Quantum Mechanics/Wave Mechanics in Chemistry Beginners Thread

I made a few more examples of resonance structures to help further reinforce the concept. It will be important later on in your OC journey that you understand the idea behind resonance stabilization, as it will help explain why certain structures are more stable than others; for example, why benzyl carbocations (second to last structure) are so much more stable than most other carbocations, even when the carbon at the benzylic position is primary. This will also help you understand why certain reactions work on one molecule and not on another, and why one pathway is sometimes preferred over another. (there are obviously many other factors that come into play as well, for instance steric hindrance, but you have to take it one step at a time. these concepts will all add up in the end)

Anyway, most of these should be pretty straightforward. I didn't show any resonance structures involving radicals, but they can also be resonance-stabilized in a similar fashion. The only one that might be a little confusing is the first step of the last example (thiazole). Just keep in mind that sulfur, like oxygen, has a third lone pair of electrons that can form bonds. So in that first step, sulfur's lone pair comes down and bonds with carbon, breaking the pi bond between nitrogen and carbon and forming a charged sulfonium ion (the sulfur equivalent to an oxonium ion) that gets countered by the negative charge it creates on nitrogen.

Quote: Originally posted by aga

Thanks for continuing to explain all this. It is very much appreciated. (p.s. what do you use for drawing the OC diagrams ?)

Quote: Originally posted by aga

p.s. what do you use for drawing the OC diagrams ?

Quote: Originally posted by Darkstar

Quote: Originally posted by aga   p.s. what do you use for drawing the OC diagrams ? ChemDraw Pro. Accept no substitutes.

Quote: Originally posted by aga

Website : http://scistore.cambridgesoft.com/DesktopSoftware/ChemDrawPr... says $190 a year For me, that's $190 per drawing !

And you won't find a knock off on the 'Darknet' either!

Quote: Originally posted by blogfast25

Two more instalments (I might cram them into one, actually), let me know when you're ready.

When a Covalent Bond is not a Covalent Bond Anymore:

So far we’ve dealt only with covalent bonds, in which one or more molecular orbitals between the bonded nuclei form a stable ensemble. We’ve also seen that covalent bonds are often permanently polarised and in this section we’ll explore bonds that are ‘extremely’ polarised.

Below is a plot of the boiling points (BP) of the chlorides CCl₄, BCl₃, BeCl₂ and LiCl versus the difference in electronegativity of the constituent elements (e.g. Cl = 3.2, C = 2.6, difference = 1.6):



The trend is clear: higher difference in EN correlates with higher BP. This is largely (but not only) due to bond polarisation.

While CCl₄ and BCl₃ are rightly considered covalent compounds, clearly LiCl isn’t: it’s an ionic compound. BeCl₂ positions itself somewhere between covalent and ionic.

Now look at the case of NaCl.

Na valence electron: 3s¹

Cl valence electrons: 3s² 3p⁵

Bonding could seemingly occur between the Na 3s¹ and a Cl 3p_x¹ electron, forming a sigma s-p bonding molecular orbital. However, the difference in electronegativity between Na and Cl is very large and as a result the bonding MO actually resides on the Cl atom entirely, with Na having lost its 3s1 electron in permanence. Na has become Na⁺ and Cl has become Cl^-. Electrostatic attraction between both ions is an ionic and not a covalent bond. This simple model for NaCl (and other ionic compounds) isn’t very realistic (no covalent bonds form) and in the last chapter we’ll spend a little time on a more appropriate model.

We can nonetheless conclude the following:

'Covalent' and 'ionic' are relative to each other, with many compounds (or bonds) somewhere between the two extremes. The extremes can be represented by H₂ (no difference between the electronegativities of the bonded atoms at all) and CsF (maximum difference between the most electronegative atom F and the least electronegative atom Cs).

In that respect, NaCl is ionic but CsF is even more ionic!


[Edited on 9-10-2015 by blogfast25]

Quote: Originally posted by aga

Maybe, let's see. Electronegativity is a measure of how strongly an atom's nucleus attracts electrons. (unsure whether this is independant of the atom's oxidation state/current electron count or configuration) Bonds are formed by electrons occupying bonding orbitals between atomic nucleii. The difference in electronegativity of the involved atoms causes the electrons to spend more time around the nucleus with the largest electronegativity. If that difference is very small the resulting molecule isn't polarised much. If it's a very large difference the molecule is so highly polarised that it's called an ionic bond instead of covalent. (does the electron truly get 'ripped off' of the Na in the case of NaCl ?) Oh ! Covalent ! Each electronegativity is close in value. Ahhhh. The B.P. is generally proportional to the polarisation of the molecule.

Born-Haber Thermochemical cycle: Ionisation Energy, Electron Affinity and Lattice Energy:

To understand the forming of ionic bonds better, let’s take the Born-Haber thermochemical cycle for a typical binary ionic compound. It represents the stages of the reaction:

M(s) + n/2 X₂(g) === > MX_n(s)



Start at the top left, from the elements.

First phase:

M(s) === > M(g): vapourisation of metal M.

n/2 X₂ === > n X(g): dissociation of X₂ molecules into X atoms.

Second phase:

M(g) === > Mⁿ⁺(g) + n e^-: ionisation of metal gas atoms to metal gas cations.

n X(g) + n e^- === > n X^-(g): ionisation of X atoms to X^- gas anions.

Third (and final phase): condensation of the ionic gas to the solid ionic salt.

