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Quantum Mechanics/Wave Mechanics in Chemistry Beginners Thread

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gdflp - 3-12-2015 at 20:32

Has the concept of conjugation and UV absorption due to conjugated bonds been explained yet in this thread, or does that need to be covered as well? This seems like a good time to do so.

gdflp - 3-12-2015 at 23:53

<b>Conjugation</b>

Conjugation occurs when there are three or more sp<sup>2</sup> or sp hybridized atoms adjacent to each other. Another way of looking at it is that there must be three adjacent atoms which each contain at least one unhybridized p-orbital. These adjacent p-orbitals tend to overlap as shown by the following diagram:

p-orbital overlap.jpg - 23kB

This overlap of p-orbitals allows the pi electrons in the compound to be delocalized and spread across the p-orbitals of all of the conjugated atoms, rather than being confined to a single atom. Thus, conjugated systems are extremely stable and also exhibit certain special properties. One example of this is benzene, despite being an alkene, it very rarely behaves like one due to the even spread of electrons throughout the molecule. Unlike the Kekule structure suggests, all of benzene's bonds are equivalent, not alternating double and single bonds; hence the unexpected behavior. This unusual behavior is present for most aromatic systems, all of which are conjugated systems.

Conjugated systems also lead to interesting behavior in regards to light absorption. As I believe you know from one of blogfast's lectures(I hope that it's been explained) atoms can absorb light by promoting electrons to higher, otherwise unused, orbitals. These electrons, which are temporarily in this higher energy state, are called <b>excited</b> electrons. These excited electrons will then eventually drop back down to their original lower energy state due to the preferential increase in entropy. The energy which was absorbed thus needs to be released, and this is done through emitting heat, lower wavelengths of light(such molecules are called fluorescent dyes), or another form of energy. In conjugated systems, the delocalization of the electrons allows them to be promoted to an excited state by light of a lower energy. Since wavelength is inversely related to energy, conjugated compounds will absorb light of a higher wavelength than unconjugated compounds. In some highly conjugated compounds, this wavelength becomes so large it reaches the visible spectrum; thus forming one class of organic dyes.

A molecule can have more than one conjugated system, consider the following example: undeca-1,3,5,8,10-pentaene.

Undecapentaene.jpg - 6kB

It contains two conjugated systems, one of which has four adjacent carbons with p-orbitals, the other contains six adjacent carbons with p-orbitals. When determining the wavelength of light which a conjugated compound will absorb, it is based upon the number of adjacent conjugated atoms present in the largest conjugated system, not the total number of conjugated atoms. Thus a compound such as triphenylmethane, with three smaller isolated conjugated systems, will absorb light of a lower frequency than anthracene, which contains one large conjugated system.

This concept can be used to explain both the acid-base and variable coloration of phenolphthalein.

Firstly, let's look at the structure of phenolphthalein in neutral or mildly acidic solution. There are three acidic protons, the two on the hydroxyl groups, and the carboxylic acid(they may or may not be shown as free protons depending on the resonance structure, but they're all there). In a basic environment, one proton is going to be removed first preferentially, which will be the carboxylic acid. Though both the deprotonated hydroxyl and carboxylate can delocalize electrons into the aromatic ring, the carboxylate can also delocalize electrons onto the carbonyl, making that proton the most acidic.

The second deprotonation is where it becomes interesting however. When a proton is removed from one of the hydroxyls, there are two resonance structures of this di-deprotonated phenolphthalein molecule; both of which are highly conjugated as shown below, connecting all three aromatic rings. This large number of adjacent conjugated atoms thus lower the energy required to promote electrons enough that phenolphthalein becomes highly colored in basic solution. Due to the high stability of the conjugated structure, the second proton is quite acidic as well and both deprotonations are complete at a pH of around 8.2

Conjugation Phenolphthalein-1.jpg - 14kB

On the other hand, phenolphthalein is also strongly colored at very low pH's, below 0. This is due to the structure of the protonated form of phenolphthalein, shown below. This structure is also quite highly conjugated, though slightly less so; hence the color being a slightly smaller wavelength. The key here is that a carbocation is sp<sup>2</sup> hybridized, thus it contributes to a conjugated system as well.

Conjugation Phenolphthalein-2.jpg - 13kB

As always, let me know if you have any questions.

Blogfast, please feel free to correct me or expand on any points which I may have missed.

[Edited on 12-4-2015 by gdflp]

Darkstar - 4-12-2015 at 05:38

@aga

I hope you've been paying attention this semester, because I've written your final exam! The test is 25 questions plus an extra credit question. Just say the word when you're ready and I'll post it.

blogfast25 - 4-12-2015 at 08:33

Wow, thank you gdflp and Darkstar.

Darkstar, are you aware this forum is MathJax/LaTex enabled?

At the risk of having to move aga into a sensory deprivation tank for a week after this onslaught, here's why pH indicators change colour, in addition to gdflp's explanation of light absorption by conjugated structures.

pH indicators are weak acids that show different colours depending on whether to molecule is protonated or not, as so completely elucidated by gdflp. I'll represent the protonated and deprotonated forms resp. as:

$$\mathrm{HIn},\:\mathrm{In^-}$$

In water, the indicator deprotonates weakly according:

$$\mathrm{HIn}+\:\mathrm{H_2O} \leftrightarrow \mathrm{In^-}+ \:\mathrm{H_3O^+}$$

and:

$$K_{HIn}=\frac{C_{H3O}C_{In}}{C_{HIn}}$$

Bear in mind that K<sub>HIn</sub> is really small, typically in the range of 10<sup>-3</sup> to 10<sup>-11</sup> or so.

We can rearrange the last expression slightly as follows:

$$\frac{K_{HIn}}{C_{H3O}}=\frac{C_{In}}{C_{HIn}}$$

Also remember that:

$$pH=-\log{C_{H3O}}$$

And:

$$pK_{HIn}=-\log{K_{HIn}}$$

We can now see that if:

$$pH=pK_{HIn}$$

... the ratio of protonated to deprotonated indicator equals 1! So if both species have different colours, at that pH the colour will be in between the colours of the species.

But if:

$$pH < pK_{HIn}$$

... then the protonated species will dominate and the colour will be that of HIn.

And if:

$$pH>pK_{HIn}$$

... then the deprotonated species will dominate and the colour will be that of In<sup>-</sup>.

In summary we can say that:

$$pH=pK_{HIn}$$

... is the turning point of the indicator.

[Edited on 4-12-2015 by blogfast25]

blogfast25 - 4-12-2015 at 09:15

Quote: Originally posted by gdflp  
<b>Conjugation</b>

Conjugation occurs when there are three or more sp<sup>2</sup> or sp hybridized atoms adjacent to each other. Another way of looking at it is that there must be three adjacent atoms which each contain at least one unhybridized p-orbital. These adjacent p-orbitals tend to overlap as shown by the following diagram:



gdflp, is there perchance a diagram missing?

gdflp - 4-12-2015 at 09:16

Quote: Originally posted by blogfast25  
Quote: Originally posted by gdflp  
<b>Conjugation</b>

Conjugation occurs when there are three or more sp<sup>2</sup> or sp hybridized atoms adjacent to each other. Another way of looking at it is that there must be three adjacent atoms which each contain at least one unhybridized p-orbital. These adjacent p-orbitals tend to overlap as shown by the following diagram:



gdflp, is there perchance a diagram missing?

