction that I have come to believe ONLY works due to resonance. I will contrast this reaction with another reaction that, due to a lack of resonance stabilization, ends up proceeding through a completely different mechanism, despite the fact that both reactions use similar substrates under the same conditions.

So a couple of months ago I made a post in an OC thread called HI+P reduction mechanism--a thread about the mechanism for the reduction of benzyl alcohols using hydroiodic acid (HI) and red phosphorous. If this sort of reaction sounds familiar to you, it's probably because it's commonly used by clandestine chemists to reduce l-ephedrine and d-pseudoephedrine (both benzyl alcohols) to d-methamphetamine (we'll talk about what the d- and l- prefixes mean when we get to chirality), and is the reason that iodine and red phosphorous are so closely watched in places like the US. This type of reduction is extremely well-known and was first discovered more than 100 years ago; however, the exact nature of its mechanism remains a mystery to this day, and there's still no unanimous consensus as to how it actually works. In the thread mentioned above, I proposed a mechanism for it (which you will see below) that is based on the various conclusions that I have found in the literature over the years.

But before I show you the reaction between HI and benzyl alcohols, I want to show you what a typical reaction between HI and an alcohol looks like. Since the benzyl alcohol in the example reaction is 1-phenylethanol, I will show you how HI normally reacts with secondary (and tertiary) alcohols. A secondary alcohol is one where the hydroxyl group is connected to a carbon that is further connected to two other carbons, by the way. Both of the reactions below involve heating a secondary alcohol with concentrated hydroiodic acid; however, despite being similar alcohols, they give completely different products. In one reaction, the hydroxyl group gets replaced by an iodine atom, which is what we would expect from treating an alcohol with a hydrohalic acid. In the other reaction, however, something unique and unexpected happens: the hydroxyl group gets replaced by a hydrogen atom instead! (what "reduction" means in organic chemistry)

So let's take a look at the usual reaction between secondary alcohols and HI. In this example, HI is reacted with propan-2-ol, also known as 2-propanol, isopropyl alcohol or simply "rubbing alcohol." Here, the hydroxyl group gets replaced with an iodo group via nucleophilic substitution, producing 2-iodopropane and a water molecule. This type of reaction is called an S_n1 reaction (unimolecular nucleophilic substitution) and is one of the most common types of reactions in OC. Here's the reaction and mechanism:



In the interest of keeping this post as short as possible, I will just briefly sum up what's happening:

Step 1 - The nucleophilic oxygen of the hydroxyl group donates a lone pair to the electrophilic hydrogen on hydronium. (remember that HI, when dissolved in water, is really in the form of a bunch of H₃O⁺ and I^– ions) This breaks the old bond on hydronium and creates a new one on isopropyl alcohol, giving a charged oxonium ion.

Step 2 - The oxonium group on isopropyl alcohol takes the electrons with it and leaves as neutral water. This generates a positively-charged carbon (called a carbocation) at position-2, where the carbon has three bonds and is short an electron.

Step 3 to Step 4 - The nucleophilic iodide ion donates a lone pair to the electrophilic carbon, filling its vacant orbital and forming a new bond that neutralizes the charges.

This is the usual reaction between secondary alcohols and HI. Seems simple and straightforward enough, right? But when a benzyl alcohol is used in that same reaction, something strange happens. Like before, the hydroxyl group leaves and gets replaced by iodine (it's arguable whether or not it actually does, but for now we'll just assume it does); however, the molecule then reacts again and ends up getting reduced by iodide, giving the reduced alcohol, I₂ and water. This reaction is a lot more complex than the ones we've looked at so far and involves radical intermediates, so don't worry if the mechanism makes absolutely zero sense after Step 2. A radical is when an atom has a single, unpaired valence electron, by the way. (which typically makes it extremely reactive) The arrows that look like they have a fishhook on the end mean that a single electron is being moved:



If you want more details regarding this mechanism, just let me know. Basically, the reaction proceeds as follows:

Step 1 - The hydroxyl group is protonated and leaves as water, giving a resonance-stabilized carbocation at the benzylic position.

Step 2 - Iodide attacks the carbocation and forms a new bond.

Step 3 - The weak carbon-iodine bond cleaves homolytically to give a resonance-stabilized benzylic radical and an iodine radical.

