Sciencemadness Discussion Board - Quantum Mechanics/Wave Mechanics in Chemistry Beginners Thread

Quantum Mechanics/Wave Mechanics in Chemistry Beginners Thread

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Darkstar - 23-11-2015 at 14:56

No naming was requested (Darkstar will do the IUPAC incantations later on, I think).

I'll do a post when I get home later regarding the possible isomers (enantiomers included) and their respective IUPAC names.

And aga . . . you know damn well why phenol is more acidic than non-aromatic alcohols! How many times have I explicitly linked electron delocalization to acidity and ease of proton removal? This is why I keep stressing resonance!!!

aga - 23-11-2015 at 15:07

Eeeeeeek !

These last two posts have triggered my canine relflex.

I need some Basket time with my paws over my eyes.

blogfast25 - 23-11-2015 at 17:35

Alrighty: 'permission to snooze', Big Paws!

When you wakey wakey when the Big Fireball rises above the Land again and you're having your first Crunchies, remember that Old Grumps' spiel about hexachlorostannate is not a genesis story of that ion, only an explanation of its actual structure...

blogfast25 - 23-11-2015 at 19:13

Erratum:

On the colours of f-block salts, I wrote "4d" where of course it should have read 4f and "5d" where of course that should have read 5f. The correct text is below.

Quote: Originally posted by blogfast25

Important corollary:

f block elements are also often colourful for similar reasons.

The 4f orbitals also split into two groups according to orientation and thus slightly different energy levels, in a ligand field. An incompletely filled 4f orbital thus also gives rise to transitions with emissions in the VIS part of the electromagnetic spectrum. Most (+3) lanthanides show colour, as do many actinides due to part-filled 5f orbitals.

Mea culpa.

[Edited on 24-11-2015 by blogfast25]

Darkstar - 23-11-2015 at 21:34

As promised, here are all the possible position and skeletal isomers of 2-pentanol, complete with stereochemistry and correct IUPAC nomenclature. Itself included, there are a total of eight possible alcohols from 2-pentanol: three of them chiral and the rest achiral, giving rise to a total of 11 different molecules all together. I would like to point out that there are technically other isomers that are also possible from 2-pentanol, which are functional isomers like 1-ethoxypropane, 1-methoxybutane, 2-methoxybutane, 2-methoxy-2-methylpropane and so on; however, these isomers are ethers and not alcohols, so I didn't bother to actually draw them.

isomers pentan-2-ol.bmp - 749kB

And in the unlikely event that anyone's interested, here's a much higher quality version of the image above in .png format and high resolution:

isomers pentan-2-ol (high quality).png - 92kB

[Edited on 11-24-2015 by Darkstar]

blogfast25 - 24-11-2015 at 08:09

Thanks, Darkstar.

To try and give aga a bit of closure re. hexachlorostannate let's look at a proposed formation reaction mechanism of that ion.

For those in possession of the required chemicals the easiest synthesis is as follows. Cold pure SnCl4 is added to a stoichiometric excess of cold 37 w% HCl with intense stirring. The overall reaction equation is:

2 HCl(aq) + SnCl4(l) < === > H2SnCl6(aq)

The excess HCl serves to push the equilibrium fully to the right and prevent ligand substitution by OH-(aq) (discussed above).

H2SnCl6 is a strong acid (its salts react neutrally) and thus more or less completely dissociated in water.

Now let's look at the structure of SnCl4. It's the same structure as methane or CCl4: Sn's outer atomic orbitals containing 4 electrons in total, hybridise to four sp3 orbitals each containing 1 unpaired electron. These sp3 each then interact with 1 lone electron in the outer shell of a chlorine atom, to form SnCl4 which like its methane analogue is a tetrahedron.

So how do we get from SnCl4 to SnCl62-? Obviously two choride ions have to be added.

Successful three-species collisions are very rare for obvious reasons so it's reasonable to expect the addition process to be a two step process. A first successful collision between a SnCl4 and a chloride ion would give rise to an intermediate and probably excited SnCl5-. A second successful collision with a second chloride ion would yield SnCl62-, probably in an excited state. Mild rearrangement of the formed MOs to minimise their mutual electrostatic repulsion then leads to the octahedral structure of ground state SnCl62- with electronic structure pointed out in the higher post.

An interesting point to note here is that chloride ions here play the part of Lewis bases while SnCl4 plays the part of a Lewis acid. SnCl4 is indeed used as catalyst in Friedel-Crafts addition reactions.

The reaction product of a Lewis base and a Lewis acid, like SnCl62-, is known as a Lewis adduct.

[Edited on 24-11-2015 by blogfast25]

aga - 24-11-2015 at 13:31

Thank you both for the tummy tickles.

I feel like such a Good Boy that i've got to wag my tail for a while.

Hope i don't get arrested again like last time.

blogfast25 - 25-11-2015 at 06:43

aga, for my 'next trick': do you know what amphoterism is?

[Edited on 25-11-2015 by blogfast25]

aga - 25-11-2015 at 06:57

ISTR that amphoteric is to do with a closed system with nothing going in/out.

In other words, No, i dont know.

blogfast25 - 25-11-2015 at 09:17

Erm... no, that's an adiabatic system!

Ligand Exchange Theory: Explaining Amphoterism

We should by now be familiar with the concept of coordination complexes and their structure. A general coordination complex can be written as:

MLnz

Where M is the central metal atom, L the ligand, n the coordination number and z (positive or negative) the overall charge of the complex ion.

The formation reaction for two generic complexes, MLn and MNn (for ease of notation the overall charges and state symbols are not shown) are shown below:

The Kf constants are the so-called complexation constants for both complexes respectively. As with all equilibrium reactions, the larger the Kf, the more the equilibrium leans to the right. We say that complexes with high Kf are strong complexes.

The Kf values of many common complexes have been measured and the following table lists quite a few of them.

It follows mathematically that if one complex is stronger than another, the latter can be displaced by the former by adding ligands of the former. The ligands of the weaker complex are then released. This phenomenon is called ligand exchange and can be quantified by the ligand exchange constant KLN.

A commonly known and colourful ligand exchange reaction takes place when NaCl solution is added in sufficient quantity to a copper sulphate solution. The latter is a blue solution which turns gradually green due on addition of the NaCl solution, to the ligand exchange reaction:

[Cu(H2O)6]2+(aq, BLUE) + 4 Cl-(aq) < === > [CuCl4]2-(aq, GREEN) + 6 H2O(l)

In Lewis acid base theory we’d say that chloride ions are ‘harder’ Lewis bases than water (strength of Lewis acids and bases is referred to as being ‘hard’ or ‘soft’). The harder base thus displaces the softer one.

Aluminium’s amphoterism:

It’s well known that aluminium salts tend to be highly soluble in acid solutions as well as in strongly alkaline solutions. In between there’s a pH zone where Al is insoluble (and precipitates as hydrated alumina).

Al(+3)’s solubility at both low and high pH is due to a ligand exchange reaction:

[Al(H2O)6]3+(aq) + 4 OH-(aq) < ==== > Al(OH)4-(aq) + 6 H2O(l)

The relative concentrations (α) of various species of Al(+3) in the full pH interval are shown below:

Note that by Al3+ in this figure is really meant [Al(H2O)6]3+(aq) and that the intermediate species noted as AlOH2+ and Al(OH)2+ are the result of partial ligand exchanges.

The amphoterism of Zn, Pb, Sn, Cu, Sb, As and others can be explained in analogous ways.

[Edited on 26-11-2015 by blogfast25]

[Edited on 26-11-2015 by blogfast25]

aga - 25-11-2015 at 11:16

Quote: Originally posted by blogfast25

Erm... no, that's an adiabatic system!

Ah yes. Of course.

Amphoteric are bisexual ions.

blogfast25 - 25-11-2015 at 19:43

@aga:

It's really up to you where we go from here. I can suggest a few avenues to further explore if you like.

Darkstar - 25-11-2015 at 20:18

@aga:

First of all, LOL at "bisexual ions." Second of all, do you think you're ready for another round of organic chemistry, albeit a bit more advanced this time around? You may have noticed that I've slowed down on the OC a bit these last couple of weeks. This was partly to give you time to catch up and to prevent information overload. Between all of the QM lectures and extracurricular OC excursions, a lot of material has been covered since July. So before we move on, do you have any questions regarding the concepts discussed thus far? Just to recap, here's what we've covered (not including the additional OC contributions by blogfast):

aga - 26-11-2015 at 11:47

I have many questions, and feel that the answers to most of them are in the body of this thread.

If it would be OK, i'd like some time to re-, re- then re- read all of the Education you kind gents have posted here.

The day job tends to get in the way, and eats up far too much valuable Chemistry time.

Without being 100% sure if this is possible, perhaps QM and OC can be combined with some practical experiment at some point.

I got some pthalic anhydride today and there's phenol on it's merry way too.

Making phenolpthalein looks fairly straightforward, so perhaps if i make an attempt at the reaction mechanisms and the underlying QM, then discuss, before doing the actual synthesis, perhaps that would be a good way to bring it all together into an actual physical manifestation of all these wonderments.