Funny you should mention that, I just noticed that and added it a few minutes ago.

blogfast25 - 4-12-2015 at 09:22

gdflp:

A point that needs to be stressed with regards to that orbital overlap (aka resonance) is that these π MOs do not in some way "merge", as that would constitute a gross violation of the Pauli Exclusion Principle, as pointed out higher up in the seminar.

[Edited on 4-12-2015 by blogfast25]

Pok - 4-12-2015 at 09:45

Wouldn't the "Beginnings" forum be a better place for this thread? People interested in the "Chemistry in General" forum are not necessarily interested in this thread.

gdflp - 4-12-2015 at 10:05

The discussion level of this thread is significantly beyond that of <b>Beginnings</b>. Considering that this started out as a quantum mechanics thread, the placement in <b>Chemistry in General</b> was entirely appropriate. Now, considering the shift in the subject matter, it could fit in <b>Organic Chemistry</b> just as well, but there's no reason to move it.

blogfast25 - 4-12-2015 at 10:08

Quote: Originally posted by Pok  
Wouldn't the "Beginnings" forum be a better place for this thread? People interested in the "Chemistry in General" forum are not necessarily interested in this thread.


I'm not sure how people who are interested in general chemistry cannot be interested in this thread, pok. Chemistry is quantum chemistry, period.

Also, the beginner's thread is mainly for unreferenced material. And I can't see many total beginners being interested in QM/QC.

blogfast25 - 4-12-2015 at 10:10

Quote: Originally posted by gdflp  
Now, considering the shift in the subject matter, it could fit in <b>Organic Chemistry</b> just as well, but there's no reason to move it.


I've been contemplating asking moderators to splice off the part that follows the main QM/QC thread but like you I can't really see how that improves things.

aga - 4-12-2015 at 10:50

Quote: Originally posted by Darkstar  
@aga

I hope you've been paying attention this semester, because I've written your final exam! The test is 25 questions plus an extra credit question. Just say the word when you're ready and I'll post it.

Please !

Can't say i'm 'ready' per-se, however fire away.

aga - 4-12-2015 at 11:05

Quote: Originally posted by Pok  
Wouldn't the "Beginnings" forum be a better place for this thread? People interested in the "Chemistry in General" forum are not necessarily interested in this thread.

Beginnings would be fine with me, however In-depth Quantum Mechanics and Organic Chemistry reaction mechanisms don't fit very well in Beginnings.

Given the sheer Range of the material discussed, it is rather 'General Chemistry' in essence.

This thread can simply be ignored by anyone who finds the material of no interest, same as any other thread.

Appologies if this thread pops up to the top of the list too often for your liking.

aga - 4-12-2015 at 11:22

Quote: Originally posted by gdflp  
Conjugation occurs when there are three or more sp<sup>2</sup> or sp hybridized atoms adjacent to each other.

Another way of looking at it is that there must be three adjacent atoms which each contain at least one unhybridized p-orbital. These adjacent p-orbitals tend to overlap

So, essentially the electrons in the bonding orbitals have 'options' on which orbital to occupy next, regardless of which nucleus that orbital is bound to ?

I can imagine an arrangement where a nearby electron could tip the charge balance causing an electron to 'hop', starting the cycle.

In a molecule like Benzene, if the electrons are truly swapping places continuously, surely that would create some magnetic field effect, albeit tiny.

Is this something that is measurable ?

[Edited on 4-12-2015 by aga]

gdflp - 4-12-2015 at 11:36

Yes, the electrons have some freedom to move around in the p-orbitals.

And not that I am aware of. In conjugated systems such as benzene the electrons are free to circulate, but that doesn't necessarily mean that they flow in one direction. This would create an electrical potential and thus require some input of energy, otherwise it would violate the conservation of energy. When aromatic rings are placed in strong magnetic fields however, a current is produced, called the <b>aromatic ring current</b>. This current has an effect on certain things, most notably in NMR(Nuclear Magnetic Resonance) spectra.

[Edited on 12-4-2015 by gdflp]

Pok - 4-12-2015 at 11:43

Quote: Originally posted by blogfast25  
I'm not sure how people who are interested in general chemistry cannot be interested in this thread, pok. Chemistry is quantum chemistry, period.

Yes. And chemistry is physics. But I personally am not interested in physical chemistry and this thread deals with theoretical aspects.

The "Chemistry in General" forum deals with practical aspects, not theoretical considerations. That's the difference.


@gdflp: If you have a questions-answer thread this is something like a "seminar" and thus indeed belongs to the beginnings. Since there is no "theoretical/physical chemistry" forum another option would be the "Miscellaneous" forum.

gdflp - 4-12-2015 at 12:03

Quote: Originally posted by Pok  

Yes. And chemistry is physics. But I personally am not interested in physical chemistry and this thread deals with theoretical aspects.

The "Chemistry in General" forum deals with practical aspects, not theoretical considerations. That's the difference.


@gdflp: If you have a questions-answer thread this is something like a "seminar" and thus indeed belongs to the beginnings. Since there is no "theoretical/physical chemistry" forum another option would be the "Miscellaneous" forum.

Why do you believe that <b>Chemistry in General</b> deals with only practical aspects of chemistry? A lack of theoretical topics on this forum as a whole doesn't mean that they don't have a rightful and important place, despite their rarity. It seems as though you want this topic moved because you simply lack an interest in it; evidently there are others who feel differently based on the 13,000+ views. You also don't visit this forum too often, I don't necessarily think that you are as familiar with the placement of topics here. In the end, it's the mods decision and they have evidently felt that its placement is just fine.

aga - 4-12-2015 at 12:05

Quote: Originally posted by Pok  
... I personally am not interested in physical chemistry and this thread deals with theoretical aspects.

The "Chemistry in General" forum deals with practical aspects, not theoretical considerations. That's the difference.

Are the practical aspects of the (currently) failed phenolpthalein synthesis not sufficient to qualify as 'practical' ?

Also gdflp's explanation of the theory behind what Should happen, are they not relevant to the Practical Application of the theory ?

If more Practical Applications and documented reactions using the discussed theories were to be posted (as was My plan, at least) would that bring this thread back under the Pok radar ?

I smell sour grapes somewhere nearby.

[Edited on 4-12-2015 by aga]

blogfast25 - 4-12-2015 at 12:30

Quote: Originally posted by Pok  


The "Chemistry in General" forum deals with practical aspects, not theoretical considerations. That's the difference.



Says who? You?

Now please go bore someone else with your pigeon holing. :(

There's was always going to be some *diot having to bring up the 'correct placement' of the thread. Sadly, it was you.
Quote:
But I personally am not interested in physical chemistry and this thread deals with theoretical aspects.


