Quantum Mechanics/Wave Mechanics in Chemistry Beginners Thread

Pages:  1  ..  6    8   blogfast25 - 11-12-2015 at 11:39

 Quote: Originally posted by aga Is the "Oxygen group" All elements on the O column (#16) of the periodic table?

Yes, all of them form bivalent hydrides(H2X) and have two lone electron pairs on the X. They're all Lewis bases, to varying degrees.

[Edited on 11-12-2015 by blogfast25]

blogfast25 - 11-12-2015 at 12:04

And now I have to add the following.

Hybridisation also depends on atom size and difference in electronegativity, see e.g. the highly stable SF6. It's shape can be found here.

The hybridisation is sp<sup>3</sup>d<sup>2</sup>, six equivalent half-filled AOs, ready for bonding to the F atoms.

[Edited on 11-12-2015 by blogfast25]

blogfast25 - 11-12-2015 at 12:37

Also:

For double bonds in C-C bonds, both C atoms hybridise to sp<sup>2</sup> and the p<sub>z</sub> isn't involved. The sp<sup>2</sup> AOs bond with each other and the H atoms (σ bonds), the two p<sub>z</sub> form a π bond (above and below the C-C σ bond).

Will add a figure (not to scale):

All half-filled sp<sup>2</sup> AOs and σ bonds in the xy plane. Two ungerate half filled p<sub>z</sub> (z axis) AOs form a π bond (above and below the C-C σ bond).

Note: no hydrogen atoms shown.

[Edited on 11-12-2015 by blogfast25]

aga - 11-12-2015 at 12:44

Enlightenment has finally occurred.

I think i finally Get It.

Well taught teach !

Now to see if i can apply the teachings ...

blogfast25 - 11-12-2015 at 13:05

 Quote: Originally posted by aga Now to see if i can apply the teachings ...

Assignment: draw a triple bond: 1 σ + 2 π bonds. The hybridisation is sp.

[Edited on 11-12-2015 by blogfast25]

aga - 11-12-2015 at 13:26

Feck
aga - 12-12-2015 at 12:35

Tried in ChemSketch and got nowhere.
Ditto MS Pain.
Best i could manage was in Fireworks :-

Blue is the sigma bond, green one pi bond orbital, magenta/red the other pi orbital, with magenta behind, red in front.

[Edited on 12-12-2015 by aga]

blogfast25 - 12-12-2015 at 12:35

We're like sips passing in the night.

That's correcto!

I'll show you mine.

[Edited on 12-12-2015 by blogfast25]

aga - 12-12-2015 at 12:40

Sops pissing to the right ?

blogfast25 - 12-12-2015 at 12:55

Ships. LOL.

I had a bit of trouble with it in ChemSketch too.

Ignore the acetylene and pent-2-yne for a minute.

My second pair of π lobes are in (parallel to) the xy plane and painted as red solid lines. So your structure is correct.

With one H on each of the two sp orbitals we get acetylene (IUPAC: ethyne), which is an interesting case. Due two the high electron density between the to C, both H are very slightly acidic (but less than ethanol, e.g.) and salts of it exist, in particular calcium carbide, CaC<sub>2</sub>. With water that generates acetylene because the acetylene is such a weak acid that even water displaces it.

Another thing about alkynes is that the triple bond imparts on the molecule some linearity not seen in alkanes, see the case of pent-2-yne, where the C atoms 1,2,3 and 4 are all on one line! Compare to the 'squiggly' structure of pentane, e.g.

And talking about double/triple bonds, I'm going to present a few more exercises here, shortly.

[Edited on 13-12-2015 by blogfast25]

blogfast25 - 12-12-2015 at 15:09

Using the electrophilic addition mechanism to alkenes and Markovnikov's Rule, predict the reaction products of the following additions:

[Edited on 12-12-2015 by blogfast25]

aga - 14-12-2015 at 13:38

Not sure about any of these, especially the last one.

It just looked stupid so it got rearranged to look better.

blogfast25 - 14-12-2015 at 15:53

Quite a few errors there. Not quite sure why.

The ones with H2SO4 are perhaps understandably harder because you have to realise that HSO<sub>4</sub><sup>-</sup> acts here as a Lewis base and latches on to the carbocation. Will add a diagram tomorrow.

Markovnikov says that the carbocation is formed on the C-atom that is already the most connected to other C-atoms.

In order of stability: primary carbocation < secondary carbocation < tertiary carbocation.

[Edited on 15-12-2015 by blogfast25]

aga - 15-12-2015 at 06:52

 Quote: Originally posted by blogfast25 Quite a few errors there. Not quite sure why.

Yes, sorry.

#5 was just idiocy and not being able to count to 4.

Seeing the mechanisms for #2 and #3 would be very helpful.

I think i've hit my limit on how much can be absorbed in a given time.