Maybe that isn't a good reaction to illustrate the OC and QM, i do not know.

Suggestions welcome.

(OC reagents on hand are salicylic acid, sulfamic acid, toluene, methanol, super dry ethanol, inositol, choline bitartrate, acetone, ethyl acetate, ascorbic acid, glacial acetic acid, plus all the usual IOC reagents as well as a ton of glassware, vac pump etc).

[Edited on 26-11-2015 by aga]

blogfast25 - 26-11-2015 at 12:54

Quote: Originally posted by aga

I got some pthalic anhydride today and there's phenol on it's merry way too.

Making phenolpthalein looks fairly straightforward, so perhaps if i make an attempt at the reaction mechanisms and the underlying QM, then discuss, before doing the actual synthesis, perhaps that would be a good way to bring it all together into an actual physical manifestation of all these wonderments.

Have you got a particular synthesis route in mind?

aga - 26-11-2015 at 13:35

utoob is all. Nile Red.

Appologies if he stole anyone's writeup.

https://www.youtube.com/watch?v=kBo5UnNRodA

Edit:

There is No reason to use that particular synth other than it has pictures.

If there is a better, or more informative synth, then groovyness all round.

More Edit:

I already have phenolpthalein so actual Product at the end is largely irrelevant.

If there are Other experiments that would be more educational, then they would be more suitable.

[Edited on 26-11-2015 by aga]

blogfast25 - 26-11-2015 at 15:00

Quote: Originally posted by aga

utoob is all. Nile Red.

Appologies if he stole anyone's writeup.

https://www.youtube.com/watch?v=kBo5UnNRodA

You would do well to transcribe the method here, in 'shorthand' so we can have a look at it. Maybe then we can suggest a QC based reaction mechanism, if one is imaginable (I haven't watched the video yet).

aga - 26-11-2015 at 15:05

OK. Gimme a few mins.

aga - 26-11-2015 at 15:30

Transcript :-

2g phenol
1.5g pthalic anhydride

Add both to what a 25ml RBF.

Add 'a few drops' of 'concentrated' hydrochloric acid.
(i detest Both of those adjectives - why not just state conc and volume ?)

Heat, unstoppered, to 150 C for two hours.

Stirring is evident, although not stated.

Mixture goes red, clear red, nearly black.

Cool to RT.
Add 10 ml distilled water + 10ml dichloromethane (DCM).
continue stirring for an unspecified time.

Transfer to to separatory funnel.
Wash remainder from the RBF with 'some' DCM. into sep funnel.

Mix/shake/vent then allow to separate for an unspecified time.

Drain off lower layer (product).

Add 10ml more DCM to the residue in the sep funnel.
Shake 'n' vent, drain off lower layer again adding to earlier lower layer.

Transfer the combined product to another sep funnel.

Wash residues into the sep funnel with a 'small amount' of DCM.

Add 5ml 2[M] NaOH solution. (goes purple)
Add 10ml water.
Cap n shake n vent, let separate.

Drain off Lower layer and discard.

Drain Upper Product layer into a beaker. Wash sep funnel with water into same beaker.

Pour into 100ml 2[M] HCl (goes white and insoluble).

Vac filter. Vac dry.

1g of product.

blogfast25 - 26-11-2015 at 16:07

Quote: Originally posted by aga

(i detest Both of those adjectives - why not just state conc and volume ?)

Thanks. With HCl when it states 'conc.' it always means 37 w% HCl (the maximum).

So it's a reaction between phthalic acid anhydride and phenol in acid conditions, the rest is basically work-up. Let's see what the Quantum Witches have to say about that brew...

And now you've transcribed it, you won't have to 'cook from video'. The number of Lobster Thermidors I've buggered cooking that straight from the teevee is frightening.

BTW: it calls for a few drops of conc. H2SO4, NOT HCl!

[Edited on 27-11-2015 by blogfast25]

blogfast25 - 27-11-2015 at 07:46

Proposed Reaction Mechanism for the Synthesis of Phenolphthalein (addition/condensation of phenol to phthalic acid anhydride):

Conc. H2SO4 acts as a catalyst. It also pushes back the deprotonation of phenol (phenoxide ions might interfere here).

Step I: protonation of one of the carboxyl groups.

This creates a carbocation (marked with a red +) where the carboxyl group was.

Step II: first addition of a phenol.

Orbital 3 latches on to the carbocation, while proton 1 latches on to orbital 2, creating water as a leaving group. Note that these movements leaves the carbocation in place, which explains why both phenols (Ph) are added to the same carbon atom and not one to each carboxyl group.

Step III: second addition of phenol and catalyst regeneration.

Orbital 3 latches onto the carbocation, proton 1 relocates to the HSO4-, there by regenerating the catalyst.

[Edited on 28-11-2015 by blogfast25]

aga - 27-11-2015 at 15:45

Forgive what follows as noob misunderstanding if you would be so kind :

I can imagine ways that the reaction intermediates occur that would be different.

Drawing it may be difficult as the notion of the orbitals' shape around each element is hard to imagine, let alone draw.

[Edited on 27-11-2015 by aga]

blogfast25 - 27-11-2015 at 16:36

Quote: Originally posted by aga

Forgive what follows as noob misunderstanding if you would be so kind :

I can imagine ways that the reaction intermediates occur that would be different.

Drawing it may be difficult as the notion of the orbitals' shape around each element is hard to imagine, let alone draw.

[Edited on 27-11-2015 by aga]

Others may have a different opinion too, it's not chiselled in granite , this proposal. So go for it, Hombre!

Use whatever means to draw stuff: manually drawn and scanned stuff works too.

I was also looking at the Enthalpy of Reaction (Standard), using bond Enthalpies as explained above. Per mol of reaction product:

1 mol of C=O broken: costs + 749 kJ.
2 mol of H-Ph broken: costs 2 X + 431 kJ
2 mol of C-Ph formed: yields 2 X - 418 kJ
1 mol of H2O formed: yields - 286 kJ

Enthalpy per mol PP formed = +489 kJ

Even allowing for some uncertainty on the estimate (data points, non-STP temperature of execution, no Entropic term calculated) this suggests a reaction that is not thermodynamically favourable.

But by carrying it out at 150 C, water is constantly removed from the reaction mix, driving the equilibrium:

PAA{s) + 2 Ph(l) ==== > PP(l) + H2O(g)

... to the right, in accordance with Le Chatelier's Principle. Endothermic equilibria are also driven right by high temperature.

[Edited on 28-11-2015 by blogfast25]

IrC - 27-11-2015 at 16:53

Quote: Originally posted by blogfast25

With HCl when it states 'conc.' it always means 37 w% HCl (the maximum).

Thanks for that. For years I have looked at procedures which never specify other than stating 'concentrated'. For HCL I always assumed 37% as fairly self explanatory, can you expand this to include other acids where that term is often used? Specifically these three: Nitric, Sulfuric, Acetic. Just in general terms where one reads a procedure and the only term used is 'concentrated'. I have wondered about this many times. One acid where I think I know what they mean is HF, where I always assumed they mean ~49%. Likewise I always assumed for Nitric they mean ~68%, but I am not really sure on either this or HF.

By the way this is one of the better threads I have seen on SCM, thanks for putting in so much effort to teach this subject.

blogfast25 - 27-11-2015 at 17:04

Quote: Originally posted by IrC

Quote: Originally posted by blogfast25

With HCl when it states 'conc.' it always means 37 w% HCl (the maximum).

Thanks for that. For years I have looked at procedures which never specify other than stating 'concentrated'. For HCL I always assumed 37% as fairly self explanatory, can you expand this to include other acids where that term is often used? Specifically these three: Nitric, Sulfuric, Acetic. Just in general terms where one reads a procedure and the only term used is 'concentrated'. I have wondered about this many times. One acid where I think I know what they mean is HF, where I always assumed they mean ~49%.

By the way this is one of the better threads I have seen on SCM, thanks for putting in so much effort to teach this subject.

Hi IrC.

Thanks for the kudos, I appreciate it.

'Conc.' is a bit of a historical misnomer, I'm afraid. Practically it means the highest concentration that an acid can be obtained as with respect to water as a solvent (or remainder or impurity).

Non-volatile mineral and stable acids like sulphuric, phosphoric and nitric acids can be worked up to about 100 w%. Even then different sub-grades tend to exist on a case-by-case basis. But the rather volatile fatty acids (formic, acetic, propionic, butyric etc etc) can also be obtained as pure substances. Glacial acetic acid means pure, anhydrous acetic acid but no one would refer to it is 'conc. acetic acid', though.

In the case of the hydrogen halides (gases in the pure form, except for HF (liquid)) 'conc.' means simply maximum solubility (at STP) which for HCl is 37 w% (approx. 12 M). The solubility limits of HF, HBr and HI you can of course find easily on teh Tinkerwebs.

Hope that helps...

[Edited on 28-11-2015 by blogfast25]

IrC - 27-11-2015 at 20:09

Quote: Originally posted by blogfast25

'Conc.' is a bit of a historical misnomer, I'm afraid. Practically it means the highest concentration that an acid can be obtained as with respect to water as a solvent (or remainder or impurity).