Step 4 - The benzylic radical is reduced via SET (single-electron transfer) from a nearby iodide ion to give a negatively-charged carbanion (essentially the opposite of a carbocation) and a second iodine radical.

Step 5 to Step 6 - The carbanion grabs a proton from hydronium to give the reduced alcohol and the two iodine radicals combine to give I₂.

This is how I believe the reaction works, which is made possible through a combination of resonance-stabilization and iodine's unique properties due to its size. This mechanism is suggested in the literature and is further supported by the fact that most other alcohols that have a pi-system directly adjacent to the hydroxyl group get reduced as well, strongly suggesting some sort of resonance-stabilized radical intermediate. The benzylic radical and iodine radical seen in Step 4 are created via homolytic cleavage of the carbon-iodine bond, which is shown in Step 3. Homolytic cleavage is where the bond breaks and each atom takes one electron with it. Many times this requires a considerable amount of energy to do. But because of iodine's size, becoming a radical isn't too difficult for it; however, for a secondary carbon atom, normally it wouldn't want anything to do with that sort of thing. And as you can see in the reaction with isopropyl alcohol, there's no breakage of the C-I bond to generate radicals and thus no reduction.

But because of the unique stability of benzylic radicals due to the adjacent aromatic ring, the already-weak carbon-iodine bond becomes even weaker. (keep in mind that iodine-carbon bonds are already weak bonds to begin with) The benzylic radical is stabilized through resonance like this:



And it is this resonance that makes the bond between the benzylic carbon and the iodine atom extremely easy to break homolytically and the unique reduction possible. The radical that would be generated in the first reaction with isopropyl alcohol if the C-I bond were cleaved homolytically would be a lot more unstable due to a lack of resonance, making the radical much more reactive and thus a lot more difficult to create. (still possible, though, just not as easy. in fact, at high enough temperatures, HI will even fully reduce carboxylic acids!)

Revisiting the Wave Function

On stuff that doesn’t oscillate, determinism and the particle-wave duality

We’ve come this far in the QM/WM/QC webinar that the time has come to revisit (and hopefully further clarify) one the most fundamental concepts of this paradigm, the wave function ψ itself. Early on we started with the following definition:



We’ve since worked out the wave functions for a number of simple quantum systems and below we find the schematised wave functions of the ground state and first two excitations for a single particle in a 1 dimensional finite well (well potential is U₀, instead of our usual notation V₀:



We’ve also seen that the wave functions ψ and associated energy values E are the eigensolutions of the time independent Schrödinger equation (here in 1D):



At this point it’s necessary to dispel one of the most profoundly and frequently held misconceptions about Quantum wave functions that new students of the quantum world often entertain: the wave function as a ‘matter wave’. This pernicious, wrong and stubborn belief stems mainly from poorly digested interpretations of the de Broglie Hypothesis, skewed readings of the wave-particle duality and teaching analogies involving ‘standing matter waves’.

Looking at the wave functions of the finite well potential it is indeed very tempting to make the analogy with the standing waves that occur in a string of a musical instrument, with the ground state the lowest base frequency (that determines the 'pitch' of the note) and the excited states the so-called harmonics (that determine the 'timbre' of the note).

But nothing could be further removed from the truth: as one physicist famously remarked about quantum wave functions, “nothing oscillates there!” As we’ll shortly see, it would have been more complete to add, “apart from probabilities!”

In the current and prevailing Born interpretation, the wave function is a probability function and the peaks and throughs probability amplitudes:



Here P(x) is the probability density distribution of the particle over its domain and the actual probability (of finding the particle) in an interval [x₁, x₂] to find is given by P( x₂, x₁.

(Note: these relationships hold for Real and normalised wave functions)

The term ‘wave function’ is therefore somewhat misleading: unlike in water or sound waves, for ‘quantum waves’ there is no oscillating medium (water, air or any other elastic medium) and no propagation, so typical of water and sound waves.

Another misconception about QM that needs dispelling is: ‘QM is non-deterministic’, which largely stems from its seemingly probabilistic nature. But at least at the most basic level one can show this isn’t true.

We saw earlier on that the expectation value < x > can be calculated from a (Real and normalised) wave function as follows:



By the expectation value < x > has to be understood the average position of the particle, taken over a very large number of measurements. For a single particle in an 1D infinite potential well of length L, we found that:

< x > = L/2

If using a detector we measure this value over a very large number of individual observations (or by observing a very large number of identical systems simultaneously) we always find < x > = L/2. Perfectly deterministic, in other words.