Each step is accompanied by relevant Enthalpies, marked in red. Added together with Hess’ Law the Enthalpies of the three phases give the Enthalpy of Formation ΔH_f⁰ of MX_n(s).

Let’s look specifically at the Second Phase again.

M(g) === > Mⁿ⁺(g) + n e^- is the ionisation Enthalpy of M(g). In reality there are n ionisation energies: 1_st: M === > M⁺ + e, 2_nd: M⁺ === > M²⁺ + e, etc etc. These Enthalpies become progressively larger with more electrons removed: it costs more energy to remove an electron from a positively charged ion than from a neutral atom.

These could also be considered the oxidation energies in the gas phase but mustn’t be confused with oxidation potentials gleaned from SRP tables.

X(g) + e^- === > X^-(g)

The Enthalpy released when, say a gaseous halogen captures an electron is known as the electron affinity. Although related to the electronegativity value it not identical to it. Electron affinity is a real energy but electronegativity is an empirically derived parameter but it’s not an energy.

Lattice Enthalpy:

Below are two representations (‘packed spheres’ and ‘ball and stick’) of the cubic ionic lattice of NaCl (the simplest of all ionic lattices).



Simply put, the lattice Enthalpy of NaCl can be represented by:



The change in Coulombic energy when the Mⁿ⁺ ions and the X_- are brought from infinity (the gas) to their position in the lattice and is given by equation:



Which can be summarised as:



The Madelung constant is lattice dependent and tabled values can be found easily.

Understand that ionic compounds aren’t made of MX_n molecules but of vast ionic lattices. There are no covalent bonds between the constituent ions.


[Edited on 11-10-2015 by blogfast25]

[Edited on 11-10-2015 by blogfast25]

Quote: Originally posted by aga

Gulp !

Quote: Originally posted by aga

Gimme a sleep cycle and it'll all be assimilated.

Let me know of any problems, I know it's been the end of a long, hard slog. Well done for staying the course!

Quote: Originally posted by aga

(so long as calculators and downright cheating are allowed)

Quote: Originally posted by aga

But ... if the answers have to be U2U'd that means no Copying either ! You're making this hard on us lying cheating weasels. Edit: As it's 10 points per question, can i assume that the workings out are required as well to get the full 10 points ? [Edited on 18-10-2015 by aga]

Quote: Originally posted by aga

Quote: 1 member supplied 2 correct answers so far (20/20). Was it you ?

Over 80 % score so far by respondents on the first 2 questions, despite the fact that the answer to Question 2 was far harder to find in the seminar than I intended. I will now add two graphics that should clarify the matter, so newcomers can earn themselves some really easy points, so what are you waiting for? Both questions remain open till Sunday coming.

Hydrogen Energy Levels:








Again, apologies for not having included this material earlier on. The Table of Content will also be updated shortly to list this post.


[Edited on 20-10-2015 by blogfast25]

Quote: Originally posted by aga

You gotta publish a table of results at the end. Oooo yeah.

Quote: Originally posted by aga

Then disconnect from the Fixed Viewpoint of Linear Time and it all goes ape (in a good way) a.k.a. agaspace.

Sorry, but my daughter bought me 'Quantum Mechanics' by F.Mandl (WILEY) for my bufftday and I can't find none of your stuff in there! She's taking the subject as a 3rd year astrophysics student this year, so I guess we'll be able to discuss your Tourette's, I mean theory.

[Edited on 21-10-2015 by blogfast25]

Quote: Originally posted by fluorescence

So blogfast suggested that I should put up the question here. As some may know when it's about coordination chemistry many theories meet and combine, Valence-Bond, Molecular Orbitals, Crystal Field and so on. But depending on what you are looking for you will mostly get only an answer from one of these theories and thus it's quite hard to combine all of them in a single theory or explain certain behaviour without changing between theories too often. The most common Coordination Numbers are probably 4 and 6 with the Tetrahedron / square planar and the Octahedron. Besides the well known ligand field splitting we know about stuff like "Hybrids", too. An Octahedron will form a d²sp³ - Hybrid. If you look that shape up that's totally looking like an octahedron. A tetrahedron should actually form an sd³ - Hybrid ( some books say sp³ for a tetrahedron but that is for main group elements only). Then there are some like trig. pyramidal (sd²). Now I was looking for high Coordination numbers and the highest one known that is formed by a monodentate ligand would be a Nona-Coordinated Compound. [XH9]2- where X = Tc, Re The compound is perfectely stable forming a tricapped trigonal prism. Analysis shows that 6 H - Ligands seem to be equal and then there is a second set of 3 H -Ligands. So the question is how that hybride is formed and whether it's really only related to d-Orbital Coordinations. If you look at the Structure of that compound it looks a bit like an octahedron that is distortet by 3 other Hydrides. So is it a d²sp³ and then there are 3 more Ligands that will sort of change the shape or is there a hybride for 9 Ligands ? I can't really find anything on that topic. [Edited on 23-10-2015 by fluorescence]

Firstly, have a look at the 'ball and stick' model of [ReH₉]^2-:



Which is indeed a tricapped trigonal prismatic.

Now have a look at this:



According to it, the structure is explained by an sp³d⁵ hybridisation, the involved AOs are shown too. Re in +7 oxidation state.

Source.

As regards,

Quote: Originally posted by fluorescence

Just saw that the chart says sp³, too with the Tetrabromo-Complex. But shouldn't there actually be d-Orbitals involved in that ?

Question 3: (for 30 points)

A particle finds itself in a one dimensional triangular potential well as shown below.