Sort of helpful but I think it kind of confused me, the part where you say "can be worked up to about 100 w%". Does this mean no water at all or the most that water will hold. I often use these charts:

http://www.sigmaaldrich.com/chemistry/stockroom-reagents/lea...

http://www.csudh.edu/oliver/chemdata/acid-str.htm

Where Nitric is 70.4, Sulfuric is 96.0 as example. My problem stems from over years only studying enough chemistry for specific things I was working on at the time. Reading this thread I can see catching up is a long way off. What I was trying to figure out was what they on average mean in terms of percentage when they say 'concentrated'. I know I am kind of going off your real topic but when you talked about HCL percentage I figured it was a chance to jump in and get an answer to something that has always given me trouble. Taking the quote from:

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

"Commercially available nitric acid is an azeotrope with water at a concentration of 68% HNO3, which is the ordinary concentrated nitric acid of commerce.

Nitric acid of commercial interest usually consists of the maximum boiling azeotrope of nitric acid and water, which is approximately 68% HNO3, (approx. 15 molar). This is considered concentrated or technical grade, while reagent grades are specified at 70% HNO3."

and

"A commercial grade of fuming nitric acid contains 90% HNO3"

When they say concentrated (using Nitric as my example) I never know do they mean 68%, or 90%. I was trying to make flash paper once that didn't work at all and what I had was the typical 68% that a supply house in Phoenix carried yet the bottle had concentrated on the label. So I guess what my real question is, how do you tell in a procedure when they state concentrated but fail to mention numbers. I ask because this failure to specify is incredibly common in many things I have read. I won't go further as I do not wish to get you too far off the topic of 'wave mechanics in chemistry' asking something better suited for beginnings, but this question has really bothered me for years.

aga - 27-11-2015 at 23:37

The vagueness of 'conc', 'hot', 'cold' could easily balls up the experiment (especially with me doing it !).

Nobody needs panic about aga and acid, as i have some Warm, Dilute NaOH on hand ...

blogfast25 - 28-11-2015 at 07:36

IrC:

100 w% here means pure acid, no water.

Sulphuric acid goes up to 98 w% which is the azeotropic concentration. But 96 and 95 w% are routinely sold and referred to as 'conc. H2SO4' too. And Oleum is kind of 200 %: it's anhydrous H2S2O7 (a solution of 1 mol SO3 in 1 mol H2SO4) but rarely available commercially.

Nitric acid is sold often as azeotropic NA, at 68 w% but I've seen the term used also for 75 w%! And > 95 w%, so called fuming nitric acid, is also available. 'White fuming nitric acid' is 99.9 w%.

Whatever the acid used, technical and academic papers should ALWAYS specify actual concentration, sadly they don't always.

blogfast25 - 28-11-2015 at 08:05

Quote: Originally posted by aga

The vagueness of 'conc', 'hot', 'cold' could easily balls up the experiment (especially with me doing it !).

Nobody needs panic about aga and acid, as i have some Warm, Dilute NaOH on hand ...

Brilliant! A recipe for even deeper burns! Heat of the NaOH solution + solvation heat of acid + neutralisation enthalpy + (in some cases) lattice energy of formed salt = skin graft!

Hint: best practice for most acid spills on skin: treat with copious amounts of cold, fast flowing tap water. Simples...

[Edited on 28-11-2015 by blogfast25]

aga - 28-11-2015 at 08:54

Quote: Originally posted by aga

I can imagine ways that the reaction intermediates occur that would be different.

Imagining is one thing !

Spent hours trying to make them work in terms of orbitals & charges and none of the 'imaginings' can physically work.

Makes great doodles though.

Came across (and tried to apply) Lewis Dot Diagrams in the process of proving myself wrong.

Quote: Originally posted by blogfast25

Brilliant! A recipe for even deeper burns!

'twas merely jest, sirrah.

Naturally one wears an all-in-one bullet proof rubber incontinence suit when handling acids.

blogfast25 - 28-11-2015 at 09:21

Quote: Originally posted by aga

Makes great doodles though.

I'm quite curious to know what Darkstar thinks of mine, actually...

Darkstar - 28-11-2015 at 09:54

@aga

If it helps, here's blogfast's mechanism with arrow pushing and possible intermediates. Like before, H–A represents the acid catalyst and A– is its conjugate base. Unfortunately, I'm a little busy with school work at the moment, so I can't do a step-by-step analysis of the mechanism right now.

aga mech.bmp - 1.2MB

Also, if you're interested, here's the mechanism in its original format and resolution:

aga - 28-11-2015 at 11:54

Beautiful Darkstar ! Thanks.

The mechanism is so far removed from anything i could think of, one would expect to disappointed.

The Opposite is true !

It's so amazing to be able to even Attempt stuff like this.

(i think he likes it bloggers. I reckon you dropped 2 marks: one for omitting the curly arrows, the other for eschewing the snakes-n-ladders format)

blogfast25 - 28-11-2015 at 15:02

Quote: Originally posted by aga

I reckon you dropped 2 marks: one for omitting the curly arrows, the other for eschewing the snakes-n-ladders format)

Aaaarrrghgh!!! Rarely has my honour been besmirched more than on this wretched day... You leave me no choice Sir but to tear all my hair out.

aga - 28-11-2015 at 15:09

Keep your hair on.

2 marks were deducted, so 8/10

Then the extra marks for starting/teaching the thread were added, so 998/10.

9980% is a pretty good result.

Edit:

10 marks were deducted from the max 1000 available for the Thread Award due to some drunkard being allowed on the course.

[Edited on 28-11-2015 by aga]

aga - 28-11-2015 at 16:16

Once again, please forgive me if this is something i've simply not understood properly.

Surely an electron transfer imparts an increase in -ve not +ve, so i can safely assume that the '+' in the circle should be a '-' (i.e. a typo)

For the Phenol, would the electronegativity of the O in the OH group Not pull electron density towards it, creating a slight nett +ve charge on the opposite carbon of the benzene ringy bit ?

blogfast25 - 28-11-2015 at 17:26

Quote: Originally posted by aga

I'm not enamoured with that bit of Darkstar's rendition either and prefer mine.

Note that -(-) = +: take away a negative (from zero) and you end up with a positive.

As regards you last paragraph, it sure remains a little mysterious as to why the C opposite the C-OH is the attacker here, no contest from me. Do bear in mind that the electron rich (and delocalised) π rings are quite nucleophilic and suitable for attack on a carbocation.

Steric hindrance may have something to do with the 'choice' of carbon No 4 because it's the most accessible of all.

The electronegative O does make the whole ring slightly less negative, correct.

Oh, and look at the way Darkstar wrote the first step of the esterification reaction (higher up), which is also a protonation step:

So he prefers to write it that way but when one of these C=O-H (double) orbitals moves, then the positive charge moves onto the C. And it's the cabocation that's crucial here because it's so electrophilic! So I prefer my simpler notation (nerner nenenener!)

I'm sure the occasional murder has been the result of disputes about OC mechanisms!

[Edited on 29-11-2015 by blogfast25]

Darkstar - 28-11-2015 at 20:35

I leave for a couple of hours and this is what I come back to!? Alright, let me 'splain.

Quote: Originally posted by aga

Once again, please forgive me if this is something i've simply not understood properly.

Please, by all means, keep asking questions. This is how you learn.

Quote:

Surely an electron transfer imparts an increase in -ve not +ve, so i can safely assume that the '+' in the circle should be a '-' (i.e. a typo)

Your eyes deceive you not. There's supposed to be a + in that circle. The positive charge is on phthalic anhydride's carbonyl oxygen, which got protonated by the acid. I always draw activated carbonyl groups like that. I discussed what "activated" means in detail in my Fischer esterification post.

Quote:

For the Phenol, would the electronegativity of the O in the OH group Not pull electron density towards it, creating a slight nett +ve charge on the opposite carbon of the benzene ringy bit ?

It's actually the exact opposite. If oxygen's lone pair drops into the aromatic ring and becomes delocalized, the meta-directing inductive effect (pull of oxygen) that would normally direct electrophilic attacks to the 3-carbon (meta) will get over powered. If you were to draw the resonance structures, you would see that phenol's oxygen has a lone pair that can be donated into the benzene ring, which creates a negative charge at the 2-position (ortho) and 4-position (para). Thus phenol's -OH group is actually a rather good electron-donating group and strongly activates the ring for electrophilic attack at the 2- and 4-position (called ortho-/para-directing).

Inductive effects are not always the most significant contributor. Always take into consideration resonance if applicable.

Quote: Originally posted by blogfast25

As regards you last paragraph, it sure remains a little mysterious as to why the C opposite the C-OH is the attacker here, no contest from me. Do bear in mind that the electron rich (and delocalised) π rings are quite nucleophilic and suitable for attack on a carbocation.

Steric hindrance may have something to do with the 'choice' of carbon No 4 because it's the most accessible of all.

I think it's a combination of steric hindrance and phenol's strong ortho-/para-directing effect (see above). If the electrophilic attack happened at the meta-position, the oxygen would be unable to donate a lone pair into the ring to form an oxonium ion and stabilize the intermediate. This makes the ortho- and para-positions much more favorable. And since the ortho-position is physically blocked by the -OH group, the para-position ends up being the more favorable of the two.