Certainly need to revise the past several lessons again (a few times).

blogfast25 - 15-12-2015 at 08:04

 Quote: Originally posted by aga #5 was just idiocy and not being able to count to 4. Seeing the mechanisms for #2 and #3 would be very helpful. I think i've hit my limit on how much can be absorbed in a given time. Certainly need to revise the past several lessons again (a few times).

#2:

Note that H2SO4 === > H+ + HSO4(-)

Although the hydrogen sulphate ion (bisulphate ion) is a Bronsted Lowry acid (quite a strong one too!) it has unbonded electron pairs on the O atom, one of which bonds to the hard Lewis acid that is the t-butyl carbonium cation. That bisulphate ion is negatively charged helps, of course.

The resulting structure is itself a strong acid by virtue of that dangling -OH group.

In ChemSketch the structure can be easily drawn as follows. Draw your C structure. Add an O atom where you want the sulphate to be. Choose HSO<sub>3</sub> (right hand side of structure menu) and add it to the O.

A well known acid in this category is the following:

It's a strong, non-oxidising acid, often used in OC.

************

Why don't we build in a pause and you can work on your empirical dissertation: the conversion of alpha-pinene to alpha-terpineol (and beyond)?

I've found another potentially doable reaction with alpha-pinene: the preparation of bornyl chloride, so the money spent on a bottle of high quality turpentine would be well spent!

[Edited on 15-12-2015 by blogfast25]

aga - 15-12-2015 at 11:13

A Pause before further enlightment rains down would be appreciated.

Some practical would be welcome !

blogfast25 - 15-12-2015 at 12:11

 Quote: Originally posted by aga A Pause before further enlightment rains down would be appreciated. Some practical would be welcome !

Agreed.

blogfast25 - 19-12-2015 at 07:54

'Blast from the past':

To see actual calculated versions of the most common molecular orbitals, use this latest version version of the Java applet. They are for the H2<sup>+</sup> molecule (no inter-electronic repulsions).

Only works in IE (NOT Chrome) and you'll probably have to gratis upgrade your Java version.

A π (ungerate) 2px molecular orbital:

[Edited on 20-12-2015 by blogfast25]

Fugitive Fermions

Metacelsus - 15-1-2016 at 12:40

I explain quantum tunneling to my younger brother.

He says:

You mean, El Chapo electrons?

aga - 15-1-2016 at 14:22

 Quote: Originally posted by blogfast25 A π (ungerate) 2px molecular orbital

Each half is remarkably similar in shape to a Haemogoblin-containing cell.

HeYBrO - 16-1-2016 at 22:33

Blog the structure you show is not p-TsOH. p-TsOH forms via electrophilic aromatic substitution as the active electrophile is HSO3/SO3 where the sulphur is attacked by the C=C in the arene.

however the below mechanism would in practice be completed using oleum, rather than H2SO4 as with toluene. This is as the methyl group is electron donating which activates the ring system.

source: http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch12/ch12-4...

blogfast25 - 17-1-2016 at 06:39

 Quote: Originally posted by HeYBrO Blog the structure you show is not p-TsOH. p-TsOH forms via electrophilic aromatic substitution as the active electrophile is HSO3/SO3 where the sulphur is attacked by the C=C in the arene.

Ok, thanks.

HeYBrO - 17-1-2016 at 15:39

I gave the correction not so much for you but just so no one becomes confused with your structure, hence gave an explanation of the mechanism. great thread btw
blogfast25 - 17-1-2016 at 18:37

 Quote: Originally posted by HeYBrO I gave the correction not so much for you but just so no one becomes confused with your structure, hence gave an explanation of the mechanism. great thread btw

No worries.

That mechanism you showed seems for the sulphonation of benzene with oleum, though, as it starts with an attack on the π-ring by SO3, not by H<sup>+</sup>.

HeYBrO - 17-1-2016 at 20:46

Quote: Originally posted by blogfast25
 Quote: Originally posted by HeYBrO I gave the correction not so much for you but just so no one becomes confused with your structure, hence gave an explanation of the mechanism. great thread btw

No worries.

That mechanism you showed seems for the sulphonation of benzene with oleum, though, as it starts with an attack on the π-ring by SO3, not by H<sup>+</sup>.

I am aware that it is benzene and oleum in the mechanism and hence stated so in my original post. Further, what does H<sup>+</sup> have to do with sulphonation/what i posted? it is not required for sulphonation of aromatic rings in what i have posted (toluene and benzene). I don't recall of any sulphonation in this context that has H<sup>+</sup> as the active electrophile.

edit: i must be misunderstanding what you mean.. Could you please explain?

[Edited on 18-1-2016 by HeYBrO]

blogfast25 - 18-1-2016 at 05:39

Take no notice: it was me getting the structure of p-TsOH wrong!

Thanks for pointing it out.