For x = or < 0, V = +∞ ('x equal to or smaller than zero')

For x > a, V = 0

For 0 < x < a, 0 > V > - V₀

Assume that in the region 0 < x < a, the Schrödinger Equation has three eigenfunctions ψ₀, ψ₁ and ψ₂ with corresponding eigenvalues E₀, E₁ and E₂.

Sketch what the three wave functions are likely to look like. Briefly explain your reasoning. Sketches do not have to neat, smooth or to scale but must show zero points and tunnelling where applicable.

Hint: look for parallels with the wave functions for P1DB and Quantum Harmonic Oscillator.

Question 4: (for 10 points)

Below are the structures (hydrogen not shown) of 2,3 dichloro naphthalene:



They would suggest two such compounds (two isomers) exist but in reality only one 2,3 dichloro naphthalene is known.

Explain why this is so.


[Edited on 26-10-2015 by blogfast25]

Electrophilic Addition Reaction Mechanism

On aga’s request I’ll extend the applications of QM/WM in chemistry to a number of explanations of simple reaction mechanisms (mainly drawn from Organic Chemistry). I’ll start with an Electrophilic Addition Reaction (EAR).

As we’ll see in much greater detail, in an EAR a permanent dipole molecule is added to a double bond.

The permanent dipole molecule X-Y can be represented as below:



Before we proceed I want to drive home what makes a dipole a dipole a little harder, using QM arguments.

Below is depicted the ground state wave function ψ₁ (idealised, not to scale and reduced to 1 dimension) of a bonding molecular σ between two X atoms (one on the left, one on the right), so there’s no difference in electronegativity:



The wave function is perfectly symmetric with respect to the line of symmetry of the molecule. Remember also that the electron density (probability density distribution) is given by:



Which in the diagram is represented by the blue ‘bell’. It is perfectly symmetric and the electrons show no preference over the left or right side of the orbital.

Now consider the case where the left hand side is X, the less electronegative atom, and the right hand side is Y, the more electronegative atom:



It can be argued that the difference in electronegativity introduces a potential V(x) across the orbital, represented by the white sloped line. This potential affect the potential energy term of the Hamiltonian operator:



So, the consequence of this potential is that it changes both ψ₁ and P(x): the electron density (green ‘bell’) is now skewed much more to the right and the X-Y molecule is a permanent dipole.

The mechanism of the EAR is so well explained (at least at the beginner level) in the link below that there nothing to add to it:

http://www.chemguide.co.uk/mechanisms/eladd/whatis.html#top



[Edited on 27-10-2015 by blogfast25]

Quote: Originally posted by blogfast25

Question 4: (for 10 points) Below are the structures (hydrogen not shown) of 2,3 dichloro naphthalene: They would suggest two such compounds (two isomers) exist but in reality only one 2,3 dichloro naphthalene is known. Explain why this is so.

Quote: Originally posted by aga

Which one is then 2,3 form please ?

Quote: Originally posted by aga

Hardest thing is Drawing the damned thing !

Sure. It takes a steady hand. It might be bit early for you.

A glimpse into the reaction mechanism of the synthesis of Indole-3-acetic acid

Previously I indicated that the reaction between indole and glycolic acid was a simple EAR (as described above) but it’s not. The orgsynth page:

http://www.orgsyn.org/demo.aspx?prep=CV5P0654

... does not show water as a leaving group but that definitely happens. Also, the double bond is preserved.

The reaction proceeds in three steps:

1) Saponification of the glycolic acid with KOH:

This is a simple acid/base reaction:

HOOC-CH2-OH + KOH === > KOOC-CH2-OH + H2O.

The actual reactant in step two is thus ^–OOC-CH2-OH (the glycolate anion), not glycolic acid itself.

2) The breaking and making of molecular bonds (MOs):

In good old fashioned OC style we’ll try and work out what happens by concentrating only on the reactive sites of the respective reactants.

The glycolate can be re-written as R₁-CH₂-OH, the indole as R₂-CH=CH-R₃.

One possible mechanism is symbolised as follows:



Bond 2 is strongly polarised due to the difference in electronegativity of C and O. The C atom is partially positively charged and is thus an electrophile.

This carbon atom now ‘snatches’ orbital 1, while the proton in that bond now becomes bonded by orbital 2, thus forming water as the leaving group.

3) Neutralisation:

After cooling the resulting mix is neutralised with HCl solution. Note that R1 = -COO^-, so we get:

-COO^- + H₃O⁺ === > -COOH + H₂O.

With the rest of the structure, that is indole-3-acetic acid which is water-insoluble and precipitates out.

Note that the reaction conditions are fairly harsh, which indicates fairly high activation energy. Tentatively we can explain this by the fact that R1, R2 and R3 are fairly bulky groups that greatly reduce the number of reactive collisions.

[Edited on 28-10-2015 by blogfast25]

Quote: Originally posted by aga

Wow ! This is actual Science APPLIED and not just random bullshit. Could this be the Beginning of SM returning to being an actual Science & Chemistry forum ?

No comment.

For some excellent MO based explanations of OC mechanisms, I can only recommend this page:

http://www.chemguide.co.uk/mechmenu.html#top

I'll be looking for some really amazing applications of QM/WM in chemistry/biochemistry that should really blow anyone's top, so stay tuned!

The Quantum Theory of Hydrogen Bonding

What?!? More stuff about the H-H bond? Haven’t we already covered that? Yes but here we will be talking about something very different but no less important.

Consider the following structures and their positions respective to each other:



The right hand structure contains an N atom with a lone electron pair (non-bonding orbital) and in principle could bond a proton by donating its non-bonding orbital (thereby acting as a Lewis base). Of course the left hand structure contains a proton.