Quote:

Technically, your way is no more correct than mine! I draw it all in a single step (nucleophilic attack and pi-bond cleavage) mostly because it saves a step. But in reality, the positive charge is neither completely on the oxygen nor completely on the carbon. So whether you draw it like I do with the charge on the oxygen, or like you do with the charge on the carbon, in reality it's on both.

I do think it helps beginners to see the charge on carbon so they understand why it gets attacked by the nucleophile, though; however, these days I just put the charge on oxygen and imagine it as oxygen pulling harder on carbon because that extra proton is now pulling on it (i.e. the inductive effect). In other words, the extra proton causes oxygen to pull even more electron density away from carbon than it normally does, activating it for nucleophilic attack.

[Edited on 11-29-2015 by Darkstar]

blogfast25 - 29-11-2015 at 09:19

Thanks for the clarifications Darkstar, especially the influence of the oxygen on the delocalised π ring in phenol, my bad on that one. And just after explaining why phenol is a weak acid too! Tsk...

Coming up next: the Markovnikov Effect!

[Edited on 29-11-2015 by blogfast25]

blogfast25 - 29-11-2015 at 10:58

Before explaining the Markovnikov effect, I want to go back to my proposed mechanism for the synthesis of indole-3-acetic acid, summarised here:

At the time no one cottoned on to the question ‘why does the addition occur on C No3 and not on the one below’?

Well, let’s look at the full structure of indole:

The No3 C is closer to the aromatic π ring and some orbital overlap increases the electron density (an effect sometimes described as ‘electron pushing’) on the No3 C, making it more susceptible to nucleophilic attack than the one below it.

We do well to remember that the neat little bars that represent MOs in our various notations create the illusion that all MOs are equal but due to various effects this is not true: due to differences in electronegativity and/or resonance/orbital overlapping, and these effects can determine outcomes significantly.

The Markovnikov Effect:

Let’s look at the electrophilic addition of HBr to 1-butene:

We’ve already covered this type of addition higher up.

In essence the highly polarised H-Br bond splits into a proton and a bromide ion and the proton snatches the π orbital and ends up on the No1 carbon. That leaves the No2 carbon as a carbocation. The bromide ion Lewis base then shares one of its (four) lone electron pairs with the carbocation et voila, 2-bromopropane is born!

However, this begs the question, ‘why does the carbocation not form on the No1 carbon?’, which would give rise to 1-bromopropane?

The reason is that the No2 carbocarbon is bound to an ethyl group (-C2H5

and a methyl group which are both ‘electron pushers’, whereas the No 1 carbocarbon is only bound to two hydrogen atoms and one propyl group. The electron pushing effect of the ethyl and methyl group lowers the charge and thus the energy of a carbocation on No2 carbon somewhat and this is preferred to a carbocation on the No1 C.

The Markovnikov effect can thus be summarised as follows:

If there is a choice of carbocations to be formed, the carbon atom bound to the most/strongest electron pushers is the preferred one. We can even simplify further: the carbocation forms preferably on the carbon atom that’s bound to the higher number of other carbon atoms.

[Edited on 30-11-2015 by blogfast25]

gdflp - 29-11-2015 at 11:02

Blogfast, aren't you referring to the Markovnikov effect? I've never heard of this phenomenon referred to as Markovsky effect.

[Edited on 11-29-2015 by gdflp]

aga - 29-11-2015 at 12:14

gdflp could be right Prof.

Seeing as he's not enrolled on this course i called Security and had him thrown out of the building.

We can cover it up no problem - nobody will ever know, just alter the records and it will all be OK.

[Edited on 29-11-2015 by aga]

blogfast25 - 29-11-2015 at 13:34

Old age, poverty and fading memory: it's been corrected.

Thanks gdflp and aga.

aga - 29-11-2015 at 13:57

Erm, there's a slight problem.

Security immediately did as asked.

Thing is, they're not too bright, and we're on the Fourth floor, and gdflp turns out to be the Bursar ...

blogfast25 - 29-11-2015 at 14:05

Quote: Originally posted by aga

Erm, there's a slight problem.

Security immediately did as asked.

Thing is, they're not too bright, and we're on the Fourth floor, and gdflp turns out to be the Bursar ...

Give him a large mince pie. T'is the season of good will, shortly.

aga - 29-11-2015 at 14:19

It was mice pies that saved the day.

The bio-chem testing lab had something of a surplus of ex-test-subjects and decided to team up with class 6C doing food science, and made a very large number of pies.

The lab should really have mentioned the Substances they were testing ...

gdflp - 29-11-2015 at 14:52

Blogfast and Darkstar are doing an excellent job of explaining theoretical organic chemistry, but so far there has been no discussion of practical organic chemistry such as extractions, explanation of fractional distillation and vapor phase diagrams, etc.(unless I missed it). Would you like me to jump in and contribute several sections on these topics aga, as they're quite important in organic chemistry? They're also important in getting high proof alcohol, which I know you're interested in

I can also throw in some labs based on the theory if you'd like.

aga - 29-11-2015 at 15:11

I think your Input would be extremely valuable gdflp.

If blogfast25 and Darkstar have no objections, please join the Party !

(Do not expect too much of your student(s) though)

Edit:

For the Record : i drink only Beer <= 4.5w%

More than that and i go all nuts.

Distilling ethanol was interesting to prove/disprove blogger's Thumper question.

Near 100% EtOH was interesting when deltaH's K₃PO₄ idea worked out well.

To distill and Drink is not one of my hobbies.

Perhaps that's one reason why homedistiller.org didn't like what i said (at all).

[Edited on 29-11-2015 by aga]

blogfast25 - 29-11-2015 at 16:35

Quote: Originally posted by aga

I think your Input would be extremely valuable gdflp.

If blogfast25 and Darkstar have no objections, please join the Party !

Sure, no objections!

In fact, it's much welcomed from my side.

[Edited on 30-11-2015 by blogfast25]

blogfast25 - 29-11-2015 at 18:02

Primary, Secondary and Tertiary Carbon Atoms:

In the spirit of Markovnikov's Rule, look at the following hydrocarbon compounds in which an exo-atom (or some group) X replaces a hydrogen atom:

In the top row, the C to which the X is bound is itself only itself bound to zero or one other C. We call that C a primary carbon atom.

In the middle row, the C to which the X is bound is itself only bound to two other C. We call that C a secondary carbon atom.

In the bottom row, the C to which the X is bound is itself bound to three other C. We call that C a tertiary carbon atom.

Note that in all three classifications the nature of the groups bonded to the C atom in question is of no importance: they may be long, short, contain double bonds (but not to that C atom itself) or even exo-atoms or functional groups without affecting its classification.

In general, in analogy with Markovnikov's rule, we can say that the electron density about the C-atom will vary broadly as: Tertiary > Secondary > Primary.

And this, together with steric hindrance considerations, has consequences, in particular for nucleophilic substitution reactions (next up!)

[Edited on 30-11-2015 by blogfast25]

Darkstar - 29-11-2015 at 20:54

Quote: Originally posted by blogfast25

Quote: Originally posted by aga

I think your Input would be extremely valuable gdflp.

If blogfast25 and Darkstar have no objections, please join the Party !

Sure, no objections!

In fact, it's much welcomed from my side.

No objections from me, either. I also welcome it.

Darkstar - 29-11-2015 at 23:12

Quote: Originally posted by blogfast25

Quote: Originally posted by aga

Drawing it may be difficult as the notion of the orbitals' shape around each element is hard to imagine, let alone draw.

Use whatever means to draw stuff: manually drawn and scanned stuff works too.

If you guys are interested, BKChem and ChemSketch are both freeware molecular drawing programs. The former is what I used before I switched to ChemDraw Pro.

Learning how to quickly draw molecules and mechanisms and then save them as various kinds of images will take a little while at first, but once you master it, you'll never want to draw them any other way. For instance, in ChemDraw, if you wanted a benzene ring, instead of having to draw it manually, all you'd have to do is just click the benzene tool on the left toolbar and then click in the drawing space where you want to put it. Presto! And if you wanted to add a second benzene ring, say one that is fused to the first one (i.e. naphthalene), all you'd have to do is just click again, this time while hovering over one of the bonds on the side of the first benzene ring that you want the second one to connect to. So in other words, you can literally draw a 10-carbon naphthalene molecule with five double bonds in just three mouse clicks. Compare that to drawing it in MS Paint-in-the-ass!

Plus it's also nice to be able to just quickly clone molecules by highlighting them and then dragging the cursor away while holding ctrl (as opposed to manually redrawing them). This is especially useful when drawing mechanisms that involve larger molecules and intermediates (like in the phenolphthalein mechanism). I know you can clone in MS Paint the same way, but the difference is that you're cloning an image of the molecule, not the molecule itself. This means everything in the little dotted-line box gets cloned, background included.