Tough shit. Don't read it then.

[Edited on 4-12-2015 by blogfast25]

Pok - 4-12-2015 at 12:36

@gdflp: I said that this forum (not the topic) deals with practical aspects. I don't think that you have any idea how often I visit the forum. I'm interested in the "chemistry in general" forum because there are alway new things. But this thread is just a summary of course book, nothing new. I have to scroll down every time due to the long list of "sticky" topics and this topic here to find the new and interesting topics.

I didn't want to delete this thread but just to get it moved. But okay, I will try to ignore it and scroll down every time like I'm used to ignore boring pop-ups.

aga - 4-12-2015 at 12:37

All this pointless Blart is digressing from the far more interesting subject matter of actual Chemistry, both theoretical and Practical.

Please take all Forum related issues to the "Forum Matters" topic.

Thanks.

blogfast25 - 4-12-2015 at 12:40

Quote: Originally posted by Pok  
I have to scroll down every time due to the long list of "sticky" topics and this topic here to find the new and interesting topics.



You lead a hard life.

You're making a fool of yourself, pok.
Quote:
I'm interested in the "chemistry in general" forum because there are alway new things.


Now you've just made me p*ss myself with laughter: this forum, whatever the section, drowns in duplicates, 'kewls', crap and general flotsam and jetsam.

[Edited on 4-12-2015 by blogfast25]

aga - 4-12-2015 at 12:54

Quote: Originally posted by blogfast25  
this forum, whatever the section, drowns in duplicates, 'kewls', crap and general flotsam and jetsam.

How DARE you !?!?

Not ALL of my posts are utter garbage ;)

blogfast25 - 4-12-2015 at 12:55

Quote: Originally posted by aga  


In a molecule like Benzene, if the electrons are truly swapping places continuously, surely that would create some magnetic field effect, albeit tiny.

Is this something that is measurable ?



Remember that electron movement in quantum systems is completely non-Classical.

In the older Bohr model of the hydrogen atom the electron did orbit around the nucleus. One defect that Bohr et al were very aware of was that this would lead to the electron constantly losing energy (synchrotron radiation) and thus inevitably spiral into the nucleus.

That's, simply put, where Schrodinger/de Broglie came into it: electrons don't move in orbits, they have quantum states. No magnetic field due to classical orbiting. But quantum systems often do have angular momentum and thus a magnetic dipole. They cause the various forms of macroscopic magnetism in materials.


[Edited on 4-12-2015 by blogfast25]

aga - 4-12-2015 at 13:13

Quote: Originally posted by gdflp  
however, a current is produced, called the <b>aromatic ring current</b>. This current has an effect on certain things, most notably in NMR(Nuclear Magnetic Resonance) spectra.

So, theoretically, inducing a strong magnetic field in any reaction where a 'ring current' may be present, the reaction could be influenced by the magnetic field.

I have Sodium Benzoate and some Huge Nd magnets.

Suggestions for a Practical reaction that generally goes One route, and might go another under such circumstances ?

Perhaps a large magnetic field plus electrolysis.

gdflp - 4-12-2015 at 13:38

That's an interesting idea, I've never heard of a reaction being affected by ring currents. Using a Helmholtz coil would produce a significantly more constant and stable magnetic field, rather than a fixed magnet. I can't think of any reactions right which would necessarily be worth a try, perhaps blogfast has an idea.

aga - 4-12-2015 at 13:46

Being a mere novice, i have no idea.

Is there an OC reaction (involving resonance/ring structures) that takes a particular pathway/mechanism every single time, despite the possibility of other pathways ?

If so, then some experiments may be worthwhile.

blogfast25 - 4-12-2015 at 14:09

Quote: Originally posted by aga  

I have Sodium Benzoate and some Huge Nd magnets.

Suggestions for a Practical reaction that generally goes One route, and might go another under such circumstances ?

Perhaps a large magnetic field plus electrolysis.


Erm... have you seen the magnets on these NMR gismoz:

https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance_spe...

:D

IrC - 4-12-2015 at 14:12

This last page has me wondering, does this mean those gas magnet inventions may actually have some basis in theory for the claimed ability to increase combustion efficiency (ring currents).


[Edited on 12-5-2015 by IrC]

aga - 4-12-2015 at 14:35

With/without magnetic field experiments on OC with ring structures is kinda the idea ...

Do not be too hard on Pok.

He's not posted any new topics recently, so it may just be jealousy.


[Edited on 4-12-2015 by aga]

aga - 4-12-2015 at 14:44

Quote: Originally posted by blogfast25  
Erm... have you seen the magnets on these NMR gismoz:

Inefficiency is not impressive.

blogfast25 - 4-12-2015 at 14:53

Quote: Originally posted by IrC  
This last page has me wondering two things. One is does this mean those gas magnet inventions may actually have some basis in theory for the claimed ability to increase combustion efficiency (ring currents).


I don't really get your question. 'Splain?
Quote: Originally posted by aga  
Quote: Originally posted by blogfast25  
Erm... have you seen the magnets on these NMR gismoz:

Inefficiency is not impressive.


The magnetic fields needed to do NMR are far, far higher than your neomags can deliver.

[Edited on 4-12-2015 by blogfast25]

aga - 4-12-2015 at 15:05

Quote: Originally posted by blogfast25  
The magnetic fields needed to do NMR are far, far higher than your neomags can deliver

Of course.

Now consider why that would be the case.

Why would an Enormous magnetic field be required to affect the passage of a single atom, when the self-evident fact is that a much much weaker magnetic field can stick a magnet to Iron so hard that a human lacks the strength to remove it ?

QM was invented to solve Other problems.

IrC - 4-12-2015 at 15:35

"I don't really get your question. 'Splain?"

I have read many articles on those devices sold to apply a strong local magnetic field to gas going into an engine, both pro and con. Some of the pro articles did quite a bit of testing MPG over the same track both with and without the device installed. Since it sounds as if what is being said was molecules with ring currents can be affected by magnetic fields, it makes me wonder if similar to a catalyst altering the activation energy level, or some type of molecular rearrangement may be induced by the field which improves the combustion process. Up to this point I had always considered the magnet on the fuel line to be more or less snake oil. This ring current discussion where magnetic fields may cause some change sparked my memory of looking into those devices decades ago. Is it possible the oil may not all be snakes or could some process little understood and yet to be researched fully actually be involved? In other words should I buy snake traps or N52 magnets?

blogfast25 - 4-12-2015 at 15:56

Well, according to the SE the magnetic dipole moment for a hydrogen atom in the eigenstate (n,l,m) is quantised and given by the following equations:


$$\mu_L=-g_L\frac{\mu_B}{\hbar}\langle \Psi_{n,l,m}|L|\Psi_{n,l,m} \rangle=-\mu_B\sqrt{l(l+1)}.$$

$$(\mu_L)_z=-\mu_B m.$$

$$n=0,1,2,3,...$$

$$l=0,1,..., n-1$$

$$m=-l,-l+1,...,0,...,+l-1,+l$$

$$\mu_B=\frac{e\hbar}{2m_e}=9.274 \times 10^{-24}\:\mathrm{JT^{-1}}$$

(μ<sub>B</sub>;)<sub>z</sub> is the z component and μ<sub>B</sub> the Bohr magneton.