If the (left) proton were to cross the gap between the two structures (and become bonded by the right hand lone pair), we could symbolically write this as:

N:H-----:N === > N:-----H:N, where the dotted line represents the gap between the structures.

Quantum mechanically the situation can be represented by a double potential well coupled by a barrier:



(Note that the wave function ψ here is the wave function of the proton!)

The proton can travel from one well (left N atom) to the other (right N atom) by tunnelling through the barrier (the gap).

But once it’s travelled from NH to N , symmetry dictates that it can also travel back the same way, so we can write:

N:H-----:N <=== > N:-----H:N

The wave function of the proton indicates that it has high probability to the left of the gap, also to the right of the gap but also limited probability in the gap.

This essentially described hydrogen bonding and it occurs wherever protons and unbounded electron pairs are present. Apart from the N-based example above, hydrogen bonding contributes to the comparatively high boiling points of water and simple alcohols.

Hydrogen bonding also plays an important part in the structure of ice:

http://www1.lsbu.ac.uk/water/hexagonal_ice.html

It also part explains the structure of DNA, more about that in a following segment.

[Edited on 28-10-2015 by blogfast25]

Typical Lewis bases include NH3, amines, water, hydroxide anions and many other ligands encountered in the formation of coordination complexes. NH3 for example has one lone electron pair to donate, water (potentially) has two.

Perhaps the ‘ultimate’ Lewis acid is the naked proton itself, so keen to receive a lone electron pair that it almost never occurs as a free species. In water the concentration of free protons is exceedingly small due to:

H⁺ + H₂O === > H₃O⁺

Other common Lewis acids include many metal cations because when they form coordination complexes they receive lone electron pairs from the Lewis base ligands.

Of particular interest as Lewis acids are a number of metal halides, MXn, like AlCl3, FeCl3, SnCl4, SbCl5 and SbF5 (list is far from exhaustive). Due mainly to cost reasons AlCl3 is probably the most popular of that array.

AlCl3 is one of these metal halides that rides the cusp between covalent and ionic compound (just look at its low MP and BP, for instance). It arises when Al’s sp² hybridisation forms sigma bonding MOs with Cl’s lone 3p_z electrons, forming a planar triangular structure with minimised inter-orbital repulsions. See diagram below left:



In chloride ions (Cl^- the Cl nuclei are surrounded by four full lone pair orbitals (as well as the inner electrons, of course), making the chloride ion a potential Lewis base.

A chloride ion can donate a lone electron pair to AlCl3, provided the latter undergoes a transition to an sp³ hybridisation, leaving one of the sp³ MOs empty. The chloride ion then donates one of its lone electron pairs into the empty sp³ MO and an AlCl₄^- (tetrachloro aluminate anion) is born. It has a tetrahedral structure and carries one unitary negative charge (diagram above right). Salts of it, like KAlCl₄, are known.

This property of AlCl3 (and other suitably Lewis acidic metal halides) is exploited in OC to create highly electrophilic species for use as ‘attackers’ in specific OC substitution reactions.

For example, if we take an alkyl chloride like propyl chloride (1-chloro propane, IUPAC, CH₃-CH₂- CH₂-Cl ) the chlorine atom is still surrounded by three lone electron pairs and acts as a Lewis base:

CH₃-CH₂- CH₂-Cl + AlCl₃ < === > CH₃-CH₂- CH₂+ + AlCl₄^-.

The last carbon atom now carries a positive unitary charge and we call it a carbocation or carbonium ion. Such a species is extremely electrophilic.

The case of AlCl₃ catalysed alkylation of benzene is well explained in the link below (but the choice of alkylating agent is a little odd because methyl chloride is a gas – BP = -24 C):

http://www.chemguide.co.uk/mechanisms/elsub/fcalkyl.html#top

[Edited on 30-10-2015 by blogfast25]

Quote: Originally posted by blogfast25

Lewis acid base theory ... lone electron pair.

Quote: Originally posted by aga

Quote: Originally posted by blogfast25   Lewis acid base theory ... lone electron pair. For the dimmest of us, could you please explain what is meant by 'lone electron pair'. This part of the whole Lewis thing has been confusing for ages. Is it a Pair that are alone, two seperate Lone lectrons that come together as a Pair that were Lone before meeting etc ?

A lone electron pair is a full molecular orbital (2 electrons) that is not involved in any bonding:



In the diagram the three short lines left and above/below the Br symbol symbolise the lone electron pairs.

Lewis isn't just known for his acid/base theory, he also 'invented' a notation in which electron pairs are symbolised by : (two dots), for example in this representation of water:





The pairs that aren't located between atoms are lone, unbonded electron pairs.

I refer also to an earlier post about water and the oxonium ion.

In its most basic form a Lewis acid base reaction can be represented as:

B: + A === > B-A, where the hyphen stands for a bonding MO between A and B. B is the electron pair donor (base) and A is the receptor (acid).


[Edited on 30-10-2015 by blogfast25]

Quote: Originally posted by aga

Oh ! 1) So two electrons, one upspin, one downspin in a single orbital that is Currently not involved in any bonding ? 2) So a Lewis acid/base could be a molecule that happened to have either a full orbital Not involved in some bondage, or an Empty orbital equally Not involved (although it wouldn't be, seeing as it would have to be empty) ?

Woohoo ! I got something right !

Gotta be worth another point in the scoring

Yout got to try, no matter what.