Speaking of which, I can tell that blogfast used the clone feature (or copy and paste) in his phenolphthalein mechanism to make phenol. If you look closely at the phenol molecules, you'll notice that the two carbons on the right side of the benzene ring have tiny perpendicular lines protruding out from them. My guess is that they used to be part of phthalic anhydride, which he drew first. And then to make phenol, he highlighted the benzene section of phthalic anhydride and cloned/copied it.

aga - 30-11-2015 at 01:10

I got the free ChemSketch program installed now.

Seems relatively easy to use.

myfirst.png - 909B

The program stuck the + on all by itself when i added the H.

It was drawn all at right angles, then changed when i clicked on the 'Clean Structure' tool.

I cannot find a tool for drawing curved arrows. Any clues ?

Edit:-

myfirst.png - 1kB

Found it, but it's a bit of a faff.

Click Tools (top menu)
Click Pen Style Panel.
Choose a colour and thickness, then click Apply.

Click Draw (second button from top left, next to Structure)
Click the curved line on the left hand menu (chose how curved)
Click the straight right arrow

Now click on where you want the arrow to start and drag to where you want it to end (It will look silly).
Click the 'select' arrow under the top left Structure button.
Click on your new arrow.
Mess with the resizing dots until it looks as you want it.

[Edited on 30-11-2015 by aga]

blogfast25 - 30-11-2015 at 07:20

Quote: Originally posted by aga

I got the free ChemSketch program installed now.

Seems relatively easy to use.

Aha. Drawback is that we now have plenty assignments for aga coming up!

Drawback 2 is that I'll have to install it too, to not be left behind.

Darkstar was right about my use of the PITA. I could have covered my tracks better, I guess... The main problem with PITA is drawing the orbital movements (curly arrows).

aga - 30-11-2015 at 07:36

I got mine here :

http://www.acdlabs.com/resources/freeware/

gdflp is being quiet, which is ominous, and worrying.

gdflp - 30-11-2015 at 07:41

Don't worry, I'm working on the first lecture. Might be another day or two though. Topic is Liquid-Liquid Extractions

ChemSketch is great for drawing structures, I've used it for a couple of years. It's a pain drawing arrows though, I copy the structures if I'm drawing mechanisms wherever possible.

blogfast25 - 30-11-2015 at 08:11

Just installed mine. Seems very powerful for freeware!

Currently struggling to create a hydroxide ion! You get stuck on the stupidest things when you're new to it...

Good UToob on ChemSketch here:

https://www.youtube.com/watch?v=rRdD95DhPiQ

[Edited on 30-11-2015 by blogfast25]

aga - 30-11-2015 at 09:34

Excellent video.

I got some ions by clicking on H, clicked in the sketch area and it made H₂.

Then click O, and click on the H₂ and it makes H₂O.

Then click on the + symbol down near the bottom left hand side (actually there's a triangle to select +,-, .+ or .-) then click on the H₂O

It automatically makes it into H₃O+ or HO-

blogfast25 - 30-11-2015 at 11:20

Quote: Originally posted by aga

Excellent video.

I got some ions by clicking on H, clicked in the sketch area and it made H₂.

Then click O, and click on the H₂ and it makes H₂O.

Then click on the + symbol down near the bottom left hand side (actually there's a triangle to select +,-, .+ or .-) then click on the H₂O

It automatically makes it into H₃O+ or HO-

Yeah, just discovered that. It's very intuitive use once you get the basics.

gdflp - 30-11-2015 at 11:30

It can also name compounds by the IUPAC nomenclature, with Tools>Generate>Name for Structure. There's a limit of 50 atoms unless you buy the full version, but it is still quite a useful tool to check if you're naming compounds correctly as you're learning(or if you learned and forgot

aga - 30-11-2015 at 12:01

ChemSketch is awesome !

One of the best software packages i have ever had.

Believe me, i've had a few ...

blogfast25 - 30-11-2015 at 12:58

First serious attempt:

Not too shabby!

aga - 30-11-2015 at 13:39

Vunderbar !

The ChemSketch people should be congratulating themselves.

The time from Download to producing that image can be measured in mere Minutes.

[Edited on 1-12-2015 by aga]

blogfast25 - 30-11-2015 at 14:26

Nucleophilic Substitution Reactions (general case):

In the most general context, a generic case of such a reaction can be written as:

R-Y + :Nu < === > R-Nu + :Y

In that sense it can be regarded as the displacement of a Lewis base (:Y) by a harder one (:Nu).

(Note that if one of the reaction products (or by-products) is volatile or insoluble, a displacement of a harder LB by a softer one may still be possible due to Le Chatelier)

Below is a simple example, the hydrolysis of bromoethane:

It’s known as an SN2 reaction.

But to substitute some Lewis base on a tertiary C atom a different reaction mechanism is called for because steric hindrance prevent the lone pair on :Nu from accessing the central, tertiary C atom since as it is completely surrounded by methyl groups and :Y itself.

Instead, a carbocation is formed on the tertiary C atom (stabilised by electron pushing), which then joins up with :Nu, as shown below:

This reaction mechanism is known as SN1.

[Edited on 1-12-2015 by blogfast25]

blogfast25 - 30-11-2015 at 14:32

Quote: Originally posted by aga

Vunderbar !

The ChemSketch people whould be congratulating themselves.

The time from Download to producing that image can be measured in mere Minutes.

And thanks to ChemSketch, aga can now indulge in his favourite pass time: playing with AOs and MOs:

Structure > Templates > Template window > orbitals

:cool:

blogfast25 - 30-11-2015 at 17:14

I mean, who can seriously forget the formation of a full π molecular orbital from two half-filled ungerate pz atomic orbitals, huh?

Halcyon days...

That 3D bit in ChemSketch is also awesome and I don't usually resort to superlatives.

[Edited on 1-12-2015 by blogfast25]

aga - 1-12-2015 at 13:43

Not sure 'awesome' is a superlative, however ChemSketch is Awesome.

Try the 3d modelling.

Edit:

Phenol turned up today, so i can do the phenolpthalein synth.

Would it be better to spin it off into a separate thread or post the method/photygraphs here ?

On the one hand, spinning it off elsewhere would lead to less pollution in this thread.
On the other hand it will add some photos.

Surprising how few participants this thread has attracted.

Perhaps just the Best felt able to get involved.

That can't be right : i'm still here and still fascinated.

[Edited on 1-12-2015 by aga]

blogfast25 - 1-12-2015 at 14:46

aga:

I think on this occasion and assuming the synth works more or less as planned, then posting it here as an illustration of a mechanism in practical conditions would be a Good Thing, pics included.

The 3D bit is awesome. It did get the 3D shape of AlCl4- wrong, yet not SnCl62-... Incredible value for no money.

[Edited on 1-12-2015 by blogfast25]

aga - 1-12-2015 at 14:50

OK. We get to go Technicolour !

[Edited on 1-12-2015 by aga]

Darkstar - 1-12-2015 at 14:57

Quote: Originally posted by aga

Surprising how few participants this thread has attracted.

Any thread with "quantum mechanics" in the title is bound to scare a few people away.

blogfast25 - 1-12-2015 at 14:59

Quote: Originally posted by Darkstar

Any thread with "quantum mechanics" in the title is bound to scare a few people away.

Yet look at the page view numbers: not bad at all!

aga - 1-12-2015 at 15:02

Agreed, however the majority ?

Early on there were Others, yet they seemed to Fade pretty fast.

Shame that just a drunkard actually wants to get educated.

Still, an excellent body of Work for Posterity if nothing else.

Soon to have beguiling Photos ...

blogfast25 - 1-12-2015 at 15:10

Synthesis of Alkyl Amines:

An interesting case is the SN2 mechanism of the synthesis of alkyl amines, in the example below by reaction of ammonia with bromoethane:

In the first step, a nucleophilic attack by ammonia, a substituted ammonium bromide is formed, in this case ethylammonium bromide (a salt).

Further reaction in step 2 with more ammonia (which snatches a proton, its bonding orbital then folds back onto the N atom) yields ethyl amine and ammonium bromide as a by-product.

Note that amines (primary, secondary and tertiary, see below) are all still Lewis bases (and in water also Brønsted–Lowry bases) and can thus be used in further substitution reactions. A tertiary amine reacted with an alkyl halide yields a fully substituted, quaternary ammonium ion (here for simplicity a quaternary methyl ammonium cation).

Alkylamines are in fact harder Lewis bases and stronger Brønsted–Lowry bases than ammonia itself (pKb for ethylamine is 3.3, for ammonia 4.75 for instance), due to electron pushing by the alkyl groups.

This reaction mechanism also opens the possibility of creating asymmetric secondary (or tertiary) amines, via:

R1-X + 2 :NH2-R2 === > R1-NH-R2 + [NH3R2]X

The following SM link links to an active *.pdf on the preparation of ethylamine in 90 % ethanol as solvent.

[Edited on 2-12-2015 by blogfast25]

gdflp - 1-12-2015 at 19:01

Liquid-Liquid Extractions

In organic chemistry, extractions are used to separate compounds by exploiting the differences in their solubility in various solvents. There are varying types of extractions, but only two are commonly used in lab-scale procedures. Liquid-solid extractions can be used to isolate a compound present in a solid mixture by choosing a solvent such that the desired compound is selectively dissolved. This type of extraction has the distinct disadvantage that dissolution is a kinetic property, thus extended times and/or harsh conditions are required to ensure a good efficiency. Liquid-liquid extractions circumvent this by extracting a compound which is already dissolved. The solute merely has to migrate from one solvent to another, this is a much faster process. In fact, these extractions are quick enough that they can be performed in under a minute per extraction using a separatory funnel.