Finally, if quantum wave functions don't describe real waves, where does that leaves us with regards to the particle-wave duality we started this webinar from? Certainly moving subatomic particles and photons do sometimes behave like waves and sometimes like particles and we don't really have a good word for that. But to describe how moving particles show wave-like properties like diffraction, modern QM needs not to describe the particle as a ‘matter wave’. Instead the formalism of the particle as a probability wave is sufficient, so there is no contradiction there.

[Edited on 8-11-2015 by blogfast25]

Brief Introduction to Wedge-Dash Notation

When I have a little more time I'll do a proper lesson on chirality. For the time being, I've prepared a mini-lesson on wedge-dash notation. This will serve as a lead-in to enantiomers by familiarizing any young OC Padawans following this thread (*cough* aga *cough*) with the concept of representing three-dimensional molecules in only two dimensions.

While drawing molecules using the line-angle formula is undoubtedly the most practical way to draw molecules in OC, the problem is that molecules aren't actually two-dimensional in real life. For non-chiral molecules (we will cover chirality soon) this isn't really a problem; however, for molecules with one or more chiral centers, we need a way to represent which direction certain bonds are oriented in three-dimensional space because it makes a difference. With that said, I present to you the wedge-dash notation:



As you can see, the normal lines are oriented along the plane of the image; the wedged lines are pointing towards the viewer; the dashed lines are pointing away from the viewer. Compare the wedge-dash drawing of propane above to the ball-and-stick model of propane below:



Notice that the bonds are oriented the exact same way as the bonds in the wedge-dash drawing suggest.

Extra Credit:



In the image above, I have drawn 2-chloropropane in the top box and (1-chloroethyl)benzene in the bottom box. In each case, I have drawn the molecule two different ways: one with the chloro group pointing towards the viewer, and one with the chloro group pointing away from the viewer.

For extra credit, explain why the two molecules in the top box are the exact same molecule, while the two molecules in the bottom box are not. What have I changed by adding a phenyl ring?

Quote: Originally posted by Darkstar

I think we scared aga away.

Quote: Originally posted by blogfast25

Question 5: (for 10 points) Predict the geometrical shape of the following molecules: a) Xenon difluoride, XeF<sub>2</sub>. b) The pentafluoride xenonate anion, XeF<sub>5</sub><sup>-</sup>. Question 6: (for 10 points) Dinitrogen difluoride (N<sub>2</sub>F<sub>2</sub> has two isomers, as shown below, left trans-N<sub>2</sub>F<sub>2</sub>, right cis-N<sub>2</sub>F<sub>2</sub>: Briefly explain why two such isomers exist.

A. 5:

a. XeF₂:

Both bonds are identical. Electrostatic repulsion between the bonding MOs is minimised if the molecule takes on a linear shape: F-Xe-F.

b. XeF₅^-:

All bonds are identical and the single charge is delocalised. Electrostatic repulsion between the bonding MOs is minimised if the molecule takes on a trigonal bipyramidal shape as shown here.


A. 6:

Two of three valence electron of each N contribute to the double bond: two p_x form a σ(pp) MO, two p_y form a π(pp) MO, forming a torsionally rigid structure.

The F atoms can then bond to the remaining N valence electron on either side of the double bond. So two different structures, called 'cis' (same side) and 'trans' (one on each side) arise.

Additional note to A. 6:

Because of the abundance of double C bonds in organic chemistry, cis/trans isomerism is very common in OC, see below for example the structures for 2-butene:

Quote: Originally posted by blogfast25

Bond Enthalpies: Estimating reaction Enthalpies from bond Enthalpies: These values can be used to estimate the reaction enthalpies of simple reactions. Here, mainly for illustrative purposes, the estimated Enthalpy of formation of ethane: 2 C + 3 H<sub>2</sub> === > C<sub>2</sub>H<sub>6</sub>, ΔH<sub>f</sub><sup>298K</sup> To find this value write this equation as a sum of ‘sub reactions’ using Hess’ Law: 1) 2 C === > C – C, yields - 368 kJ 2) 3 X [H<sub>2</sub> === > 6 H.], costs 3 X + 435 = + 1305 kJ 3) C – C + 6 H. === > H<sub>3</sub>C – CH<sub>3</sub>, yields 6 X – 414 = - 2484 kJ Adding up: ΔH<sub>f</sub><sup>298K</sup> = - 368 + 1305 – 2484 = -1547 kJ per mol C<sub>2</sub>H<sub>6</sub>.