To this then has to be vector-added the intrinsic spin magnetic dipole moment of an electron (spin 'up', spin 'down', remember?).

So we're talking pretty small numbers here.

[Edited on 5-12-2015 by blogfast25]

blogfast25 - 4-12-2015 at 16:01

Quote: Originally posted by IrC  
"I don't really get your question. 'Splain?"

I have read many articles on those devices sold to apply a strong local magnetic field to gas going into an engine, both pro and con. Some of the pro articles did quite a bit of testing MPG over the same track both with and without the device installed. Since it sounds as if what is being said was molecules with ring currents can be affected by magnetic fields, it makes me wonder if similar to a catalyst altering the activation energy level, or some type of molecular rearrangement may be induced by the field which improves the combustion process. Up to this point I had always considered the magnet on the fuel line to be more or less snake oil. This ring current discussion where magnetic fields may cause some change sparked my memory of looking into those devices decades ago. Is it possible the oil may not all be snakes or could some process little understood and yet to be researched fully actually be involved? In other words should I buy snake traps or N52 magnets?


Oh yeah, I remember a bit of brouhaha around these. I'm not an expert by a long shot on these issues IrC, but my money is on smelling snake oil. And I never buy into 'it wos Big Oil wot killed it' type of arguments.

I've never really heard of magnetic fields significantly affecting reaction outcomes at all (but there's a first time for everything, maybe?)

[Edited on 5-12-2015 by blogfast25]

blogfast25 - 5-12-2015 at 07:06

And here's a suggestion for an 'aga research project' in OC...

Firstly, you could just do some off-the-shelf synths: that's safe but a bit boring.

Or you (we) could do something a bit more off the beaten track with potential usefulness.

Here's the proposal.

Preparation of some saturated tertiary alcohols from α-pinene:

α-pinene is a fairly OTC material (natural turpentine contains quite a lot of it) with a skeletal structure that is found in many natural derivatives (the result of evolutionary biology, no doubt).

Below are two α-pinene derivatives (4 and 6) that would be very well worth testing as catalysts in the KOH/Mg reduction reaction (see pok's sticky thread).

alpha pinene derivatives.gif - 12kB

Preparing them from α-pinene seems doable but definitely not run-of-the-mill and with no guarantee of success.

The project would involve:

* Literature research (precursor and intermediates properties, recipes etc)
* Reactions
* Separations
* Characterisation of intermediates and end-products
* Reaction mechanism hypotheses

Needless to say: it'd keep you/us busy for a while!

aga, don't feel bamboozled into anything and make a free choice.

If you're interested, rest assured there will be plenty of assistance from the community.

Any questions?

PS: ultimate blackmail argument: 'you'd be giving something back to the community!' :D;)

[Edited on 5-12-2015 by blogfast25]

aga - 5-12-2015 at 08:40

Cor blimey guv'ner !

blogfast25 - 5-12-2015 at 09:50

Quote: Originally posted by aga  
Cor blimey guv'ner !


An under-whelming response?

1 to 2 (one step), 2 to 3 (one step) and 3 to 4 (one step) are relatively straightforward. 3 to 5 is a bit of an unknown.

Anyroads, think it over. Perhaps other members may have an opinion too...

[Edited on 5-12-2015 by blogfast25]

aga - 5-12-2015 at 11:26

You reckon i can get anywhere near doing that chain of reactions with a > 0% probability of achieving any product at all ?

Over-awed and disbelieving is the response !

Turpentine, hmm, seeing as it's already bonkers, how about starting with pine resin ?

(lots of pine trees up in the mountains).

Edit:

Turning a Pine smell into a Lemon smell would be fun.



[Edited on 5-12-2015 by aga]

gdflp - 5-12-2015 at 11:50

The only issue I see with the suggested chain is the hydrogenation. They're a pain to do in an amateur setting as many of them need pressurized hydrogen; at the very least some pricey transition metal catalysts are necessary (Pd/C). It would be a good endeavor to gain practical experience though.

aga - 5-12-2015 at 12:00

How much hydrogen and how high a pressure ?

Blogger's reaction sequence obviously <i>has</i> to be done, now it's been suggested.

gdflp - 5-12-2015 at 12:35

If using Pd/C or Pt/C as a catalyst, you should be able to run the hydrogenation at atmospheric pressure, but you will need to generate dry hydrogen. Borohydride and acid is the standard in professional labs, but I find it a waste of borohydride. The reaction of a metal such as iron with an acid will generate hydrogen at a reasonably consistent rate; if the hydrogen is passed through a drying tube it should be of sufficient quality for this reaction. The easiest way to transfer the hydrogen is via a balloon, so no trouble there. Once you get to that phase of the reaction, we can work out a more precise procedure based on what equipment you have. If you decide to use Raney nickel as the catalyst, which can be prepared using relatively simple aqueous chemistry, you will need a hydrogen cylinder and equipment which you trust to be under at least several atm of hydrogen pressure, not a great situation if the glass fails.

aga - 5-12-2015 at 12:35

@bloggers: is the 2-(4-methylcyclohexyl)propan-2-ol (item #4) a <i>desired</i> product in this sequence, or is it a possible by-product that should be minimised ?

aga - 5-12-2015 at 12:42

Quote: Originally posted by gdflp  
will need a hydrogen cylinder and equipment which you trust to be under at least several atm of hydrogen pressure, not a great situation if the glass fails.

Does it have to be Glass ?

For high pressure/temperature i'd rather do some standard plumbing with steel.

Would the iron or other components of steel interfere ?

Edit:

"Pd/C or Pt/C" i do not know what that means.

Palladium or Platinum and Carbon is some form is the only sense i can make from the terms.

[Edited on 5-12-2015 by aga]

gdflp - 5-12-2015 at 12:57

And come to think of it, I don't know why I specifically said glass, an iron or steel pipe will work. It still seems like a pipe bomb to me though, I'd rather use hydrogen at atmospheric pressure if at all possible. You can look up "Parr Hydrogenators" to see what a laboratory grade apparatus designed for this looks like.

Yes Pd/C and Pt/C are shorthand for activated carbon impregnated with palladium or platinum metal. Typically, they are either 5% or 10% by weight precious metal, with 5% being the most commonly used.

aga - 5-12-2015 at 13:09

Activated Carbon !

I think i've heard of that somewhere ....

Given the number of variables i'd probably be better off just buying some, however this is Amateur Chemistry, so maybe better to make everything where possible.

gdflp - 5-12-2015 at 13:13

If you decide to prepare it, OrgSyn has a detailed preparation here.

aga - 5-12-2015 at 13:19

Superb !

A clear sign that What you Know can definitely be augmented by Who you know.

The choice of Route should really be up to the teaching staff, seeing as i'm just the spanner monkey, and you guys obviously know better than i.