[Edited on 18-1-2016 by blogfast25]

Darkstar - 3-3-2016 at 01:20

Dear students and faculty members of B&D University,

We are pleased to announce that the teachers' strike resulting from the recent budget cuts has officially concluded. After three long months of non-stop protesting, the lengthy standoff between B&D University and members of its chemistry department finally came to an end Wednesday evening after the successful negotiation of a new contract. Under this new agreement, the university will significantly increase all chemistry-related funding, allowing the school's laboratories to be retrofitted with the modern glassware and equipment necessary to conduct college-level experimentation. For the first time since its inception, B&D University will no longer force students to carry out their chemical reactions via makeshift condensers and flasks made from copper pipes and pickle jars.

And now that I'm back from striking, I can't help but wonder how B&D University's brightest (and only) pupil has been doing since I've been away. Is aga still on his noble journey to learn organic chemistry or has he decided to throw in the towel and go back to inorganic?

[Edited on 3-3-2016 by Darkstar]

blogfast25 - 3-3-2016 at 07:03

@Darkstar:

Carbanions, Darkstar, carbanions! Organometallics, that's one subject that's been 'studiously' avoided so far...

Magpie - 3-3-2016 at 08:50

There's a subject I would like to see B&D university cover: activity coefficients and fugacity. I know I can read about these on Wiki but what I'm looking for is how a chemist actually uses them.

If you can't look them up of what value are they? Can they be derived?

When are they of value to the home chemist?

blogfast25 - 3-3-2016 at 09:05

 Quote: Originally posted by Magpie There's a subject I would like to see B&D university cover: activity coefficients and fugacity. I know I can read about these on Wiki but what I'm looking for is how a chemist actually uses them. If you can't look them up of what value are they? Can they be derived? When are they of value to the home chemist?

Activity coefficients were briefly covered here with some worked examples, in case you missed it.

What use these concepts have to the home scientist is the same for professional chemists: it depends enormously what you're doing or are trying to achieve.

[Edited on 3-3-2016 by blogfast25]

aga - 5-3-2016 at 00:38

 Quote: Originally posted by Darkstar no longer force students to carry out their chemical reactions via makeshift condensers and flasks made from copper pipes and pickle jars.

Yay ! At last something worth stealing from the lab !

 Quote: ... wonder how B&D University's brightest (and only) pupil has been doing since I've been away. Is aga still on his noble journey to learn organic chemistry or has he decided to throw in the towel and go back to inorganic?

Got stalled in the simple (!) process of making a tertiary alcohol.

Depression set in over the senseless waste of so many innocent reagent's lives, then turned to booze and drugs, as is the B&D University tradition. Hopefully the new lab equipment will make that cheaper.

Speaking of which, exactly how does a tiny amount of water interfere in the chlorination of ethanol ?
(anhydrous=rapid formation of white trichloroacetaldehyde precipitate, any water=clear, sweet smelling solution).

Glad to see you back Darkstar !

blogfast25 - 5-3-2016 at 05:10

 Quote: Originally posted by aga Speaking of which, exactly how does a tiny amount of water interfere in the chlorination of ethanol ? (anhydrous=rapid formation of white trichloroacetaldehyde precipitate, any water=clear, sweet smelling solution). Glad to see you back Darkstar !

Do post that question with relevant data in the t-alcohol thread, old chap. We'll try and tend to it.

'Lectures' will resume very shortly here.

Darkstar - 6-3-2016 at 15:00

 Quote: Originally posted by aga Speaking of which, exactly how does a tiny amount of water interfere in the chlorination of ethanol ? (anhydrous=rapid formation of white trichloroacetaldehyde precipitate, any water=clear, sweet smelling solution).

This should probably be discussed over in blog's OC thread, but isn't the alcoholate the primary product when chlorinated under anhydrous conditions? I'm guessing the white precipitate is chloral alcoholate. That would make sense considering the acidic environment due to HCl production and the excess ethanol present, which would shift the equilibrium in favor of hemiacetal formation. Why the chloral doesn't go all the way to an acetal under those conditions, I have no idea. Perhaps the hemiacetal is unusually stable for the same reason the hydrate is thanks to the adjacent electron-withdrawing chloro groups? If you have the means, you could try a melting point test on the precipitate.

Also, I kind of doubt that water is directly interfering with the chlorination since acetaldehyde is commonly chlorinated under aqueous conditions to give chloral hydrate. Which also leads me to believe that you're probably producing the hydrate when using wet ethanol. I'd imagine that the hydrate's solubility would be somewhat increased due to the extra water present. And since the hydrate is likely more soluble in the ethanol-water mix than the chloral alcoholate is in the anhydrous ethanol, that might also explain why there's no precipitate in the case of the former.

Anyway, we should probably discuss this in the other thread. I'm also down with continuing your "education," so just let me know if you'd like to. Perhaps we can even make a lesson out of this reaction? I see that blogfast has already lightly touched on some possible mechanisms in the OC thread. I'm thinking this reaction would be a good introduction to bond homolysis, free radicals and free-radical halogenation. And considering the increasing popularity of using bleach to oxidize alcohols, that initial oxidation of ethanol to acetaldehyde by chlorine makes for a rather interesting discussion all on its own. My guess is that it goes through an unstable ethyl hypochlorite intermediate that quickly decomposes into acetaldehyde and HCl.