Liquid-liquid extractions require two separate solvents, one for the impure solute to be dissolved in initially, and the extraction solvent which extracts the desired compound from the impure mixture. There is a wide variety of extraction solvents, but there are some characteristics which they must posses. Firstly, they must be immiscible with the initial solvent, so that the two can be physically separated after the extraction. Secondly, they must show a good solubility for the compound to be extracted, for example, dichloromethane and diethyl ether are good extraction solvents for polar compounds, whereas hexane is good extraction solvent for nonpolar compounds. Thirdly, the extraction solvent ideally shows minimal solvation towards the impurities in the compound to be extracted, to achieve a better separation. Finally, the extraction solvent must have a boiling point which varies greatly from the compound being extracted, so that the two are easily separated after the extraction. For solids, solvents with low boiling points are ideally used so that they can be removed quickly and, in the case that the substrate is heat sensitive, the compound doesn't decompose when the solvent is removed. For obvious reasons, the extraction should not form an azeotrope with the target compound.

The amount of solute which is extracted in one extraction is given by the following equation :$$ \text{Partition coefficient} = \dfrac{\dfrac{x}{\text{volume of extraction solvent}}}{\dfrac{\text{amount of solute} - x}{\text{volume of solution to be extracted}}} $$ where x is the amount of solute which is extracted in that extraction. The partition coefficient, k, is defined as follows : $$ k = \frac{\text{solubility of compound in extraction solvent}}{\text{solubility of compound in starting solvent}}$$ k is unique to a solute in a certain set of solvents, and can either be determined based on reported solubilities or experimentally.

I'm going to define some variables to clean this up a bit : $$ x : \text{Amount of solute extracted} $$$$ s_0 : \text{Initial amount of solute} $$$$ s : \text{Amount of solute present at the beginning of the current extraction} $$$$ k : \text{Partition coefficient} $$$$ r : \text{Ratio of volume of extraction solvent to volume of starting solvent} $$$$ n : \text{# of Extractions} $$
So, with these new variables we can rewrite the equation above as : $$ \text{Equation 1 : } k = \dfrac{\dfrac{x}{r}}{s_0 - x} = \dfrac{x}{r(s_0 - x)}$$ This can be rearranged to yield the following equation, solved for x : $$ \text{Equation 2 : } x = \dfrac{s_0kr}{1 + kr}$$ Now let's do an example to demonstrate the use of this equation :
Assume that the partition coefficient of caffeine in a dichloromethane/water solvent system is 6.4. 200ml of coffee contains 140mg of caffeine. If the coffee is extracted with 2 x 100ml dichloromethane, how much caffeine remains in the coffee. So, to start off, we plug all of the variables into Equation 2 : $$x = \frac
{.140 \cdot 6.4 \cdot 0.5}{1 + 6.4 \cdot 0.5} = .107g$$ Thus, 107mg of caffeine are extracted in the first extraction. Now we find our new s and solve for x in the second extraction : $$ s = s_0 - .107 = .033g$$$$ x = \frac{.033 \cdot 6.4 \cdot 0.5}{1 + 6.4 \cdot 0.5} = .025g$$ Finally we add up the amount of caffeine extracted in the first and second extraction, and subtract from the original amount in the coffee to find the amount of caffeine remaining in the coffee. $$ .140 - (.025 + .107) = .008g$$ So, in those two extractions, 94% of the caffeine was removed from the coffee using 200ml total of dichloromethane. Now let's repeat the problem except this time, we're going to use one 200ml portion of dichloromethane : $$ x = \frac{.140 \cdot 6.4 \cdot 1}{1 + 6.4 \cdot 1} = .121g $$ So now, using the same amount of dichloromethane, 19mg of caffeine remain in the coffee, thus the extraction was only 86% efficient. This demonstrates an important concept in extractions, more extractions using lesser amounts of solvent are more efficient than fewer extractions using greater amounts of solvent. Of course, this is only the theoretical aspect, in reality mechanical losses provide a limit to how small the amount of extraction solvent can truly become.

Ignoring that for a moment though, let's look at a way of speeding up determing the amount of solvent extracted in the nth extraction in a series of extractions. Firstly, I am going to define one more constant, b, which will make the following equations more aesthetically pleasing : $$ b = \frac{kr}{1 + kr} $$ Now consider an arbitrary number of successive extractions a, the total amount of solute extracted can be described by the following equation : $$ \sum_{i = 1}^{i = n} bs_i$$ where si is the amount of solute remaining in the beginning of that extraction. If we expand this to a series, we get : $$ \sum_{i = 1}^{i = n} bs_i = bs_0 + bs_1 ... + bs_n$$ We are given s0 in the original problem, but we need to define the rest. We can define s1 however, in terms of bs0 by realizing that s0 - bs0 is equal to the solute remaining after the first extraction which is equal to s1. If we make the same argument for the subsequent terms, we get the following : $$ bs_0 + bs_1 ... + bs_a = bs_0 + b(s_0 - bs_0) + b((s_0 - bs_0) - b(s_0 - bs_0)) = bs_0 + bs_0(1 - b) + bs_0(1 - b)^2 ...$$ Thus, the amount of solute extracted in the nth extraction can be defined by : $$ \text{Equation 3 : } bs_0(1 - b)^{n - 1}$$ and the total amount extracted by n extractions is : $$ \text{Equation 4 : } \sum_{i = 1}^{i = n} bs_0(1 - b)^{n - 1} $$This can be used to directly compute something quite interesting. As mentioned previously, a greater number of extractions with smaller amounts of solvent is more efficient than fewer extractions with greater amounts of solvent, even if the total solvent used is the same. Using the previous definitions, we can adapt the derived equation to represent the efficiency of a varying number of extractions. If we let : $$ r = \frac{r}{n} $$ then we get the following equation : $$ b = \frac{kr}{1 + kr} = \frac{\frac{kr}{n}}{1 + \frac{kr}{n}} = \frac{kr}{n + kr} $$ Finally, this can be plugged into Eq. 4. This allows us to calculate and compare the efficiency of two series of extractions which utilize different numbers of extractions, but the same total amount of solvent.

Just as a note, we can see that : $$\lim_{n \rightarrow \infty} \left( \sum_{i = 1}^{i = n} bs_0(1 - b)^{n - 1} \right) = s_0$$ thus, theoretically, as the number of extractions performed goes to infinity, the extraction efficiency approaches 100%(again ignoring mechanical losses).

Edit : Fixed Eq. 2

[Edited on 12-2-2015 by gdflp]

blogfast25 - 1-12-2015 at 19:17

Very nice lecture, gdflp. Nice to see there are still some SMers who don't shy away from a bit of math.

What method did you use to render the formulas so well?

Going from Eq.1 to Eq.2, did you lose an index 0 on s?

[Edited on 2-12-2015 by blogfast25]

gdflp - 1-12-2015 at 19:28

Thanks. I used LaTeX to display all of the formulas and what not, the forum supports raw coding using "$$" to mark the beginning and the end of the code. It takes a bit of time to type everything out, but it's definitely worth it for the clarity it provides. The forum uses MathJax as an interpreter, I believe that it supports several other similar languages as well.

blogfast25 - 2-12-2015 at 06:57

Quote: Originally posted by gdflp

Thanks. I used LaTeX to display all of the formulas and what not, the forum supports raw coding using "$$" to mark the beginning and the end of the code. It takes a bit of time to type everything out, but it's definitely worth it for the clarity it provides. The forum uses MathJax as an interpreter, I believe that it supports several other similar languages as well.

$$\hat{H}\Psi=E\Psi$$

Hell... so it does! And I never knew that... Neither I think do Darkstar and annaandherdad. I use it all the time on Physics.SE and have been cursing SM for a long time for not supporting it.

Wish I knew before: it would have made much of this thread so much clearer and more elegant...

[Edited on 2-12-2015 by blogfast25]

aga - 2-12-2015 at 10:08

Quote: Originally posted by gdflp

Liquid-Liquid Extractions

Looks like great stuff gdflp. Thanks !

There are just 3 bits i'm struggling with : the first bit, the middle bit and the last bit.

So this applies to any extraction, i.e. solid extractions as well.

Is that the case where extraction solvent = starting solvent ?

Quote:

more extractions using lesser amounts of solvent are more efficient than fewer extractions using greater amounts of solvent

Is this why a Soxhlet is said to be an efficient extraction tool ?

(Lots of cycles with small amounts of solvent each time)

gdflp - 2-12-2015 at 10:16

I will be covering liquid-solid extractions soon, much less math involved in them. Liquid-solid extractions are when solids are directly extracted with a solvent, whereas liquid-liquid extractions are when a solution with dissolved compounds is extracted with another solvent.