If it means buying Platinum etc, no problem, just takes time to arrive is all.

blogfast25 - 5-12-2015 at 14:16

gdflp:

Lab hydrogenations can be carried out without actual hydrogen with Pd/C catalysts and ammonium formate as hydrogenation agent, see here:

http://oxfordchemserve.com/palladium-on-carbon-catalyst-pdc-...

(see third pic, here with formic acid)

I wouldn't have suggested this chain of synths w/o bearing that in mind. Pd/C catalysts aren't that expensive because they don't contain much Pd anyway.


[Edited on 5-12-2015 by blogfast25]

blogfast25 - 5-12-2015 at 14:21

Quote: Originally posted by aga  
@bloggers: is the 2-(4-methylcyclohexyl)propan-2-ol (item #4) a <i>desired</i> product in this sequence, or is it a possible by-product that should be minimised ?


No 4 is an end product, as would be No 6. The desired t-alcohols are preferably fully saturated (no double bonds) because of the harsh conditions of the KOH/Mg reduction.

But even just No3 as a clean product would be a worthy goal, should hydrogenation fail/prove too difficult.

gdflp - 5-12-2015 at 14:24

Indeed, I was aware of that but it slipped my mind completely, thanks for pointing it out. The only issue arises if aga doesn't have formic acid, in which case it might be simpler using a balloon of elemental hydrogen.

aga - 5-12-2015 at 14:27

I do not have formic acid.

There are plenty of ants hereabouts, however i will not collect and distill them.

I suppose it could be purchased though.

blogfast25 - 5-12-2015 at 14:36

More references on Pd/C hydrogenations with ammonium formate as H donor:

http://www.organic-chemistry.org/abstracts/literature/250.sh...

https://www.erowid.org/archive/rhodium/chemistry/cth.af.revi...

https://www.ocf.berkeley.edu/~jmlvll/lab-reports/chemoselect...

blogfast25 - 5-12-2015 at 14:37

Quote: Originally posted by aga  
I do not have formic acid.

There are plenty of ants hereabouts, however i will not collect and distill them.

I suppose it could be purchased though.


Ammonium formate is easypeasy to obtain.


[Edited on 5-12-2015 by blogfast25]

aga - 5-12-2015 at 14:44

OK. It's going a bit crazy, so needs some parameters.

There is a lot of glassware, a single stage vac pump, some reagents (IOC + a few OC) on hand.

Locally available materials include plumbing supplies, so steel tubing/fittings are fine.

Translate Ammonium Formate into Spanish, please, then try to find it.
It could be Ant Powder, Artichoke Booster, or Alfalfa Stimulant (it is that random).

200 euros is about the limit for a single experiment at my level of learning.

blogfast25 - 5-12-2015 at 14:47

Conversion 1 to 2 is treatment with GAA, no sweat.

After separation/mild clean up of 2, the conversion of 2 to 3 is simple alkaline de-esterification, probably in MeOH. Plenty, plenty plenty recipes to emulate, no sweat.

blogfast25 - 5-12-2015 at 14:48

Quote: Originally posted by aga  
OK. It's going a bit crazy, so needs some parameters.

There is a lot of glassware, a single stage vac pump, some reagents (IOC + a few OC) on hand.

Locally available materials include plumbing supplies, so steel tubing/fittings are fine.

Translate Ammonium Formate into Spanish, please, then try to find it.
It could be Ant Powder, Artichoke Booster, or Alfalfa Stimulant (it is that random).

200 euros is about the limit for a single experiment at my level of learning.


200 Euros? You're laughing, no sweat.

NH4 formate: eBay I think. It has some uses and should be very cheap.

aga - 5-12-2015 at 15:04

Yes. Ebay. 250g is now on it's way from Poland.

blogfast25 - 5-12-2015 at 15:15

You don't hang about much... :)

aga - 5-12-2015 at 15:32

Can't.

If i delay even 1 day, the postage time to Spain is so long that i forget what the item was intended for.

blogfast25 - 5-12-2015 at 15:45

Quote: Originally posted by aga  
Can't.

If i delay even 1 day, the postage time to Spain is so long that i forget what the item was intended for.


Have you looked for alpha-pinene? There's a US Amazonner...

Anyway, let's continue most procurement chatter by U2U: no point cluttering up this thread with mundanities...

[Edited on 6-12-2015 by blogfast25]

Darkstar - 5-12-2015 at 21:51

Quote: Originally posted by aga  
Quote: Originally posted by Darkstar  
@aga

I hope you've been paying attention this semester, because I've written your final exam! The test is 25 questions plus an extra credit question. Just say the word when you're ready and I'll post it.

Please !

Can't say i'm 'ready' per-se, however fire away.


Don't worry, the exam is actually extremely easy for the most part. There is one question in particular (it's one of the first five) that might be a little tricky, though. The exam consists of 25 questions and two optional extra credit questions (added a second one). For questions 1-20 and the second extra credit problem, you can write the answers in the body of your reply post (just type "red" or "blue," "nucleophile" or "electrophile" etc); for questions 21-25 and the first extra credit problem, you will need to use either ChemSketch or MS Paint-in-the-ass.

While the exam shouldn't take very long to complete, please feel free to do it at your own pace. There's no need to rush. All I ask is that you try to do as much of it as you can without looking up the answers, particularly the last five questions. These questions are more or less the "real" test, so to speak. The last mechanism (question 25) is one that blogfast recently covered, by the way. Also, don't get too hung up on the extra credit problems if you're having trouble figuring them out, especially the first one. I did give a ton of hints for the second one, though, so it'll probably be the easier one to figure out. (I practically gave you the answers!)

Anyway, here's the test. I took the liberty of typing your name in the blank for you:

final exam2.bmp - 4.2MB

[Edited on 12-6-2015 by Darkstar]

blogfast25 - 6-12-2015 at 07:47

Smashing, Darkstar. Will be taking that test myself. Forever young! :D

aga - 6-12-2015 at 12:07

aga's OC Final Exam Answers Part 1

1 Red
2 Blue
3 Red
4 Red
5 Blue

Used the 'ARIO' rule i found on utoob.

6 Nucleophile
7 Electrophile
8 Electrophile
9 Nucleophile
10 Electrophile

11 Electrophilic
12 Nucleophilic
13 Electrophilic
14 Nucleophilic
15 Electrophilic

16 B. Phenyl
17 E. Hydroxyl
18 D. Ether*
19 C. Amine
20 A. Carboxyl*

* Had those two backwards before i checked

Drawn answers to follow (hopefully before xmas).

Takes some time to work stuff out, then longer to draw them !

[Edited on 6-12-2015 by aga]

aga - 6-12-2015 at 12:34



21-1-butanol.gif - 2kB

22-phenoxide.gif - 2kB

aga - 6-12-2015 at 13:46



This was totally stolen from your example on page 15 of this thread, just with a phenyl ring and an extra wiggly bit on the ethanol.


[Edited on 6-12-2015 by aga]

23 reaction.gif - 7kB

blogfast25 - 6-12-2015 at 14:26

Sorry, but you only went and used the so caligraphically wrong types of double arrows in that last answer. We're very pedantic about these things here at B&D. Fail!