Speaking of which, I actually made a post a while back about these sorts of oxidations if you're interested. The mechanism proposed was for the oxidation of THF to GBL using aqueous calcium hypochlorite in acetonitrile with acetic acid as the catalyst. And although the mechanism shows the active oxidizing species to be an activated hypochloronium ion (H2ClO+), these days I actually tend to lean more towards molecular Cl2 itself as the real oxidizer in bleach oxidations.

Edit: Now that I think about it, the ethanol in the chloral synthesis could very well also be oxidized via a competing radical disproportionation mechanism as well, bypassing any sort of ethyl hypochlorite intermediate. Welcome to the wonderful world of radical chemistry where no one has any real clue as to what the hell is actually going on.

[Edited on 3-7-2016 by Darkstar]

aga - 7-3-2016 at 00:31

Woohoo !

Some theory on the clorination of ethanol would be great, seeing as that's the first step in this synthesis i'm failing with.

Carbon atoms as Lewis bases:

blogfast25 - 7-3-2016 at 10:00

Recapping from what we seen so far, many reactions in OC can be summarised by the following scheme:
$$\text{A}+\text{:B} \to \text{A-B}$$

Where A is a Lewis acid, :B a Lewis base and : an unbonded electron pair and A-B the adduct. Of course this step may be preceded and/or followed up with other steps to make up the full reaction path.

So far, our Lewis acid has invariable been a partly positively charged C-atom or a carbocation.

Is it possible for a C-atom to play the part of a Lewis base, an electron pair donor? Enter organo-metal compounds.

Organo-metal compounds:

For those unfamiliar with this class of chemical compounds, there’s at least one you’ll have heard of, i.e. the now infamous lead-based anti-knocking agent tetraethyl lead:

A more interesting one is methyl lithium, CH3Li, or LiMe.

Over-simplified (the real structure is more complicated and less than fully ionic), LiMe would be made up of negatively charged CH3 so-called carbanions and positively charged Li cations. The structure of the carbanion is shown below:

The Me carbanion has a tetrahedral structure with an unpaired, filled MO sitting ‘on top’. This makes the ion a very hard Lewis base.

To the right is a generically substituted carbanion. The substitutions tend to stabilise the ion by charge distribution.

As said, the representation of LiMe as an ionic compound is a simplification but one thing is certain: due to the big difference in electronegativity between Li and C, the Li-C bond is highly polarised, with a strong partial charge residing on the C-atom. This makes LiMe a useful reagent in OC.

Organo-metal compounds like LiMe are particularly useful as reagents towards compounds containing a partly positively charged C-atom, like the C=O carbonyl group. This double bond is highly polarised due to the difference in electronegativity between C and O.

Here’s an example of the reaction between LiMe and acetone (carried out in an ether-type solvent):

In the first step the lone electron pair of the carbanion latches on to the partly positively charged carbonyl C-atom, while the double carbonyl bond opens up to the Li cation. An adduct has been formed.

To complete the synthesis, in the second step some weak acid solution is added, which hydrolyses the adduct to a tertiary alcohol.

Note that in the reaction two carbon atoms have been linked together, it’s a so-called C-C coupling reaction.

LiMe has many drawbacks though: its incompatible with oxygen, CO2 and reacts violently with water and alcohols to boot.

Safer, tamer organo-metal reagents have therefore been developed.

Next instalment: Grignard organo-metal reagents.

[Edited on 7-3-2016 by blogfast25]

aga - 12-4-2016 at 15:09

Hello ? Anybody here ? hello ? ello ?

An echo. Good. I must be the first here.

Best make some nitrogen triodide and paint it on certain things ...

blogfast25 - 12-4-2016 at 16:29

 Quote: Originally posted by aga Hello ? Anybody here ? hello ? ello ? An echo. Good. I must be the first here.

Hmmm... for some time one of the absentees here was you.

Normal service to be resumed very shortly now.

aga - 22-4-2016 at 11:48

Dear B & D professors,

OC wise, are those two Cl on 1,4 dichlorobenze not rip-offable, e.g. with conc NaOH, perhaps with a little electrolysis stimulation ?

From what little i vaguely recall, the delocalised electron nature of the benzene ring should make those Cl more vulnerable than if they were bonded otherwise.

Give them some Na to party with, and well, who knows ? (not me).

That compound came up as a napthalene substitute in mothballs while i was looking for an OTC precursor to Anthracene, which i would still like to play with seeing as it's an OC semi-conductor.

[Edited on 22-4-2016 by aga]

Darkstar - 23-4-2016 at 10:52

Welcome back to class. Here is your lesson on aryl halides for the day (using mono-chlorobenzene).

 Quote: Originally posted by aga OC wise, are those two Cl on 1,4 dichlorobenze not rip-offable, e.g. with conc NaOH, perhaps with a little electrolysis stimulation?