Soxhlet extractors are used for liquid-solid extractions, and there is a different reason why they are so efficient; using multiple portions of extraction solvent is typically only done with liquid-liquid extractions. I will explain them along with liquid-solid extractions since they go hand in hand. Would you like an explanation of rotary evaporators as well, since they fit in nicely with this topic?

aga - 2-12-2015 at 10:21

Of course !

Thanks for joining in and spending your time on this.

[Edited on 2-12-2015 by aga]

blogfast25 - 2-12-2015 at 10:26

Quote: Originally posted by aga

Quote:

more extractions using lesser amounts of solvent are more efficient than fewer extractions using greater amounts of solvent

Is this why a Soxhlet is said to be an efficient extraction tool ?

(Lots of cycles with small amounts of solvent each time)

Listen young man, I know you a bit and know that basic algebra is not beyond you. So please study that basic math in there because therein lies the answer to your question. Math compresses many logical statements into a few lines but 'unpacking' that logic into ordinary language can be time consuming.

/End of mild-slap-on-wrist.

If there is an object lesson to be taken away from gdflp's lecture, it's that multiple extractions using small amounts of solvent each time is MUCH to be preferred to one (or two) extractions with large volumes of solvent.

Soxhlets, useful as they are, don't really comply with that. If I had a pound for every beginner misusing a Soxhlet I'd be rich, BTW...

[Edited on 3-12-2015 by blogfast25]

aga - 2-12-2015 at 12:53

The multiple-small-extraction message stands out for sure.

In what way do <strike>we</strike> those sad, naieve noobs misuse a soxhlet ?

blogfast25 - 2-12-2015 at 13:21

Quote: Originally posted by aga

In what way do <strike>we</strike> those sad, naieve noobs misuse a soxhlet ?

1. Showing off the shiny glass when it's not needed.

2. Under-loading the cartridge: in the spirit of gdflp's lecture, use kind of as much of the material you want to leach something from as possible...

[Edited on 2-12-2015 by blogfast25]

blogfast25 - 2-12-2015 at 13:51

Industrially, L/L extractions can also be carried out in continuous mode: a rising column of the lighter solvent with the heavier solvent falling through it. The solute then transfers from one solvent to the other.

And fractionated distillations were in my engineering course (a million moons ago!) considered counter-current liquid-vapour extractions.

aga - 2-12-2015 at 14:06

Under-loading a Soxhlet ?

Confuseled.

Stuff some cotton wool in the pipe.
Add some solvent to prevent an airlock in the syphon.
Stuff the tube with the 'stuff' up near the top of the syphon pipe.
Add some more cotton wool if the 'stuff' floats.

Is there more to it than that ?

This did not work with my rig + DCM as the cooling water was not iced.

The DCM boiled & refluxed happily Above the stuff i was trying to dissolvificate.

blogfast25 - 2-12-2015 at 14:22

Quote: Originally posted by aga

Under-loading a Soxhlet ?

Confuseled.

Not putting enough coffee in the cartridge (if you're extracting caffeine from coffee, for instance). You want a decent ratio of solid (from which to extract) to solvent.

Two more sets of reaction mechanisms coming up but I'll give you a breather till tomorrow...

[Edited on 2-12-2015 by blogfast25]

aga - 2-12-2015 at 15:09

Phew ! Many thanks !

I got :-

blogfast25's brain-melting QM info via repeated hammering on the head.
Darkstar's OC Explained via arterial injection.
gdflp's liquid sep Maths via brain stem insertion.

Could take a few moments for the Noggin to Heal a bit.

Also would like to do the phenolpthalein thing and post fotygrafs.

A good Practical would be a liquid-liquid sep (hint-hint).

Edit:

(Hint-hint-hint)

Maybe a synth that includes QM, OC and the Sep things, all in one amazing technicolour Pop.

[Edited on 2-12-2015 by aga]

blogfast25 - 2-12-2015 at 16:12

Quote: Originally posted by aga

A good Practical would be a liquid-liquid sep (hint-hint).

For now I suggest to follow the vid's recipe: it seems well made.

aga - 3-12-2015 at 11:35

Well, i attempted the phenolpthalein synthesis and failed.

All went absolutely fine up to the point where the insoluble product was supposed to vac filter out.

The filter paper broke (a single hole broke through) and sucked the product+solution into the vac flask.

Uncertain where the product was, i added NaOH solution to go purple-hunting and found it everywhere.

One thing led to another and i ended up with 400ml of purplish basic solution,

In an effort to discover what PP is NOT soluble in, i added various reagents to samples in test tubes. I even tried salting out with K₃PO₄.

An amazing result happened with ~1g K₃PO₄ + 2ml toluene + 2ml ethanol.

A 12 phase mixture !?!?!

Help us out here gdflp !

blogfast25 - 3-12-2015 at 12:20

Aww, that kind of sucks, especially because you failed to recover the product, post-disaster ('keep calm and carry on').

The cause is almost certainly too high acidity of your slurry: conc. HCl ruins filter paper real easy, so another filter medium or separation method would be needed. Try repeated filtering over a PP or PE tea sieve or non-woven glass fibre. Even better here would be a high porosity glass frit filter.

aga - 3-12-2015 at 12:35

As it happens, i salvaged the product, albeit in the form of 400ml of milky liquid.

Hopefully, with tonight's low temperature and a bit of Time, there'll be some product to recover tomorrow or the day after.

If not, it'll get centrifuged - it will not escape.

Edit:

Guess i should post the glorious techncolour ...

[Edited on 3-12-2015 by aga]

blogfast25 - 3-12-2015 at 13:58

Will we be able to oxidise Xenon (or Krypton) with super acids soon?

Meant to be a bit of ‘Awe! Shock!’ light relief from an overdose of reaction mechanisms I realised after writing this up it might be a bit heavier on theory than I anticipated. Oh well: bear with me anyway...

Acids and Bases Recap:

We can define an acid’s strength as its capacity to protonate bases:
$$\mathrm{HA} + \mathrm{:B} \leftrightarrow \mathrm{A^-} + \mathrm{HB^+}$$

In Brønsted–Lowry theory we use water as a base (and standard), allowing to determine an objective constant known as the Ka:

$$K_a=\frac{C_AC_{H3O}}{C_{HA}}$$

An important corollary is that the protonated base BH+, in Brønsted–Lowry theory called the conjugated acid of B, is itself an acid.

Symmetrically, the deprotonated acid A-, in Brønsted–Lowry theory called the conjugated base of HA, is itself a base.

Most crucially the theory shows also:

The stronger the acid, the weaker its conjugated base.

The stronger the base, the weaker its conjugated acid.

Carborane Super Acids:

Super acids, with acid strengths millions of times higher than that of the most common strong mineral acid H2SO4, are nothing new and the first carborane acid was synthesized in 1967. Since then the technology and its applications have, needless to say perhaps, come a long way.

The general structure of carborane super acids is shown below (draw that in ChemSketch!):

As can be seen from the possible substituents, it’s rather a big family.

Protonate anything?

In Brønsted–Lowry theory we would never consider benzene, substituted benzenes or other so-called arenes, as bases simply because the protonating species H3O+ is far from strong enough to protonate these species. Pure H2SO4 is far from capable of protonating that stuff either: they are simply far, far too weak bases for that to happen.

But specific carboranes achieve just that and the resulting cation in the case of benzene is called benzenium - C6H7+, with the following structure:

benzenium ion.png - 3kB

Attentive readers might now expect the benzenium cation and the conjugated base of the carborane acid to form a Lewis base acid adduct in the usual, here schematised way:

$$\mathrm{A^-} + \mathrm{HB^+} \to \mathrm{A-HB}$$

where a covalent bond between the Lewis acid and Lewis base forms.

But that doesn’t happen here: the beauty of carborane super acids is that their conjugated bases are extremely weak Lewis bases that don’t form adducts, instead they form real salts: A-BH+ ionic lattices. Below is the structure of a toluenium carborate salt:

Because of the near-absolute inertness of carborate conjugated bases, carborane super acids have been referred to as “Gentle Giants” because despite their immense strength they are non-corrosive to glass.

Trialkylsilylium ions:

The cation Si4+ doesn’t exist because its small size and high charge surround it with a very powerful electrical field that would snatch hydride or oxide ions immediately, in order to lower its charge and thus its energy.

Even silicon equivalents to more traditional (electron pushing stabilised) tertiary carbocations were a bit of a Holy Grail of chemistry until the advent of carborane super acids. But at least one tri(isopropyl)silylium carborate salt has been obtained, see structure below:

Protonating alkanes:

Even plain old alkanes, known for their general chemical inertness, aren’t safe from carborane super acids, witness below the structure of a t-butyl carborate (which is stabilised by the electron pushing methyl groups}:

Oxonium Salts:

When we dissolve HCl in water we know that the hydrochloric acid deprotonates almost completely acc.:

$$\mathrm{HCl} +\mathrm{H_2O} \to \mathrm{Cl^-} +\mathrm{H_3O^+}$$

The solution can thus in a sense be considered as a solution of oxonium chloride. Evaporating the solvent does not however yield a salt and this is in part due to the fact that Cl- is a very, very weak base but still too strong.