(just kidding)

aga - 6-12-2015 at 15:07

Are you drunk ?

If not, try harder !

Look again.

aga - 7-12-2015 at 02:03

Not sure about this one at all.

24reactionmech.gif - 7kB

aga - 7-12-2015 at 03:12

25 Reactionmech.gif - 4kB

aga - 7-12-2015 at 04:02

extra-1.gif - 3kB

aga - 7-12-2015 at 04:31

Extra #2

a)
The electron density is more concentrated around the Nitrogen due to :-
1 Resonance

2 Small difference in electronegativity between N and Cl

3 the presence of the electron density pulled away from the H bound to the N

4. Cl has nowhere to pull electron density from apart from the N, pulling stronger on the resonance electrons through the inductive effect.

b)

Do not really know - it seems that a) negates the answer i first thought of for b).

Stab in the dark #1 :-
The C attached to O and N gets strongly +ve, allowing the A- to a come in with a nucleophilic attack, pushing electron density up to the O, allowing the H+ to perform an electrophilic attack on the O.

S.I.T.D. #2 :-
The N is better protected in terms of physical shielding from the Cl/C/H and the pull of the Cl imparts a less -ve charge on it than that of the O.

S.I.T.D #3:-
the incoming H+ approaching the O tips the balance of the resonant electron density to be more strongly around the O, allowing the H-O bond to form.

Phenolpthalein product finally got recovered.
Yield = 0.31g = 9.2 % based on phenol (pitiful).

[Edited on 7-12-2015 by aga]

blogfast25 - 7-12-2015 at 09:07



Phenolpthalein product finally got recovered.
Yield = 0.31g = 9.2 % based on phenol (pitiful).

[Edited on 7-12-2015 by aga][/rquote]

Due to the filtering mishap, I guess? Ah well, out it down to experience and move on, I guess...

aga - 7-12-2015 at 12:35

Yes, it was over-vacuum that caused the filter to pop.

Similar to Nile Red's result, this is a beige powder, so also contaminated and not pure phenolpthalein.

I suspect that the workup could be improved in that process.

[Edited on 7-12-2015 by aga]

blogfast25 - 7-12-2015 at 12:57

Quote: Originally posted by aga  
Yes, it was over-vacuum that caused the filter to pop.

Similar to Nile Red's result, this is a beige powder, so also contaminated and not pure phenolpthalein.

I suspect that the workup could be improved in that process.

[Edited on 7-12-2015 by aga]


Mine is commercial and it's also grey/beige.

Try thermal recrystallisation from EtOH?

aga - 7-12-2015 at 13:52

Nah.

0.31g is a bit small to mess with.

Goes nicely purple in basic conditions, so job (badly) done.

Maybe re-do it one day if there's nothing else to do.

Darkstar - 7-12-2015 at 15:00

@aga:

That was faster than anticipated. I will grade your test when I get home from skewl. It looks like you did fairly well for the most part; however, there do seem to be a few areas that you're still struggling with judging by the last three mechanism questions (questions 24, 25 and the first extra credit problem. though to be fair, the latter did involve a ring closure via intramolecular substitution, something we haven't actually covered yet, which is why it was extra credit).

Anyway, when I get home later I'll post the correct answers along with explanations for the ones that you missed. But overall you did very well!

Since I already have it saved on my desktop, here's the answer to the first extra credit problem, by the way:

extra credit problem1.bmp - 253kB

blogfast25 - 7-12-2015 at 17:32

As a minor clarification to that last diagram, the first red arrow shows the lone electron pair (unbonded MO) on the nitrogen atom 'on the move'. Later the nitrocation get's its lone pair back.

As I said to aga, lone pairs as Lewis dots : can be added in ChemSketch via Templates > Template Window > Lewis structures. Overlook the lone pairs at your peril!

Darkstar - 8-12-2015 at 08:57

Here is your graded test. I made sure to use lots of red ink for dramatic effect. I also turned a blind eye to the fact that you cheated on at least two occasions (and probably more I don't know about!).

By the way, we received the brib--I mean donation you sent us through PayPal. Maybe now B & E University's chemistry department can finally afford some decent pickle jars and copper condensers. Hell, we may even have enough for the name-brand pool chemicals now! Not that any of that has anything to do with your grade, of course.
Quote: Originally posted by aga  
aga's OC Final Exam Answers

Questions 1-5 (2 points each)

1. Red ✔
2. Blue ✔
3. Red ✔
4. Red ✔
5. <strike>Blue</strike> ✘ (-2 points)

Questions 6-10 (2 points each)

6. Nucleophile ✔
7. <strike>Electrophile</strike> ✘ (-2 points)
8. Electrophile ✔
9. Nucleophile ✔
10. Electrophile ✔

Questions 11-15 (2 points each)

11. Electrophilic ✔
12. Nucleophilic ✔
13. Electrophilic ✔
14. Nucleophilic ✔
15. Electrophilic ✔

Questions 16-20 (2 points each)

16. B. Phenyl ✔
17. E. Hydroxyl ✔
18. D. Ether ✔
19. C. Amine ✔
20. A. Carboxyl ✔

Questions 21-22 (7.5 points each)

21. 1-Butanol ✔ 2-Butanol ✔ Isobutanol ✔ Tert-butanol ✔
22. Structure 1 ✔ Structure 2 ✔ Structure 3 ✔ <strike>Structure 4</strike> ✘ (missing, -2 points)

Questions 23-25 (15 points each)

23. Fischer Esterification: Correct Products ✔ Correct Mechanism ✔ (cheater!)

24. Base-Catalyzed Ester Hydrolysis: Correct Products ✔ Correct Mechanism ✘ (missing bond cleavage in first step and partially incorrect arrow-pushing in second step, -5 points)

25. SN2 Reaction: Correct Products ✔ (will accept) Correct Mechanism ✘ (incorrect nucleophilic attacks and incorrect formation of primary carbocation and amide anion, -9 points)

Extra Credit Questions (10 points each)

Extra Credit 1. Rearrangement: ✘ (incorrect nucleophilic attacks and incorrect formation of primary carbocation and amide anion, -10 points)

Extra Credit 2. ✔ (too much to list--will give partial credit. a detailed explanation will follow soon. -5 points)

Total Missed (out of 100) = 20 Points
Extra Credit Earned = 5 Points

Final Score = 85/100 = 85%


I'll post a detailed answer guide later.

[Edited on 12-8-2015 by Darkstar]

aga - 8-12-2015 at 09:36

Woohoo !

85% puts me at the top of my class !

(also the bottom but i'm an optimist)

I guess you were right : it isn't too hard to get started with Quantum Mechanics and Organic Chemistry, so long as you find the Best teachers !

The info may be out there on the web, but without someone to guide you, the pieces would never fall into place.