Quote: Originally posted by aga

Gotta be worth another point in the scoring

Quote: Originally posted by aga

Wow ! This is actual Science APPLIED and not just random bullshit. Could this be the Beginning of SM returning to being an actual Science & Chemistry forum ?

In light of blogfast's recent posts, I will also post a reaction mechanism and explanation to further help you apply what you're learning here. If you recall, you made a thread on the synthesis of ethyl acetate earlier this year. Since you weren't able to fully understand the synthesis back then due to your lack of organic chemistry knowledge at the time, I think now might be a good time to give it another go; however, before we do, we need to make sure that you understand a few things first:

I know this is a lot to take in, and I'm sure at least some of it you already know, but it's very important that you fully grasp this stuff before seriously attempting to make sense of the mechanism below (or any OC mechanism, for that matter). I literally intend to explain, in detail, what's going on in every single step for you, so the more you understand going into my explanation, the more you'll get out of it in the end.

A. You need to know what skeletal structures are and what they represent. In OC, molecules are drawn using the line-angle formula. This is basically just a short-hand way to draw carbon chains, as drawing them in this manner is extremely quick and convenient, especially when the chain is long. With the exception of certain functional groups, carbons and hydrogens are not shown because they are already implicit within the "lines" of the skeletal structure itself. An exception are hydrogens that are directly bonded to non-carbon atoms. In those cases, it is correct to show the hydrogens explicitly. But other than that, the only atoms that should be explicit are heteroatoms (atoms other than hydrogen and carbon). Where the lines begin and end are where the carbons are. The lines connecting those carbons to other atoms represent bonds. The number of lines indicates the number of bonds. By counting the number of bonds that are connected to a given carbon atom, the number of implicit hydrogens on it can also be determined.

B. You also need to know how many bonds each atom needs in order to become a neutral species. In this case, we're only dealing with hydrogen, carbon and oxygen:

- Hydrogen needs one bond.
- Carbon needs four bonds.
- Oxygen needs two bonds.

Thus an oxygen that only has one bond (like a hydroxide ion, OH^– becomes negatively charged because it has one more electron orbiting it than it has protons in its nucleus. Likewise, an oxygen with three bonds (like a hydronium ion, H₃O⁺ has a positive charge because it is deficient one electron. Also, keep in mind that a single bond counts as one bond, a double bond as two, and a triple bond as three. Thus an atom that needs four bonds could either have four individual single bonds, two double bonds, one double bond and two single bonds, or one triple bond and one single bond. (quadruple bonds technically exist as well, but they're not exactly common) So assuming that it's a neutral species, a carbon that is bonded to another carbon through a single bond as well as an oxygen through a double bond must then have one implicit hydrogen. Such a group is known as an aldehyde group. (just as an FYI, even though it's connected to a carbon atom, it's actually correct to show the hydrogen in an aldehyde group explicitly if you want to. personally, I always show it because aldehydes look goofy without it for some reason)

C. While I didn't bother to actually show the lone pairs on the atoms in the mechanism below, keep in mind that, like the implicit carbons and hydrogens, the lone pairs are there even though you can't see them. Remember that oxygen can form up to three bonds to give a positively-charged oxonium ion, so a neutral oxygen with two bonds still has another lone pair that can participate in bonding. Understanding this is going to be a central part of understanding the mechanism below.

D. Lastly, as I've mentioned before, the red arrows indicate the flow of electrons. This is called "electron pushing," and is how you follow reaction mechanisms in organic chemistry. The arrows begin on the electrons that are being moved, which are usually a pair of electrons that are either currently participating in a bond, or a lone pair on an atom that are being donated to another atom to form a new bond. The other end, the one with the arrowhead on it, is where the electrons are ending up. So if the arrow starts on a neutral oxygen and ends up pointing to a hydrogen atom, it means the oxygen is donating its lone pair to the hydrogen to form a new bond (which usually also involves cleaving the old bond between the hydrogen and whatever it was bonded to heterolytically). And just for future reference, there are also two kinds of arrowheads: full arrowheads and half arrowheads. A full arrowhead means that it's a pair of electrons that are being moved; a half-sized, "hooked" arrowhead means that it's a single electron being moved. For now, we will only be concerned with the former kind.

So with all of that in mind, let's now take a look at the acid-catalyzed reaction between acetic acid and ethanol to give ethyl acetate and water. I have prepared the following mechanism:



This reaction is called a Fischer esterification, and is actually rather simple. As I mentioned in your original thread back in March, this reaction is 100% reversible, so the reactants and products are in equilibrium. The reaction is acid-catalyzed both ways, so the removal of water is necessary to prevent protonation and hydrolysis of the ester back into a carboxylic acid.

Quick Note: the H–A in the mechanism above is the acid catalyst. The "H" represents the removable proton, and the "A" is the rest of the molecule. The hyphen in-between them is the bond. The negatively-charged A^– anion between steps 3 and 4 and 6 and 7 is the deprotonated acid. And while I show the A^– anion as being the base that removes the acidic protons from the protonanted intermediates, in reality it could be anything with a free lone pair (like an oxygen on a water, acetic acid, ethanol, or ethyl acetate molecule, or even an oxygen on one of the various intermediates).