But oxonium carborates have been isolated, like the one below:

Less well known is that besides H3O+, there’s a whole zoo of similar (but less abundant) oxonium ions out there, like: $$\mathrm{ H_5O_2^+},\:\mathrm{ H_7O_3^+},\:\mathrm{ H_9O_4^+} ...$$

Oxidising Xe (or Kr) with Carborane Super Acids?

A traditional method of oxidising metals is by means of Brønsted–Lowry acids, here for a generic bivalent metal:
$$\mathrm{M} + 2\:\mathrm{ H_3O^+} \to \mathrm{M^{2+}} + \mathrm{H_2} + 2\:\mathrm{H_2O}$$

... for metals with an SRP < 0 V.

Is it possible that the following reaction will be possible with future super acids?

$$\mathrm{Xe} + 2\:\mathrm{HCB} \to \mathrm{Xe^{2+}} + 2\:\mathrm{CB^-} +\mathrm{H_2}$$
Source.

[Edited on 4-12-2015 by blogfast25]

Phenolpthalein Synthesis

aga - 3-12-2015 at 15:00

As per Nile Red's process https://www.youtube.com/watch?v=kBo5UnNRodA

2g Phenol
1.5g Pthalic Anhydride
1ml of 96 w% H₂SO₄
were added to a 25ml RBF with a magentic stir bar.

Immediately some orange colour was seen.

The RBF was suspended in a mineral oil bath (SS dog bowl + baby oil) on a hotplate/stirrer.

The stirrer and heating were switched on and the bath allowed to heat to 150 C.

The oil begain to smoke at 150 C so the temperature was reduced to 130 C. Steam was still seen to be escaping at 130 C.

Any future attempt at this method will use a small beaker rather than an RBF and an extraction fan to more efficiently remove the steam from the reaction water.

After 1 hour the mixture took on a more Orange colour.

The RBF was removed and swirled to re-incorporate the pthalic anhydride stuck higher up the flask.

At 2 hours the mixture was a very dark purple mobile liquid.

The white crystal growth around the neck of the RBF became very noticeable, with crystals around 10cm long protruding down from the glass neck. These were scraped with a glass rod back into the reaction mixture.

On Cooling the reaction mixture became a thick tar, trapping the stirbar.

Adding 10ml distilled water + 10ml dichloromethane (DCM) did nothing to dissolve the tar.

(In future experiments the 10ml water would be added to the hot mixture. The larger volume of water should (hopefully) not flash-boil yet keep the product mobile.)

Freeing the stirbar using a glass rod was difficult, however once free, and turned to 100% stir speed the tar began to dissolve.
By angling the RBF over the more stubborn tar, most of the tar was dissolved after 15 minutes.

When stirring was stopped, the mixture formed two layers within 4 minutes.

The RBF contents were transferred to a 100ml separatory funnel, then ~2ml of DCM were used to wash the residues from the RBF into the sep funnel..

The layers separated within 5 minutes.

The Lower Layer was drained into a 100ml beaker.

10ml of DCM was added to the sep funnel, which was capped, then shaken and vented 5 times.

After 5 minutes the layers were separated, and the Lower Layer was drained into the same beaker as the previous lower layer.

The Upper layer was drained into a beaker for later disposal.

After washing the sep funnel with DIW twice, the combined Lower Layers were added back to the sep funnel, and 5ml of 2[M] NaOH solution was added.

This was capped, then shaken and vented 10 times.
It went purple, and the layers separated.

The Lower layer was drained into a 100ml beaker for later disposal.

The product-rich Upper layer was drained into a fresh 100ml beaker, then the sep funnel was washed with a small amount of DIW to flush the remainder into the same beaker.

The collected product was poured into 100ml 2 [M] HCl solution.

At this point the product was vac filtered, and the filter paper devolped a hole, creating a whole new world, including 12 phase mixtures.

(it didn't work)

The Product simply WILL be recovered somehow.

Resistance is futile.

blogfast25 - 3-12-2015 at 16:52

aga:

Looks well done except for the filtering mishap, which is at 2 M HCl is not really explained by acidity attacking the filter. Perhaps it was a faulty filter or vacuum was too high?

BTW, if you actually want to use your phenolphthalein (as a pH indicator) you'll have to wash it: ice cold DIW, a few small aliquots of it should do it, to rinse off the acid. Or some solvent in which it is insoluble but that is miscible with water.

[Edited on 4-12-2015 by blogfast25]

gdflp - 3-12-2015 at 17:08

I agree with blogfast, as long as you're using some sort of laboratory filter paper, it shouldn't break due to 2M HCl. Typically, you only run into issues with more concentrated acids, such as the solution from quenching a nitration. What are you using as a vacuum source? If you're using a rotary pump, I would try introducing a small air bleed into the system to reduce the vacuum; since even a vacuum around 150 torr is plenty for filtrations with reasonably coarse precipitates such as this. Acidic solutions at these concentrations may weaken the filter paper, and the pressure from a strong pump might cause it to tear. Worst case, you could try a gravity filtration, it will just be many painful minutes/hours longer.

As for the rest of the rest of the procedure, nice job! The initial dissolution of the crude phenolphthalein is an example of a liquid-solid extraction, and the migration of the phenolphthalein from the dichloromethane layer into the alkaline aqueous layer is an example of a liquid-liquid extraction, albeit a slightly special case. Both are going to be covered very soon.

blogfast25 - 3-12-2015 at 18:15

I also suggest to deep-clean your glassware used: my experience with PhPht is that it sticks to anything quite enthusiastically, especially glass. Use warm, alkaline water, so you can detect the last traces of it and scrub them off.

And no OC synth lab report is complete without a yield determination.

[Edited on 4-12-2015 by blogfast25]

gdflp - 3-12-2015 at 18:19

Small traces also quite happily remain on stir bars. More than once I've dropped a seemingly clean stir bar into an alkaline reaction mixture, only to have the entire thing turn bright pink. It's quite aggravating, it pissed me off enough that I bought a few extras which I keep separate and use solely for pH sensitive dyes like phenolphthalein.

blogfast25 - 3-12-2015 at 18:29

Quote: Originally posted by gdflp

Small traces also quite happily remain on stir bars. More than once I've dropped a seemingly clean stir bar into an alkaline reaction mixture, only to have the entire thing turn bright pink. It's quite aggravating, it pissed me off enough that I bought a few extras which I keep separate and use solely for pH sensitive dyes like phenolphthalein.

Yeah, it's really stubborn, isn't it? You can have some real mystifying and very pink surprises if you don't remember/understand PhPht was involved...

[Edited on 4-12-2015 by blogfast25]

gdflp - 3-12-2015 at 19:02

Liquid-Liquid Extractions : pH Manipulation

Liquid-liquid extractions are a powerful tool for separating organic compounds, but there are several tricks to increase the number of situations in which they can be used. One of these tricks is modifying the pH of the aqueous portion of an extraction to manipulate the solubility of one of the components. By either raising or lowering the pH of the solution, the aqueous and organic solubility of compounds can be changed quite drastically, for example a saturated aqueous solution of triethylamine at 25°C is 0.5M whereas a solution of triethylammonium chloride, the salt formed when triethylamine is precisely neutralized with hydrochloric acid, is 10.5M. The two common classes of organic compounds which experience these drastic solubility changes are amines, which become soluble in acidic solutions, and carboxylic acids, which become soluble in basic solutions. There are other compounds with low pKas, but these are the most commonly encountered.

Amines and Carboxylic Acids

As you can see by the above diagram, in acidic solutions, amines become significantly more water soluble due to the ionic bond present in the new compound, a salt. Likewise, in basic solutions, carboxylic acids become significantly more water soluble as their corresponding ionic salt. This technique can be used to separate amines and carboxylic acids from neutral organic compounds. Separation is accomplished by first creating a solution of an amine or carboxylic acid and a neutral organic compound in an organic solvent such as diethyl ether or dichloromethane, then washing with either a dilute(~1 - 2M) solution of an acid, typically hydrochloric acid, or a base, typically sodium hydroxide, depending on the substrate.

An example of this is the phenolphthalein synthesis which you did; since phenolphthalein contains acidic protons, it can be treated as a carboxylic acid with respect to this method. The dichloromethane solution of phenolphthalein containing various byproducts was washed with 2M sodium hydroxide, which allowed the phenolphthalein to migrate from the organic layer to the aqueous layer, leaving behind impurities. The solution was then reacidified to reduce the water solubility of the phenolphthalein and allow it to be recovered, as is typical with these types of extractions.

[Edited on 12-4-2015 by gdflp]

blogfast25 - 3-12-2015 at 19:43

Quote: Originally posted by gdflp

3An example of this is the phenolphthalein synthesis which you did; since phenolphthalein contains acidic protons, it can be treated as a carboxylic acid with respect to this method.

Someone will have to explain that to aga, using Brønsted–Lowry (BL) theory with regards to pH indicators, their 'pH-chromism' and that they are invariably weak acids... to get that message across. Unless gdflp wants to oblige now, I will do it tomorrow.

Very nice post, BTW. :cool:

[Edited on 4-12-2015 by blogfast25]

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