Quote: Originally posted by aga

Am i answering QM Q 7 with the revised or original methodology ?

As per our U2U use the 'revised'^* method if you can. Scoring will reflect my error, so it should be an easy 10 smackaroons!



* Revised number = original + 717 kJ/mol. Easier than decimalisation!!

[Edited on 9-11-2015 by blogfast25]

Quote: Originally posted by aga

Are there any OC Questions i've missed ? Still a Whopping chunk of OC to re-read and digest some more (he said, hoping for mercy)

Embroidering slightly more on bond Enthalpies:

In the post on bond Enthalpies I indicated that the listed values should be considered as averages, in the sense that the exact value of a bond Enthalpy depends on its molecular environment. The following link to a *.pdf illustrates that very well (besides also listing a far greater number of values):

https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociat...

(the values are bond dissociation enthalpies, so almost always positive)

Here's a sample of values for various C-C bonds in different molecules to illustrate the point of bond Enthalpy variability:

Quote: Originally posted by aga

Quote: Step 1 - The nucleophilic oxygen of the hydroxyl group donates a lone pair to the electrophilic hydrogen on hydronium. Am i being Thick or does 'Lone Pair' mean TWO electrons ? Is ONE electron being donated or TWO ?

I've drawn the two electron pairs (MOs) on the hydroxyl group:



It's the one marked green '1' that now attacks the proton on the hydronium (oxonium) and forms a bond with it. So it's not a question of 'donating ONE or TWO electrons': the lone, non-bonding MO 1 now becomes a bonding MO, bonding that second proton to the O atom.

Personally I would describe that step more easily as the protonation of that hydroxyl group by a H₃O⁺ cation (hydronium = oxonium): it's really a classic acid/base reaction:

R-OH(aq) + H₃O⁺(aq) < === > R-OH₂⁺(aq) + H₂O(l)

But in terms of movement of MOs, what I said above is what happens.

I prepared 2-chloropropane by that method but using ZnCl₂(anh.) as a catalyst, not so long ago:

http://www.sciencemadness.org/talk/viewthread.php?tid=30180#...


[Edited on 10-11-2015 by blogfast25]

[Edited on 10-11-2015 by blogfast25]

Quote: Originally posted by aga

Sorry, still not Clear. It's the term 'Lone Pair' that i'm struggling with. Does that mean 'An orbital with between 0 and 2 electrons in it' or does it mean 'An orbital with two electrons in it' or does that mean 'An orbital with 0-2 electrons that is currently not involved in bonding' ? The confusion arises from previous descriptions of electrons being swapped, and now we're into bonding orbitals where the electrons in it (or not) are somewhat abstracted.

In the context of Step 1:

An orbital with two electrons in it, that is currently not involved in bonding

But also in the context of Step 3:



The iodide ion has 4 lone electron pairs and uses one of them to bond with the carbonium ion with: one of the carbonium MOs is empty. The iodide acts as the Lewis base, the carbonium ion as the Lewis acid.

[Edited on 10-11-2015 by blogfast25]

Quote: Originally posted by aga

Thanks for that. 1) So 'Lone Pair' is a misnomer and should be interpreted as 'electron orbital' irrespective of it's apparent electron configuration, due to hybridisation ? 2) If you would not mind, please explain it in terms of your recent illustrated example (the one with the OH, red arrow, H3O) etc)

1) Yes! 'non-bonding molecular orbital' is probably the most accurate term, actually.

2) I prefer the other example:



The structure of the iodide anion is an sp³ hybridised tetrahedron, all four MOs are full (2 electrons, Pauli conformant).

The electronic structure of the central, charge carrying C atom in the carbonium cation is:

* sp³ hybridisation
* two of the sp³ bond to two carbons
* one of the sp³ bonds to one hydrogen
* one of the sp³ is EMPTY!

Iodine then 'donates' one of its full sp³ orbitals to the empty C sp³ MO and a C-I covalent bond is formed. Et voila, everybody's happy.

aga too?


[Edited on 11-11-2015 by blogfast25]

And here's a question for Darkstar and/or Annaandherdad.