Without a strong electron-withdrawing group (EWG) ortho or para to the halogen, the only way you're going to get a chloro group off of a benzene ring with NaOH is by reacting them under extremely harsh conditions (like 300 °C at 200 bar/3,000 psi). Cathodic reduction in an electrochemical cell will remove a halogen from a benzene ring, though, albeit through an entirely different mechanism.

As far as the reaction between aryl chlorides and NaOH goes, if an EWG like -NO2 is present, either in the ortho- or para-position (or both), NaOH will remove the chlorine by first adding to the benzene ring and breaking aromaticity. The unstable intermediate then collapses back into a much more stable aromatic system by eliminating either the hydroxyl group or the chloro group. If the hydroxyl group gets eliminated, a nucleophilic attack by hydroxide can occur again. If the chloro group gets eliminated, the reaction stops. This kind of reaction is a type of nucleophilic aromatic substitution, and proceeds via the usual SNAr mechanism:

While not shown in the simplified mechanism above, the intermediate of the second step is stabilized by the nitro group like this:

Notice that not only does the nitro group help to further delocalize the negative charge by spreading it out across more atoms, but the nitrogen also has a positive charge on it, increasing the stability of the adjacent negative charge on the carbon that the nitro group is connected to. And because the nitro group is strongly electron-withdrawing, it also significantly increases the electropositivity of the carbon opposite of it as well, making the chloro group carbon much more susceptible to a nucleophilic attack by hydroxide in the initial step of the substitution.

Also, as mentioned above, at high enough temperatures and pressures it's possible for substitution by NaOH to take place even without an EWG; however, substitution in this manner proceeds through a different mechanism that involves the formation of an extremely reactive benzyne intermediate:

 Quote: From what little i vaguely recall, the delocalised electron nature of the benzene ring should make those Cl more vulnerable than if they were bonded otherwise.

It's actually the other way around. Because the benzene ring's carbons are sp2-hybridized and the C-Cl bond has partial double bond character (due to the partial delocalization of one of chlorine's lone pairs into the aromatic ring), the bond length between chlorine and carbon in chlorobenzene is actually shorter than in alkyl chlorides, making it much stronger and harder to break.

Look at the resonance structures for chlorobenzene:

Notice that three of the structures have a double bond between chlorine and carbon. This means that, in reality, the C-Cl bond isn't a true single bond, and the actual structure of chlorobenzene has a partial double bond between chlorine and carbon.

 Quote: Give them some Na to party with, and well, who knows ? (not me).

Depending on the solvent and conditions employed, sodium metal can indeed remove aryl chloro groups. What happens is that the sodium transfers an electron to the substrate to give a radical anion, which quickly decomposes into a chloride anion and a phenyl radical. A second sodium atom then transfers an electron to the phenyl radical to give phenylsodium (an organometallic compound):

Subsequent reaction of this superbase with a proton source will replace the sodium with a hydrogen.

aga - 23-4-2016 at 11:08

So OC is just as hard as Calculus then. OK.

Some bits to clearup if you would be so kind : 'strong EWG' is still a bit fuzzy.

Is that relatively strong, or strong enough to make All the difference reaction-behaviour-wise ?

For the final mechanism you posted, is this more-or-less the same for dry-distilling any benzene-based compound with NaOH to get benzene ?

blogfast25 - 23-4-2016 at 11:46

 Quote: Originally posted by aga So OC is just as hard as Calculus then. OK.

Based on this particular chapter I'd say it was the other way around!

aga - 23-4-2016 at 11:56

 Quote: Originally posted by blogfast25 Based on this particular chapter I'd say it was the other way around!

Yes, you're right. Sorry. I'll reverse it:

.KO .neht suluclaC sa drah sa tsuj si CO oS

Darkstar - 23-4-2016 at 12:24

 Quote: Originally posted by aga Is that relatively strong, or strong enough to make All the difference reaction-behaviour-wise ?

The stronger the electron-withdrawing group is, the easier it will be to substitute the halogen and the less harsh the reaction conditions will need to be. Having more EWGs on the ring ortho and para to the halogen will also make it easier. For example, substituting chlorobenzene with NaOH requires reaction temperatures of 300+ °C and pressures exceeding several thousand psi, while substituting 2,4,6-trinitrochlorobenzene can be done under normal atmospheric pressure at nearly room temperature.

 Quote: For the final mechanism you posted, is this more-or-less the same for dry-distilling any benzene-based compound with NaOH to get benzene ?

I think there may have been a misunderstanding on that last part. By "give them some Na to party with," I thought you meant elemental sodium. But looking back, I'm guessing you really meant sodium ions. That last reaction I posted is between chlorobenzene and actual sodium metal. The sodium metal reduces off the R-X halide group to give an organosodium compound, which is an extremely strong base that could then be protonated to give the desired R-H group. Blogfast did a lesson on organometallics just a few posts up.

aga - 23-4-2016 at 12:31

Yes, my language was vague because i really do not know what i'm talking about.

Apologies.

I must have missed that reaction mechanism lesson and will go back and read it.

Thanks for the pointer, and we'll call it quits on the Epcot Centre Incident.