So let's get started. The reaction mechanism is shown in seven steps, so I'll just go through each step one at a time:

Step 1 - In the initial step, acetic acid becomes protonated by the acid catalyst. The carbonyl oxygen (the one with the double bond) donates its lone pair of electrons to the hydrogen atom on the acid, breaking the weak H–A bond and forming a new bond between oxygen and hydrogen. This creates a positively-charged oxonium ion where oxygen has three bonds. One of the reasons that this happens is due to our old friend, resonance stabilization. Remember him? By alternating the double bonds like I've shown you before, you can change which atom has the positive charge on it. Thus the charge gets spread out across three different atoms (the carbon and both oxygens), making protonation quite favorable under acidic conditions because the charge isn't completely localized, making it somewhat weak. Also, the reason the oxygen with the double bond gets protonated and not the other oxygen (the hydroxyl group oxygen) is because if the other oxygen were protonated, resonance would not be able to get the positive charge off of it. No matter what you did, you wouldn't be able to move the charge to a different atom. Not to mention that the carbon it's attached to is already extremely electron-deficient (will explain in the next step) with a strong partial positive charge, which would create a highly unfavorable electrostatic repulsion between the two. Thus the oxygen with the double bond is what gets protonated under strongly acidic conditions.

Step 2 - Since the carbon atom in acetic acid is bonded to two highly electronegative oxygen atoms, it naturally has a strong partial positive charge on it; both oxygens are strongly pulling electron density away from carbon at all times. And while the partial positive on carbon is already quite strong even without the protonation in Step 1, when that proton gets slapped onto the carbonyl oxygen, it starts pulling electron density away from carbon FAR more strongly than it was before. Think about it: the oxygen is already highly electronegative and electron-hungry even when it's neutral, right? So how hard do you think it's going to pull with a POSITIVE charge on it? Because now you've got a nucleus that's already highly electronegative even when it has the same number of protons and electrons, only now that nucleus has become electron-deficient as well. Another way to think about it is that the hydrogen is now pulling strongly on the oxygen, which, in turn, begins pulling that much harder on carbon to make up for the loss of electron density that hydrogen is causing. This is actually an important concept in OC called the inductive effect.

And because the partial positive charge on carbon is now exceedingly strong, the carbon becomes extremely vulnerable to a nucleophilic attack by the oxygen on ethanol. In this state, the acetic acid molecule is said to be "activated." And as such, the oxygen on ethanol, being a nucleophile (an electron-hungry species with a free electron pair that can be donated to an electrophile), immediately attacks the positively-charged carbon atom, an electrophile (a species with an empty orbital that attracts nucleophiles). The attack breaks the pi bond between the carbonyl oxygen and neutralizes the positive charge by turning the carbonyl group into a hydroxyl group.

Step 3 to Step 4 - There isn't much to say about these two steps. The charged intermediate gets deprotonated by a base to give the neutral intermediate seen in Step 4. I didn't bother to draw the arrows in these steps, but I think it's fairly obvious what happens. The acid catalyst grabs the proton back and is regenerated. Then the bottom oxygen gets protonated by the acid to give another charged oxonium ion. The protonation is no different than before--a lone pair on oxygen is donated to the hydrogen on the acid, breaking the H–A bond and forming a new bond between the bottom oxygen and hydrogen. This creates the charged oxonium intermediate seen in Step 5.

Step 5 - In this step, a lone pair from the top oxygen comes down and forms a pi bond between it and carbon, breaking the bond between carbon and the bottom oxygen. Since the bottom oxygen is positively charged, like before, it is pulling very strongly on carbon. The top oxygen then donates a lone pair to the electron-deficient carbon, allowing the bottom oxonium group to leave as water. This is favorable because the leaving group, water, is a neutral species. There's no charge on it to attract it back to the oppositely-charged carbon as it tries to leave. And the protonated carbonyl group that is created when the lone pair comes down is also quite stable for the reasons mentioned in the first step (our friend, Mr. Resonance, once again), and thus its formation is favorable.

Step 6 to Step 7 - In the final step, the protonated ester gets deprotonated by a base to give ethyl acetate.

Quote: Originally posted by Darkstar

I cannot believe that I just typed all of this... Man, I better get some SERIOUS extra credit or something.

Quote: Originally posted by MrHomeScientist

This thread is full of great information. Thanks to everyone for contributing this! This brings me back to QM in college when we went over the hydrogen atom, and I had one of those epiphany moments where I saw how physics makes chemistry work. Amazing!

Quote: Originally posted by Darkstar

I cannot believe that I just typed all of this... Man, I better get some SERIOUS extra credit or something.

Quote: Originally posted by aga

Can this be stickied in any way mods, Please ? This is such amazingly good Chemistry/Physics Education that i absolutely refuse to see it buried in garbage posts.

Quote: Originally posted by aga

Quantum Tunnelling might be worth a try. I'll set up a particle with a wavefunction they cannot refuse (unless they fight back with n=0).

Quote: Originally posted by aga

Quantum Tunnelling might be worth a try. I'll set up a particle with a wavefunction they cannot refuse (unless they fight back with n=0).

Quote: Originally posted by aga

OC seems a Lot less impenetrable now.

We will get you there eventually; however, I noticed that you said, "maybe phenyl means phenol-shaped-bit-stuck-to-another-bit" in your competence advice thread. That is simply unacceptable. We must address this issue immediately. So it is time now for a brief introduction to moieties and functional groups. I drew some random molecules with a bunch of functional groups on them and wrote their names next to them. I also threw in a little refresher on skeletal structures as well. I hope you're not color blind...

Thanks, Darkstar. Very clear presentation.

Answers to Questions 3 and 4 tomorrow.

The questions for next week:

Question 5: (for 10 points)

Predict the geometrical shape of the following molecules:

a) Xenon difluoride, XeF₂.

b) The pentafluoride xenonate anion, XeF₅^-.