I found it on a physics forum I'm a member of and got stuck trying to solve it.

Here's the question:



Another member part-answered the question as follows:



Because I'm not firm on my legs with Dirac notation, I wanted to calculate the energy expectation value using:



I first calculated the normalisation constant (the wave function is zero everywhere but in -a/2 < 0 < +a/2) and found it to be 10/√(61a⁵.

But then it dawned on me we don't know V(x), so we don't know the Hamiltonian operator?

I'm stuck.

[Edited on 11-11-2015 by blogfast25]

Quote: Originally posted by annaandherdad

The problem is probably trying to say that the potential is 0 for -a/2 < x < +a/2, and infinity outside that range.

Quote: Originally posted by aga

OK. Exhausted my Butane lighter testing that out. Can i bill the B&D Education Store for the replacement butane ?

Quote: Originally posted by annaandherdad

Blogfest, I got your message. The problem is not very well posed. The problem is probably trying to say that the potential is 0 for -a/2 < x < +a/2, and infinity outside that range. The energy eigenfunctions are u_n(x) = sqrt(2/a) sin( 2n pi x/a) and the energy eigenvalues are E_n = (2/m) (n pi hbar /a)^2 However, the first question, what is the energy value you get on the first measurement, cannot be answered. The wave function is a linear combination of all of the energy eigenstates, so there are probabilities for measuring the different energy eigenvalues. You can't say which one will be measured, only the probabilities. For the second question, the number of times you get the ground state is approximately the probability of being in the ground state times 1000. The probability of being in the ground state is an integral that is a bit of a pain to do.

So there's no way to determine the amplitudes in the superposition? Fourier transform, anyone?

Can the superposition be be normalised? After all, the particle is bound...

Thanks!

Moderators: when will this site adopt MathJax (Latex rendering of math formulas) like below:



It's truly bizarre that on a forum about science in the 21^st Century we still have to render the language of science with sub and sup tags...

[Edited on 12-11-2015 by blogfast25]

Quote: Originally posted by blogfast25

So there's no way to determine the amplitudes in the superposition? Fourier transform, anyone? Can the superposition be be normalised? After all, the particle is bound... Thanks!

Quote: Originally posted by Darkstar

For extra credit, explain why the two molecules in the top box are the exact same molecule, while the two molecules in the bottom box are not. What have I changed by adding a phenyl ring?

The molecules in the top box are the same thing rotated 180 degrees around the Y axis.

The ones in the bottom box are mirror images of each other.

I was about to triumphantly blurt 'steroisomer !' then had the sense to look that up.

The mirror-image thing makes them Enantiomers.

Edit:

If it's right, take the extra credits in liu of the butane


[Edited on 12-11-2015 by aga]

Quote: Originally posted by aga

The molecules in the top box are the same thing rotated 180 degrees around the Y axis. The ones in the bottom box are mirror images of each other. I was about to triumphantly blurt 'steroisomer !' then had the sense to look that up. The mirror-image thing makes them Enantiomers.

Quote:

If it's right, take the extra credits in liu of the butane

Quote: Originally posted by Darkstar

[...] while drinking and doing unauthorized experiments with the undergrads.

I heard it was experiments ON the undergrads.

Quote: Originally posted by Darkstar

Well, I didn't want to bring this up, but I'm afraid you have forced my hand. There's a small matter of a room in one of the freshman dormitories that you completely trashed last weekend while drinking and doing unauthorized experiments with the undergrads. I've already received over two dozen complaints regarding broken glassware in the sink, chemical spills/splatters/spatters/puddles/sprays all over the walls, floors and counter tops, stains, holes in the carpet from what looks like a large acid spill, complaints of foul and obnoxious odors, hazardous waste that was improperly disposed of (halogenated compounds, toxic heavy metals etc), empty beer cans all over the place...

Almost forgot the squigly diagram for QM question 8




Edit :

Seeing as we can't send images via U2U it might be easier to do the DarkStar thing and post the Answers in the thread, where images can be included.

OK with you bloggers ?

[Edited on 12-11-2015 by aga]

Fur aga:

The normalisation procedure for the problem higher up:



Note that A has two roots, one positive, one negative, which represent the wave function's parity (remember those fellows gerate and ungerate?)