All photos erased.

Darkstar - 1-5-2016 at 12:10

 Quote: Originally posted by aga All photos erased.

How can we be sure that you won't try to recover them for use in future blackmail attempts? Surely you're aware that nothing's ever truly deleted simply by emptying the recycle bin. So just to be safe, we are sending someone from IT over to your dorm to scrub your hard drive. A three-pass wipe ought to be enough.

aga - 1-5-2016 at 12:44

Three-pass wipes are strictly reserved for the day after the Michlemas Dinner, as Sir should be aware.

Edit:

What happened to the alkycarboxywotsits and all that Electron pushing ?

More Edit:

Strange, but earlier today i went back and re-read a lot of this thread and copied out those group/name structures you posted many moons ago.

[Edited on 1-5-2016 by aga]

Darkstar - 1-5-2016 at 15:10

Whenever you're ready for more alkycarboxywotsits, red squiggly lines, equilibrium arrow thingies, snakes and ladder diagrams, nomenclature mumbo-jumbo etc, just say the word.

Speaking of which, I've been wondering if perhaps these lessons would be more effective if you were the one deciding the curriculum. I was thinking that maybe you could pick some topics that you think you might be interested in learning more about, either simply out of curiosity or for actual practical application, or even just as a refresher on stuff we've already covered. We could always go back over the material you're still struggling with.

Anyway, where we go next is up to you. There's definitely no shortage of OC topics to choose from. If you're still having trouble predicting the products of some of the more common types of reactions or proposing plausible mechanisms for them along with proper arrow pushing, we could always go back to some of the reactions we've covered in the past.

We could even go over the OC exam that you took back in December. Unfortunately I was so busy with school at the time that I never got the chance to really go over the questions you missed like I had intended to. You did appear to have a pretty good grasp on electrophiles and nucleophiles from what I could tell, which are the very foundation of organic chemistry and the majority of its reactions; however, like gdflp pointed out back in December, you also seemed to be having trouble drawing the actual mechanisms through which they react with one another, particularly the electron pushing parts.

aga - 1-5-2016 at 23:37

gdflp is right !

Yes, it would be great if we could go over the mechanisms again, especially the ones i got totally wrong.

Edit:

It would be awesome if we could do the Theory behind a particular reaction, then do the Practical.

There's still the TCA synthesis so we can convert pinene to turpineol, although that may be a series of difficult cases.

[Edited on 2-5-2016 by aga]

Darkstar - 2-5-2016 at 04:36

 Quote: Originally posted by aga It would be awesome if we could do the Theory behind a particular reaction, then do the Practical.

I think blogfast has a lesson on Grignard reagents already typed up that he's been waiting to post. You want to continue on with the organometallic theme and try a Grignard reaction? They aren't too terribly complicated or anything, but everything must be as dry as possible. A small amount of iodine to kick off the reaction between the magnesium metal and the alkyl halide will help speed things up as well.

aga - 2-5-2016 at 06:30

Cool ! Let's go !

My new fume hood is about a week from completion tho ...

aga - 2-5-2016 at 08:48

Hmm. A week should be almost enough time to understand the reaction mechanisms !

Grignard reagents and reactions:

blogfast25 - 2-5-2016 at 09:24

This post is the follow up to Carbon atoms as Lewis bases, please refresh as what follows might not make a lot of sense otherwise.

Grignard reactions make use of organo-metallic compounds known as Grignard reagents, obtained by reacting an organo-halide with magnesium metal. The resulting complex R-CH<sub>2</sub>-Mg-X carries a partial negative charge on the Mg bonded C atom due to the strong difference in electronegativity between C and Mg, making it a strong nucleophile (Lewis base).

Below’s an example of the full sequence (three distinct steps) of a Grignard reaction between 1-bromopropane, Mg and pentan-2-one:

Step 1: formation of the Grignard reagent.

1-bromopropane is slowly added to a suspension of Mg shavings in super-dry ether of THF. The organo-magnesium complex forms.

Step 2: formation of the Adduct.

pentan-2-one is added to the solution of Grignard reagent. The partially negatively charged C-Mg atom bonds to the partially positively charged C atom of the carbonyl >C=O group. The partially positively charged Mg atom bonds to the O atom in the carbonyl group.

The reaction between two carbon atoms (one as Lewis base, one as Lewis acid) is often called a C-C coupling reaction or chain extending reaction.

Step 3: hydrolysis of the Adduct.

The adduct is rarely the desired end-product. Usually and equivalent of dilute HCl solution is added, which converts the adduct to alcohol by electrophylic attack of H<sup>+</sup> on the Mg-O bond.

After work up, the end product 4-methylheptan-4-ol is then obtained.

Carbonyl groups are very suited as Grignard substrates because of the strong polarisation of the C=O bond, as shown here.

As regards the organo-halide starting point, suitability depends on choice of halogen. In order of suitability: I>Br>Cl>F.

Water is the enemy of this reaction, as water hydrolyses the Grignard reagent. All reagents used must therefore be of suitable dryness.