Question 6: (for 10 points)

Nitrogen difluoride (NF₂ has two isomers, as shown below, left trans-NF₂, right cis-NF₂:



Briefly explain why two such isomers exist.

[Edited on 1-11-2015 by blogfast25]

Quote: Originally posted by Darkstar

a brief introduction to moieties and functional groups. I drew some random molecules with a bunch of functional groups on them and wrote their names next to them. I also threw in a little refresher on skeletal structures as well. I hope you're not color blind...

Answers to Questions 3 and 4:

A.3:



All three wave functions are zero at x < or = 0 because there V = +∞.

For the ground state ψ₀ the wave function (red) takes on a sine (or bell) shaped form with no zero-points, until the potential V > E₀, then the wave function tends to 0 for x tending to +∞.

For the first excitation ψ₁ the wave function (blue) must exhibit one zero point as shown, then when the potential V > E₁, then the wave function tends to 0 for x tending to +∞.

For the second excitation ψ₂ the wave function (green) must exhibit two zero points as shown, then when the potential V > E₂, then the wave function tends to 0 for x tending to +∞.

This particular potential shape is a crude approximation of the often encountered Morse Potential.

A.4:

There is only one naphtalene and one 2,3 dichloro naphthalene due to resonance.

Armed with Darkstar's Guide to OC Magik i've picked these three chem names to decipher (from the today's posts list)

#1 - 1,3 propanediol

#2 - Benzotriazole

#3 - Triethylamine

Had to look up what 'iol' meant, and found that 'diol' means two Hydroxyl (OH) groups. The 1,3 are the carbon positions, and it's so short there can't really be much question as to which number goes where.



With Benzotriazole, not sure if this is right, seeing as it's not 'triazo' so perhaps the 'le' means something more.

Also unclear why it's not called 'Phenotriazo' seeing as the ring looks the same.



Triethylamine was trickier.

Tried a string of ether segments in a chain with the amine on the end before looking up Nitrogen on ptable.com to find it wants 3 bonds, so the 'tri' kinda means it chucks the H for C, although that makes it not 'amine' as per the clue sheet.

The 'ethyl' instead of 'ether' also makes me suspect that this one is very wrong.



Edit:

Oops. Put Nitriles on the Benzo instead of Azos.

[Edited on 2-11-2015 by aga]

Quote: Originally posted by aga

Quote: with no zero-points, until the potential V > E0, then the wave function tends to 0 for x tending to +∞ Confused. The E0 waveform never gets anywhere near V=0, unless '0' means zero deviation from the ground state, in which case it starts there, so has at least 2 (start and where it goes flat). Are the bits past the right hand box wall tunnelling ?

Look at the reddish horizontal line (E₀, at some point it (call it x₀ crosses the V(x) line. From then on (for x > x₀ V(x) > E₀. In that region we get tunnelling. The wave function decays exponentially to ψ = 0 for larger and larger x.

The same holds for the blue and green cases but their 'crossing points' lie at higher values of x.

[Edited on 3-11-2015 by blogfast25]

Quote: Originally posted by aga

Armed with Darkstar's Guide to OC Magik i've picked these three chem names to decipher (from the today's posts list) #1 - 1,3 propanediol #2 - Benzotriazole #3 - Triethylamine

Slightly smoother version of 1,3 propanediol and triethylamine (no oxygens):



[Edited on 2-11-2015 by blogfast25]

Further confused !

Where did the oxygens go ?

Edit:

Triethylamine should be made on the Isle of Man



[Edited on 2-11-2015 by aga]

Quote: Originally posted by aga

Further confused ! Where did the oxygens go ?

Here are the molecules in question:



The first one, 1,3-propanediol (under IUPAC nomenclature rules, it would technically be propan-1,3-diol), you got right; however, it will usually be drawn like above.

The second one, benzotriazole, is a lot harder since we have covered neither nomenclature rules nor heterocyclic compounds. The molecule is basically benzene fused with 1,2,3-triazole.

The last one, triethylamine, is a nitrogen with three ethyl (not ether!) groups on it. The ethyl groups are just ethane molecules with one less hydrogen on them. The eth- prefix means that it's a chain with two carbons, and the -yl suffix means that a hydrogen has been replaced with something else.

Unfortunately, I'm a little pressed for time at the moment, so I can't elaborate further. I'll probably do a section later today covering some basic nomenclature rules and show you some of the common heterocylics. It's awesome that you're at least trying to apply what you're learning, though. Your teachers are very proud!

EDIT - Fixed a mistake I made in 1,3-propanediol's IUPAC name. Was in a hurry when making the image and just copied and pasted the "1,3-propanediol" while I was adding the text below the molecules in ChemDraw. It is all fixed now.

[Edited on 11-2-2015 by Darkstar]

Quote: Originally posted by aga

Tried a string of ether segments in a chain with the amine on the end before looking up Nitrogen on ptable.com to find it wants 3 bonds, so the 'tri' kinda means it chucks the H for C, although that makes it not 'amine' as per the clue sheet.

It might also help aga to understand that skeletal notations do have another bearing in reality.

Have a look at the diagram below. Left are the ball-and-stick representations of the first 3 alkanes and right their skeletal representations:



Note that the bond angles in the ball and stick models are to scale (true angles) and that the skeletal representation is essentially the projection of the C nuclei (the hydrogens aren't shown unless necessary) and the lines representing the bonding orbitals onto a horizontal plane, see below for butane:



Which will bring me to a later subject: molecular conformations.