Now the wave function has been normalised we are certain that the probability density distribution P(x) is entirely correct. The wave function is now ready to extract information from.

Quote: Originally posted by aga

Seeing as we can't send images via U2U it might be easier to do the DarkStar thing and post the Answers in the thread, where images can be included. OK with you bloggers ?

Hydrolysis of 1-chloropropane by sodium hydroxide:



NaOH fully dissociates in water into sodium and hydroxide ions.

The nucleophilic (Lewis base) hydroxide anion attacks the polarised carbon atom. A new bond is formed, between the OH and the C atom.

The C-Cl MO folds back onto the Cl atom, forming a chloride ion as leaving group.


[Edited on 14-11-2015 by blogfast25]

Quote: Originally posted by aga

i just saw that i got the H in the OH- attacking the 3-Carbon.

As far as nomenclature is concerned, the carbon that chlorine is bonded to is technically the 1-carbon. There are way too many nomenclature rules to attempt to explain them all in a brief post, but basically you number carbon chains based on what's attached to them, not left-to-right like you read. In 1-chloropropane (notice that it's not called 3-chloropropane!), you start on the carbon with the chloro group and count from there. On the other hand, if the chloro group were attached to the middle carbon, then you could start counting on either end since it wouldn't make a difference (either way, the carbon that the chloro group ends up on is the 2-carbon); however, you would NOT start counting on the second carbon because you always want the longest carbon chain as the parent chain. Carbon chains are named using a prefixparentsuffix structure (i.e. 1-chloropropane or 1-chloropropan-2-ol), where the parent chain is considered the "main" chain and everything else just "additions" to it.

In larger, more complex molecules with multiple functional groups, the order of precedence determines what gets prefixed and what gets suffixed. In other words, certain groups take priority over others. The most important group on a molecule gets suffixed at the end by itself, and all other groups get prefixed in alphabetical order in front of the parent chain. An exception to this are groups like chloro, bromo, iodo, fluoro and alkyl groups (methyl, ethyl, propyl, butyl etc), which never get suffixed regardless. In some cases, this priority will also determine which carbon you start counting on as well. Also, the prefix name for a functional group is usually different than the suffix name for it (i.e. when prefixed, hydroxyl groups are "hydroxy," but when suffixed, they are "ol").

But don't worry about trying to make sense of all of this right now. I will do a more in-depth lesson on IUPAC nomenclature in the future. Consider this extracurricular reading for the time being. Below are some examples I made for your viewing pleasure. Oh, and by the way, that last one is a sample of the kind of stuff I will be asking you to name on the homework questions!

Quote: Originally posted by aga

Is that Fly Spray or Slug-knackerer ?

It lacks a methyl on C No 5 for that.

Quote: Originally posted by Darkstar

But don't worry about trying to make sense of all of this right now.

Quote: Originally posted by aga

Please Sir(s), what does a Hartree-Fock equation do ? Or a 'Virtual Orbital' mean ? [Edited on 15-11-2015 by aga]

Hartree-Fock lies at the heart of several approximation methods for multi-atom multi-electron systems (usually called molecules)

For multi electronic systems the trusted Schrödinger equation:



... has no analytical solutions because the inter-electron repulsive electrostatic forces cause Potential Energy (V) terms to occur in the Hamiltonian Operator, that aren't encountered in simpler one electron systems.

To find numerical solutions for these cases a number of approximative/iterative techniques have been developed. Hartree Fock is an important element in some of these methods.

Where did you find the term "Virtual Orbital"?


[Edited on 16-11-2015 by blogfast25]

Quote: Originally posted by aga

Please Sir(s), what does a Hartree-Fock equation do ?

Question 9: (for 10 points)

Calculate the Lattice Enthalpy of NaF(s) from the following data (all in STP conditions):

Enthalpy of Formation of NaF = - 575 kJ/mol
Enthalpy of vapourisation of Na = + 97.42 kJ/mol
Ionisation energy of Na (Na == > Na⁺ + e) = + 495.8 kJ/mol
Enthalpy of bond dissociation of F₂ = + 159 kJ/mol
Electron affinity of F ( F + e == > F^- = - 328 kJ/mol

Question 10: (for 20 points)

Many d-block elements form colourful aqueous complexes with ammonia. Explain why the following are colourless:

a) Cu(NH₃₄⁺(aq)
b) Zn(NH₃₄²⁺(aq)

Oh, and these are the last questions of the current QQuiz!