[Edited on 2-5-2016 by blogfast25]

aga - 2-5-2016 at 11:47

Perhaps a week was being optimistic ...

So, first off we got the formation of the Grignard reagent.

Electronegativites of Mg =1.3, C=2.6, Br=3.0

The Br atom and the adjacent C fight over the weaker Mg and rip the s**t out of it, leaving that set of 3 atoms (the functional group ?) unstable, in that the group are electron deficient.

Reading up (in this thread) why the C ends up being a strong Lewis Base instead of the Mg or Br ...

blogfast25 - 2-5-2016 at 12:51

 Quote: Originally posted by aga [...] in that the group are electron deficient. Reading up (in this thread) why the C ends up being a strong Lewis Base instead of the Mg or Br ...

It's simpler to merely consider that the electron density (ψ2 is the light green 'bell', remember that? ) of the Mg-C bond is skewed toward the C-atom (the picture is a schematic and not to any scale):

The C-atom therefore carries a partial negative charge.

[Edited on 2-5-2016 by blogfast25]

aga - 2-5-2016 at 13:55

The electron density probability plot is pretty much in the old noggin since the first time.

An amazing feat of Teaching, considering the density of this noggin.

Inescapable is the Br's superior electronegativity, which is what makes me wonder why It is not the 'active' component instead of the C.

I get the feeling that it is an accomplice to a far more complex mechanism that a noob like me should not be asking about in public.

[Edited on 2-5-2016 by aga]

blogfast25 - 2-5-2016 at 14:34

 Quote: Originally posted by aga Inescapable is the Br's superior electronegativity, which is what makes me wonder why It is not the 'active' component instead of the C.

Good question but easily answered. Br's electronegativity means that it won't share another of its three remaining unbonded MOs with any Lewis acid. So structures like -Mg-Br-C- can never arise. Br- is an extremely weak Lewis base.

One could imagine a [H-Br-H]+ ('bromonium cation') structure but in reality that doesn't exist (bar perhaps in extreme acid conditions).

aga - 3-5-2016 at 07:55

OK. Looking at the orbitals and trying to relate that to Lewis dots, it appears to me like this :-

The Mg - Br bond is a σsp effectively filling Br's p orbitals.

The Mg - C bond is also σsp however C's p orbitals are still incomplete, therefore less stable, so it's -ve charge makes it more nucleophilic than the -ve charge on the more stable Br electron configuration.

Add to that the higher negativity of Br and it is less likely to let go of any electrons than either C or Mg.

Mg has no more 'easy' electrons to share in bonding, which leaves C as the atom most likely to engage in further bonding, making it the 'active' bit.

Poor old Mg has been robbed down to a slight +ve charge, and is so depressed it's unwilling to even talk about it.

Darkstar - 3-5-2016 at 08:32

 Quote: Originally posted by aga Perhaps a week was being optimistic ...

Just keep asking questions until everything makes sense. We are more than happy to share our knowledge with you. I'll also add some stuff on Grignards to further supplement blog's post. For consistency, I'll also use 1-bromopropane and pentan-2-one.

 Quote: So, first off we got the formation of the Grignard reagent.

Formation of Grignard reagents are thought to proceed through a SET (single-electron transfer) mechanism that takes place on the surface of the magnesium metal. For the sake of simplicity, you can envision this reaction as consisting of three steps: oxidation of the magnesium metal, reduction of the organic halide, and formation of the R-MgX complex:

For the purposes of preparing a Grignard reagent as part of your next practical, the above explanation should be more than sufficient; however, if you would like a slightly more advanced understanding of what is going on, we can further break the reaction down to elementary steps and intermediates. While the true mechanism for the formation of Grignard reagents isn't actually known (particularly the mechanism of the halide reduction step), the reaction most likely involves an initial electron transfer from the magnesium metal to the organic halide (R-X) to give a radical anion (R−X−•) and a magnesium(I) cation (Mg+).

The unstable radical anion then collapses into an organic radical (R•) and a halide anion (X), the latter of which serves as the new counterion for the magnesium(I) ion (Mg+X). The magnesium(I) ion then transfers a second electron to the organic radical to give a carbanion (R) and a magnesium(II) ion (Mg2+), which together with the halide anion make up the organometallic complex known as the Grignard reagent (R-MgX).

The mechanism described above can be conceptualized as follows:

Recall that arrows with a "fishhook" on the end depict the transfer of a single electron, while the more common "full" arrowhead represents the transfer of a pair of electrons. Also, remember that a radical is where an atom has an unpaired valence electron. In other words, one of the occupied valence orbitals only has one electron in it (called a singly occupied molecular orbital, or "SOMO" for short). This typically makes radical species extremely reactive. But don't beat yourself up if you don't completely understand everything discussed above, especially the SET stuff, as it isn't exactly introductory-level material.

 Quote: Reading up (in this thread) why the C ends up being a strong Lewis Base instead of the Mg or Br ...