Astonishing. Truly astonishing, considering :-

a) QM started 9th July this year (152 days ago)
b) OC started 6th Oct this year (63 days ago)
c) We only do this in our spare time

Thanks to both of you for your Priceless efforts - long may it continue !

blogfast25 - 8-12-2015 at 10:22

Quote: Originally posted by aga  


Thanks to both of you for your Priceless efforts - long may it continue !


Congrats! The latter part depends on you of course. Here at B&D our teachers operate on Duracell's, so they can go the extra mile!

Coming up later tonite: two simple but interesting mechanisms on similar molecules with surprisingly different outcomes...

[Edited on 8-12-2015 by blogfast25]

aga - 8-12-2015 at 14:47

85% in OC and X% (please give use a clue) in QM isn't Great.

The missing 15% will probaby turn out to be the Crucial bit at some point.

Perhaps a re-cap focussing on the Missing Link(s) would be a good idea.

... followed by The Festive Party with whole Goose, Swan, booze, drugs, prostitutes, plus the usual end of term trimmings.

[Edited on 8-12-2015 by aga]

gdflp - 8-12-2015 at 15:04

I wouldn't worry about it too much. You seem to mostly be having trouble with drawing mechanisms, and you are by no means out of the woods in that regard:cool: There are some general patterns which you get more comfortable with though practice. And remember, organic chemistry is not easy; an 85% is a good grade!

blogfast25 - 8-12-2015 at 15:34

An Elimination Reaction:

Here are two reactions with quite similar reagents but very different outcomes.

propyl chloride hydrolysis 2.png - 4kB

Which is a nucleophilic substitution we should be familiar with by now. It's achieved by treating the ethyl chloride with mild NaOH solution.

Now look at this one:

Elimination 1.gif - 4kB

Here 2-bromopropane is refluxed with strong KOH. The equilibrium is pulled rightwards due to propene being a gas, which escapes from the reactive mix (Le Chatelier principle).

[Edited on 8-12-2015 by blogfast25]

aga - 9-12-2015 at 14:21

Quote: Originally posted by gdflp  
You seem to mostly be having trouble with drawing mechanisms, and you are by no means out of the woods in that regard

OK. So more practice required.

Gimme some to attempt, please.
Quote: Originally posted by gdflp  
85% is a good grade!

No, it's a terribly bad grade.

< 95% indicates that my understanding of some fundamental principles are basically Wrong.

(~5% can be forgiven as human error).

Cannot build on a house on sand.

[Edited on 9-12-2015 by aga]

Darkstar - 9-12-2015 at 23:00

I wish I weren't so busy with school right now (last week of the semester) because I really want to go over the questions you missed one by one. I've prepared a bunch of images in ChemDraw to help aid me in explaining everything to you, but I just don't have the time to actually go into the level of detail that I want to. The good news is that I'll have plenty of free time starting next week. So maybe in the meantime blogfast and gdflp can help you understand why you missed some of those questions.

One thing I noticed is that you seem to have forgotten that nitrogen must have a free electron pair in order to be neutral. This means the ammonia in question 25 doesn't need to be deprotonated before it can act as a nucleophile. NH3 is already nucleophilic due to the free lone pair on nitrogen that can form a fourth bond with an electrophile. (that's why ammonia solutions are basic) This is also why you missed question 7, where you labled NH3 as being an electrophile. Yes, NH3 technically can act as an electrophile; however, deprotonating ammonia requires either a strong reducing agent like one of the alkali metals, or a base that is even stronger than ammonia's conjugate base, NH2<sup>–</sup>, which is already an extremely strong superbase that can't exist in water. So for all practical purposes, NH3 is considered a nucleophile for the most part.

Also, you're showing the formation of highly unfavorable primary carbocations as well. Unless stabilized through resonance by an adjacent pi-system (e.g. primary benzylic carbocations), positive charges on primary carbons are extremely unstable. Hopefully blogfast or gdflp will elaborate more on why that is.

[Edited on 12-10-2015 by Darkstar]

Darkstar - 10-12-2015 at 00:31

Here are some images that I have prepared. I originally intended to post these along with detailed explanations for what is shown, but try to make sense of them as much as you can for now. Ask blogfast or gdflp for help if there's anything you're still not clear on.

Ammonia's lone pair and geometry (trigonal pyramid):

ammonia.bmp - 179kB

Hydrogen bonding greatly increases the acidity of salicylic acid's carboxyl proton (question 3):

salicylic acidity.bmp - 424kB

More on resonance stabilization and its effect on proton acidity (question 4):

resonance final exam.bmp - 489kB

All possible resonance structures of the phenoxide anion (was only looking for the first four, though):

phenoxide.bmp - 246kB

General mechanism for the Fischer esterification (question 23):

est.bmp - 735kB

Answer to question 24:

est reverse final exam.bmp - 398kB

Answers to question 25 (SN2 reaction) that I would have accepted:

sn2 final exam.bmp - 1.4MB

Answer to the first extra credit question (rearrangement via intramolecular substitution):

extra credit problem1.bmp - 253kB

In-depth look at the N-chloroacetamide molecule from the second extra credit question:

exam amide.bmp - 221kB

amide overview.bmp - 404kB

amide2.bmp - 429kB

[Edited on 12-10-2015 by Darkstar]

aga - 10-12-2015 at 00:34

It feels like the principles covered are vital, so best to get them 100% clear before moving on.

There's no rush and i'd appreciate the errors pointing out as and when time allows.

blogfast25 - 10-12-2015 at 04:37

Aromatic substitutions:

Nitration of benzene.gif - 9kB

Substitutions on arenes require both an electrophilic (Lewis acid) and a nucleophilic (Lewis base) species.

Shown in the general case, the electrophile X<sup>+</sup> draws in one of the three resonant π MOs. The nucleophile Nu<sup>-</sup> then snatches a proton from the ring, thus restoring the π MO.

Perhaps the best known aromatic substitution know is the nitration of benzene with concentrated nitric acid (2.) Concentrated (near 100 w%) nitric acid has the property of auto-dissociating, acc. the equation shown, thus being able to supply both the electrophile and the nucleophile. Here the reaction is shown with some resonance structures.

Case 3 we have already touched on much higher up, a Friedel-Crafts arene alkylation.

Here only the preliminary step of the formation of the Lewis base, the primary butyl carbocation is shown by reaction of chlorobutane with AlCl<sub>3</sub>.

After the electrophilic attack on the arene on one of the three resonant π MOs, a proton is then extracted from the ring by the tetrachloroaluminate and HCl evolves.

Multiple substitutions:

In each case the π ring is fully restored, making the substituted arene itself the potential subject of further substitutions on the 3 and 5 carbons, see below e.g. the structure of trinitro benzene:

trinitro benzene.gif - 2kB

[Edited on 10-12-2015 by blogfast25]

aga - 10-12-2015 at 11:21

Wooooooah Hoss !

Back to the Lone Pair on ammonia if one would be so kind.

Looking at the orbitals (which i think is confusing me a lot) N starts with 1s<sup>2</sup>2s<sup>2</sup>2p<sup>3</sup> (please say if that notation is correct or not).