[Edited on 17-11-2015 by blogfast25]

Quote: Originally posted by aga

Who is bringing the booze & drugs this year ? Students or Masters ?

Quote: Originally posted by Cheddite Cheese

Question: How exactly does Russel-Saunders coupling work? Also, how does it give rise to Hund's Rule?

Quote: Originally posted by Cheddite Cheese

Question: How exactly does Russel-Saunders coupling work? Also, how does it give rise to Hund's Rule?

Quote: Originally posted by aga

What do we want ? More tests ! More QM ! More OC ! When do we want it ? Now !

Quote: Originally posted by blogfast25

Question 9: (for 10 points) Calculate the Lattice Enthalpy of NaF(s) from the following data (all in STP conditions): Enthalpy of Formation of NaF = - 575 kJ/mol Enthalpy of vapourisation of Na = + 97.42 kJ/mol Ionisation energy of Na (Na == > Na<sup>+</sup> + e) = + 495.8 kJ/mol Enthalpy of bond dissociation of F<sub>2</sub> = + 159 kJ/mol Electron affinity of F ( F + e == > F<sup>-</sup> = - 328 kJ/mol Question 10: (for 20 points) Many d-block elements form colourful aqueous complexes with ammonia. Explain why the following are colourless: a) Cu(NH<sub>3</sub><sub>4</sub><sup>+</sup>(aq) b) Zn(NH<sub>3</sub><sub>4</sub><sup>2+</sup>(aq)

Quote: Originally posted by Darkstar

I'll post some more OC lessons soon.

And one for the road.

Most alcohols react neutral with pure water. Solutions of phenol in water are slightly acidic.



Explain this difference from a structural point of view.

[Edited on 21-11-2015 by blogfast25]

Draw all isomers of this alcohol:

Quote: Originally posted by aga

It's cos they do not absorb/emit any photons in the 390nm to 700nm range, having carelessly donated their sole 4d electron so some charity case or other. Some elements are dangers to themselves.

Quote: Originally posted by aga

In MS Paint ? You're kidding. Numbering the carbons left to right, the OH can go on the 2, 3 or 4 postions. Not entirely sure why it can't go on 1 or 5, but it just feels like it can't.

Quote: Originally posted by blogfast25

Draw all isomers of this alcohol

'owzat ?

Quote: Originally posted by Cheddite Cheese

There are other isomers. Keep on trying. (And some of the compounds have two enantiomers, but I'm not sure we're counting those.) [Edited on 11-22-2015 by Cheddite Cheese]

Quote: Originally posted by Cheddite Cheese

Aren't there three (not two) more alcohols? [Edited on 11-22-2015 by Cheddite Cheese]

More quiggles

More squiggles !!! OK.




Hmm

Looks like Foothark to me.

[Edited on 22-11-2015 by aga]


[Edited on 22-11-2015 by aga]

The first three you started off with were fine and the very first one (pentan-2-ol) is an enantiomer. The others again are conformations.

Here are the other three, with the second one being an enantiomer:



The way to find enantiomers is to look for carbon atoms that are bonded to four different groups.

Take the middle one in my drawing. The C-atom bonded to the OH group is also bonded to a methyl group, an isopropyl group and finally a hydrogen atom. Four different species.

Such a molecule, when depicted in 3D always has a mirror image molecule. Together the pair are a pair of enantiomers.

[Edited on 23-11-2015 by blogfast25]

Neopentanol (aka 2,2-dimethylpropanol) hasn't been mentioned yet. That was the third one I was thinking of.



[Edited on 11-22-2015 by Cheddite Cheese]

Quote: Originally posted by Cheddite Cheese

Neopentanol (aka 2,2-dimethylpropanol) hasn't been mentioned yet. That was the third one I was thinking of.

Aga: Ok then. Now let's look at the following (fairly big) chunk.

Express recap of Brønsted–Lowry (BL) acid base theory:



The indexed Cs in the K equations represent actual concentrations (assuming a dilute solution), not what it says on the bottle (for “0.1 M acetic acid” e.g. the C_HA is always (slightly) lower than the nominal concentration of 0.1 M).

It has to be noted also that A^-, the so-called conjugated base of HA, is in fact also a weak base. See the alkaline solutions of acetates, carbonates, bicarbonates etc.