[Edited on 3-11-2015 by blogfast25]

Quote: Originally posted by blogfast25

Which will bring me to a later subject: molecular conformations.

Quote: Originally posted by Darkstar

Speaking of stereoisomerism, I was actually thinking about doing a brief introduction on chirality soon. So maybe you do conformers and I'll do enantiomers? And then whoever's up for it can do one on cis-trans isomerism and finally one on how to name all of these different types of stereoisomers using R/S and E/Z notation.

Molecular conformations:

Isomers:

Two chemical compounds are said to be isomers is they have the same molecular formula but a different structure. Consider the simple case of n-butane (“n” for “normal”, left) and isobutane (IUPAC: methyl propane, right):



Both have the empirical formula C₃H₈, yet structurally they are different. Both compounds have indeed slightly different physical and chemical properties.

Now with general alkanes – C_nH_2n+2 – we can have a lot of fun.

Take a look at some isomers of heptane C₇H₁₆:



Top: n-heptane
Middle: 2-methyl hexane and 3-methyl hexane
Bottom: 2,4-dimethyl pentane

Now imagine writing all the isomers of decane – C₁₀H₂₂ – or higher C. Thanks to carbon’s unparalleled ability to form covalent bonds with itself and hydrogen (and other elements), isomers are very common in organic chemistry.

Molecular Conformations:

Let’s go back to these ball-and-stick models of the first three alkanes:



These models suggest a certain rigidity that isn’t really there. Due to the nature of covalent single σ bonds, the carbon nuclei (and the atoms bonded to them) can be rotated around the inter-nuclear axis, leading to a number of different shapes, known as conformations. The diagram above thus really shows the molecules in only one of their conformations.

The principle of conformations is very nicely explained (in under two minutes!) in the video below:

https://www.youtube.com/watch?v=yBtMjTZrMFc

It might be tempting to see molecular conformations of the same molecule also as isomers but that’s not really true.

A particularly interesting case is cyclohexane, in skeletal notation depicted as a regular hexagon. But due to bond angles that’s far from a realistic representation:



The source page contains a java script that doesn’t work on my tooter but it might on yours.

The energy levels are caused by various ‘stresses’ introduced into the molecule as it transforms from ‘left chair’ to ‘right chair’. As the chair conformation is the lowest energy level liquid cyclohexane can be expected to be made up of mostly that conformation.

Finally, and to get Darkstar on his way to expand on chiral isomers, an interesting bicyclo compound, named alpha pinene:



A ‘distant cousin’ of cyclohexane, alpha pinene has a carbon bridge connection that creates a secondary cyclic structure (hence “bicyclo”). It has two permanent and distinct main conformations (termed (+) and(-)) that are mirror images of each other. The two forms are so-called enantiomers.


[Edited on 3-11-2015 by blogfast25]

Quote: Originally posted by aga

Hmm. Watching the video, when he twisted the left hand part (as seen on screen), that brought the hydrogens on the left-hand end black ball into much closer proximity to those attached to the green ball on the right. Presumably all molecules take on a stable form based on where the electronegativity and bonding orbitals of all the atoms are best balanced ? Assuming further, given the sheer size of some of these OC molecules, the distance between some atoms (e.g. at either end of a long chain) would make those charges irrelevant - they;d never be able to get close enough to make a difference.

Sounds fair.

Send me the bill when i'm demonstrably as knowledgeble as both of you

@aga

Sometime later today when I have a little more time I'll post your next lesson. For now, it is time for your very first OC pop quiz. I have prepared the following for you:



The reaction above is an example of base-catalyzed ester hydrolysis. In this reaction, ethyl acetate is hydrolyzed using aqueous NaOH to produce sodium acetate (a salt) and ethanol. This reaction is the opposite of esterification (which we looked at earlier) and is the basis of soap-making. Triglycerides (fats) are essentially glycerin with ester groups instead of hydroxyl groups. Since I remember you saying that you have done saponification before, I figured this reaction would be a good one to show you next. So based on what you've learned thus far, here's what I want you to tell me:

Question 1 - Identify the electrophile and nulcleophile in step one of the mechanism and the electrophile and nulcleophile in step three of the mechanism. It is absolutely critical that you grasp this concept. Being able to properly differentiate between electrophiles and nucleophiles is quite possibly the single most important aspect of understanding OC reactions!

Hint: electrophiles are electron-deficient and consequently attract nucleophiles, which are electron-hungry.

Question 2 - Briefly explain what is going on in each step of the mechanism. The level of depth I went into in my post on the Fischer esterification is not necessary. One or two sentences per step describing what is reacting with what (and preferably why and how) is all that I am looking for.

Extra Credit - For extra credit, tell me why this reaction is more or less irreversible under basic conditions. Notice that, while I used equilibrium arrows in the mechanism steps to show that they are reversible, in the overall reaction at the top, a normal arrow is used instead. This is because, after the third and last step of the mechanism, the reaction can no longer be undone without first lowering the pH. So in other words, tell me why the reaction is no longer reversible after the final step of the mechanism.

Hint: why is it reversible before the ethoxide ion is protonated but not after? An ethoxide ion is ethanol without a proton on its hydroxyl group, by the way.

Quote: Originally posted by blogfast25

Oooopsie, a typo in Question 6

Sheesh ! i almost forgot there was another question and it's mid-week already !

Edit:

WTF ?!?!

As soon as i finished typing that there's two new Darkstar Questions !

Right.

Game On.

[Edited on 4-11-2015 by aga]