Hybridised is it this :-

1s<sup>2</sup>2s<sup>1</sup>2p<sup>3</sup>3p<sup>1</sup> giving the valence orbitals 2p<sup>3</sup> and then a 3p<sup>1</sup> as that extra 'lone pair' (even though it is just one electron) ?


[Edited on 10-12-2015 by aga]

blogfast25 - 10-12-2015 at 14:03

Quote: Originally posted by aga  

Looking at the orbitals (which i think is confusing me a lot) N starts with 1s<sup>2</sup>2s<sup>2</sup>2p<sup>3</sup> (please say if that notation is correct or not).

Hybridised is it this :-

1s<sup>2</sup>2s<sup>1</sup>2p<sup>3</sup>3p<sup>1</sup> giving the valence orbitals 2p<sup>3</sup> and then a 3p<sup>1</sup> as that extra 'lone pair' (even though it is just one electron) ?


Forget about the 1s<sup>2</sup>, they're inner electrons, not valence.

Re. the remaining 2s<sup>2</sup>2p<sup>3</sup>, that's 5 electrons in 4 AOs. These hybridise to four 2sp<sup>3</sup>-type hybrid AOs, one of which contains an electron pair, the three others contain only one unpaired electron. The four sp<sup>3</sup> AOs point to the corners of a tetrahedron (like CH<sub>4</sub>;).

The three half-filled sp<sup>3</sup> can now σ-bond to whatever has a half-filled AO (H, C, X etc). So we end up with a structure like the one top right in Darkstar's last post.

The one unbonded filled sp<sup>3</sup> AO is now available for dative bonding: sharing with an electropile or D-block metal metal atom (complex formation).

That's what makes ammonia (or substituted ammonia like amines) a Lewis base (and Bronsted-Lowry bases).

[Edited on 10-12-2015 by blogfast25]

aga - 10-12-2015 at 14:13

px, py, pz appear to be the orbital choices, yet there are 4 possible hybridised orbitals.

Would it not make more sense (geometrically) to have 4 electrons in a d orbital ?

I guess not, given the energy requirement to put them there.

Clearly i do not understand hybridisation at all.

blogfast25 - 10-12-2015 at 16:23

They are not choices. They are the orbitals dictated for n = 2 (Principal Quantum Number), by the SE and the Aufbau Pricinple.

d-orbitals lie at a higher level (at least here because the nearest one is 3d), that's correct.

For the Al hexa-aqua complexes we did invoke six 3sp<sup>3</sup>d<sup>2</sup> hybrid AOs, involving 4d. But 3p and 4d aren't as far apart (energetically speaking) as 2p and 3d. In any case for N(+3) (ammonia etc) there's no need for higher than sp<sup>3</sup> hybridisations.

Hybridisations are about minimising electrostatic repulsions between the AOs/MOs, this lowers potential energy of the molecule.

[Edited on 11-12-2015 by blogfast25]

aga - 11-12-2015 at 07:56

I think this is where the problem was : the 2s and 2p energy diagrams were confused with orbitals somehow, despite there being only three 2p boxes ...

The p orbitals can contain up to 6 electrons in one of 6 'lobes'.

N hybridisation in NH3 is from the 2s<sup>2</sup> 2p<sup>3</sup> ground state to a 2s<sup>1</sup> 2p<sup>4</sup> state, giving 4 outer orbitals, 1 for each hydrogen atom and one with a lone (single) electron in it looking for a nucleus to snaffle.

This one, single, solitary electron is bewildering called a 'Lone Pair'.

Hopefully this is correct. If not, please stamp it flat immediately.

blogfast25 - 11-12-2015 at 08:31

Quote: Originally posted by aga  
I think this is where the problem was : the 2s and 2p energy diagrams were confused with orbitals somehow, despite there being only three 2p boxes ...

The p orbitals can contain up to 6 electrons in one of 6 'lobes'.

N hybridisation in NH3 is from the 2s<sup>2</sup> 2p<sup>3</sup> ground state to a 2s<sup>1</sup> 2p<sup>4</sup> state, giving 4 outer orbitals, 1 for each hydrogen atom and one with a lone (single) electron in it looking for a nucleus to snaffle.

This one, single, solitary electron is bewildering called a 'Lone Pair'.

Hopefully this is correct. If not, please stamp it flat immediately.


You do seem to have a liiittle trouble counting the fingers on your hand! :(

4 hybrid orbitals. A total of 5 electrons. "Mirror, mirror on the wall: how do we fit that in, all?"

Simples: three hybrids AO contain only 1 electron, one hybrid AO contains 2 electrons. Hence, in the latter case we call it a lone (unbonded) electron pair. And 3 + 2 = 5 (in Roman numerals: III + II = V, :D)

I think you're still grappling with dative bonding, so I'm gonna put up something to kill those gremlins. Watch this space!

[Edited on 11-12-2015 by blogfast25]

aga - 11-12-2015 at 08:47

That's making more sense.

So how do you predict the hybridised configuration ?

blogfast25 - 11-12-2015 at 09:36

Refresher on dative bonding:

I'll take, merely as an example, the case of the adduct formed by the interaction of the Lewis base (nucleophile) ammonia and the Lewis acid (electrophile), here a secondary propyl carbocation. (This is the second step in the reaction mechanism for the substitution of the halogen of isopropyl halide by ammonia - formation of amine(s)). Not to scale:

Dative bonding.gif - 5kB

On the positively charged carbon atom there's an AO that is empty (zero electrons). That 'space' is filled up by ammonia 'donating' its lone electron pair, which now becomes shared between the C and N nuclei. The latter is of course the definition of a simple σ bond.

The positive charge of course doesn't disappear: it merely relocates to the N atom because it has 'lost' half of an electron pair, i.e. one electron (thus 0 - (-1) = +1).

The resulting ion is called isopropyl ammonium cation.

[Edited on 11-12-2015 by blogfast25]

blogfast25 - 11-12-2015 at 09:52

Quote: Originally posted by aga  


So how do you predict the hybridised configuration ?


Good question!

I could put up a complicated algorithm here but it's not really necessary.

1. Oxygen group: sp3 hybridisation. 4 hybrid orbitals. Two of these will contain lone electron pairs.

2. Nitrogen group: sp3 hybridisation. 4 hybrid orbitals. One of these will contain a lone electron pair.

3. Carbon group: sp3 hybridisation. 4 hybrid orbitals. None of these will contain a lone electron pair.

4. Boron group: sp2 hybridisation. 3 hybrid orbitals. None of these will contain a lone electron pair.

5. Beryllium: sp hybridisation. 2 hybrid orbitals. None of these will contain a pair electron pair. The rest of that group is too electropositive to form hybrid orbitals: they lose their valence electrons in reactions.

[Edited on 11-12-2015 by blogfast25]

[Edited on 12-12-2015 by blogfast25]

aga - 11-12-2015 at 10:51

Incredibly helpful !

All is clearer by a very large margin.

Just 1 question :-

You say "Oxygen group: sp3 hybridisation. 4 hybrid orbitals"

Is the "Oxygen group" All elements on the O column (#16) of the periodic table ?
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