If B is the conjugated base of HA then we can even derive (derivation not given here) that:

pK_a + pK_b = 14

K and pK values thus offer an objective criterion by which we can rank acids and bases from strong to weak.

Stabilisation of Phenoxide anion:

The deprotonation reaction of phenol in water is of course:



We know that non-aromatic alcohols react neither acidly nor basic, so what prompts phenol to be such a (very!) weak acid? The answer lies in the stabilisation of the phenoxide anion due to de-localisation of the negative charge. The full explanation can be found here. A de-localised negative charge is less attractive to protons.

Why strong acids are strong acids:

Let’s recall the case of HBr:



The H-Br bond is highly polarised due to the large difference in electronegativity between Br and H. In the presence of a willing proton receptor like water the deprotonation reaction of HBr runs to completion (equilibrium fully to the right). HBr solutions contain almost no free, undissociated HBr, containing only bromide and oxonium ions (and solvent water).

Cross-paradigm agreement:

In a decent country with reasonable laws and rules these regulations apply from north to south and east to west.

The same is true of a good scientific paradigm (set of consistent theories). And here we have quite an interesting case of correspondence between Brønsted–Lowry acid base theory and its complementary Lewis acid base theory.

Consider the case of an interesting chloro-complex, called hexachlorostannate:



Without going into MO structure (which is simple) the black orbitals lie in the xy-plane, the red ones align with the z-axis. The structure is in accordance with a typical EX₆ molecule, an octahedron.

Cl^- can be seen as the conjugated base of HCl, an acid slightly stronger even than HBr. By definition higher up Cl^- must be an extremely weak base, as of course evidenced by the neutrality of NaCl solutions.

So if Cl^- is a weak BL base, is it also a weak Lewis base? Actually, yes.

Firstly, is it a Lewis base at all? Yes. In hexachlorostannate, the bonds are formed by six Cl^- donating one of its four lone pair electron to empty hybrids (sp³d², to be precise). That’s the essence of a Lewis base: an electron pair donor.

Hexachlorostannate salts (ammonium and alkali metals) can be crystallised easily from water. But they’re not hugely stable:

1. Adding a little ammonia to their solutions immediately precipitates Sn(OH)4. Adding stronger alkali leads to formation of stannate: Sn(OH)₆^2-. This due to ligand exchange: hydroxide ions are a far stronger Lewis base.

2. Heating K2SnCl6 to above the BP of SnCl4 causes the latter to distil off.

Many other chloro-complexes are far less stable: tetrachlorocuprates and tetrachloroferrates exist in solution but cannot be isolated from them (at least not by simple evaporation of solvent), KAlCl4 does exist but can’t be recrystallised from water. All three are subject to ligand exchange with hydroxide ions (ammonia solution, for instance).

So Cl^- is a very weak BL base and isn’t all too great as a Lewis base either.


[Edited on 24-11-2015 by blogfast25]

Quote: Originally posted by blogfast25

... Without going into MO structure (which is simple) the black orbitals lie in the xy-plane, the red ones align with the z-axis. The structure is in accordance with a typical EX<sub>6</sub> molecule. ... In hexachlorostannate, the bonds are formed by six Cl<sup>-</sup> donating one of its four lone pair electron to empty hybrids (sp<sup>3</sup>d<sup>2</sup>, to be precise).

Quote: Originally posted by aga

A step-by-step explanation of what the electrons are doing in the formation of this ion would be very very helpful. Pretty pretty please.

Alright, but only 'cos you ask so nice...

Sn ground state: [Kr] 4d¹⁰5s²5p².

The central Sn ion in the hexachlorostannate ion is in oxidation state +4, so it has lost these four 5s²5p² valence electrons, these orbitals are now empty.

The empty 5s, three 5p (one p_x, one p_y and one p_z and two 5d orbitals (also empty) now hybridise into 6 sp³d² orbitals, all empty.

The chloride ions are soft Lewis bases because they each have 4 lone electron pairs: electron configuration of Cl^- is [Ne] 3s²3p_x²3p_y²3p<sub>z</sub&g t;².

Each chloride ion now donates one of its lone electron pairs into one of the 6 empty sp³d². Each molecular orbital resulting from this is full. These MOs are essentially σ MOs.

Clearish, squire?

[Edited on 23-11-2015 by blogfast25]