As blogfast already pointed out, most of the electron density in the C-Mg bond is shifted towards the carbon. But something else to consider, which hasn't been mentioned yet, is the leaving group you'd get if Br attacked the carbonyl group. Think about it: why is it that when you add concentrated HBr or a bromide salt to a solution of acetone, you don't end up with this (at least not in any significant quantity):

Surely the nucleophilic bromide ions will try to attack the electrophilic carbonyl carbon, right? And if the solution is strongly acidic, the oxygen on acetone will also be protonated, making the carbonyl carbon even more electrophilic than it normally would be. So why don't we get the molecule shown above?

The reason is primarily due to the unusual instability of geminal halohydrins. Because the carbon is bound to two extremely electronegative groups, it is highly electron-deficient and will immediately attract any nearby electrons. Since the oxygen has an extra pair of electrons that it can form bonds with, and the bromide ion makes a rather good leaving group (i.e. it isn't very strongly attracted back to the carbon as it leaves), the oxygen will just donate a pair of electrons to the electron-deficient carbon and kick out bromine, returning the unfavorable geminal halohydrin back to a much more favorable carbonyl configuration.

Also, if the media is strongly basic, like in Grignard reactions, the oxygen will be deprotonated and carry a negative charge, meaning TWO free lone pairs available to donate to carbon. This makes the elimination of bromine in order to return to a neutral and more stable configuration even MORE favorable. With carbon as the attacking nucleophile, however, elimination in this manner cannot happen, as the leaving group (a negatively-charged carbon atom) is way too strongly attracted back to the electrophilic carbon to leave.

 Quote: Inescapable is the Br's superior electronegativity, which is what makes me wonder why It is not the 'active' component instead of the C.

There's also a flip-side to that as well: the more electronegative an atom is, the less it will want to share its electrons should the opportunity arise. That's one of the reasons ions like F and Cl are such weak bases.

 Quote: I get the feeling that it is an accomplice to a far more complex mechanism that a noob like me should not be asking about in public.

Now you've gone and opened Pandora's box. While Grignard reagents are usually taught at the introductory level as simply being this R-MgX complex floating around in some ethereal solvent (usually diethyl ether or THF), the truth is that they actually exist as tetrahedral adducts with the solvent (which is why they dissolve in it). The reason ethers are so commonly used as solvents in Grignard reactions is because not only are they unreactive towards the Grignard reagent, but they also stabilize the R-MgX complex by coordinating with the magnesium atom and giving it a full octet:

Notice that the magnesium is surrounded by electronegative atoms, which provides a sort of stabilizing effect. Similarly, the Grignard reagent also coordinates with the carbonyl compound to form a six-membered cyclic ring, which may also help to explain why bromine isn't the active attacker. It is really through this structure that most Grignard reactions are thought to actually proceed, though the mechanism is often simplified when first introduced at the lower levels. I haven't done a great deal of research regarding the specifics of these Grignard reaction mechanisms, but my guess is that the electrons move like this during the attack on the carbonyl carbon:

Anyway, I hope this has been of at least some help to you. Don't be discouraged if some of this seems a little overwhelming. The level of depth gone into here far exceeds what is typically taught in introductory-level courses, so don't be put off of trying to learn organic chemistry because of it. And don't be afraid to continue to ask questions if you're still not clear on anything.

blogfast25 - 3-5-2016 at 08:42

That was very interesting, Darkstar. Thanks.
aga - 3-5-2016 at 09:20

discouraged ? No way !

This is cool stuff, and the greater the depth of my confusion and ignorance the better !

To be honest, it's all making more sense now than it did in the lesson before calculus class started.

aga - 3-5-2016 at 12:24

@darkstar: "coordinating with the magnesium atom and giving it a full octet"

What does 'co-ordinating' mean ?

if it is not forming bonds then erm, what ?

Darkstar - 3-5-2016 at 12:50

The oxygens on the ether/THF molecules are forming bonds with the magnesium atom. "Coordination" just means that both electrons of the bond come from the same atom. In this case, both electrons of the O-Mg bonds come from oxygen, which are dropped into empty orbitals on magnesium.
aga - 4-5-2016 at 10:30

Ooooh ! the Empty Orbitals thing again.

So the co-ordinating O just jumps in and lobs an electron-filled orbital at Mg, which is so happy to see electrons again that it just 'has it'.

Now there's a question : in this case, would the Mg do any hybridisation of the remaining two electrons in the s orbital ?

aga - 3-9-2016 at 14:36

Seeing as the Dark Lord is back, how about an actual do-able OC synth that demonstrates an OC mechanism/principle ?

Edit:

And stacks of 250ml beakers for some reason.

[Edited on 3-9-2016 by aga]

blogfast25 - 3-9-2016 at 14:52

 Quote: Originally posted by aga Seeing as the Dark Lord is back, how about an actual do-able OC synth that demonstrates an OC mechanism/principle ? Got loads of reagents now

It's been said before but I'll say it again: a decent Fischer esterification is a very doable OC synth with a multi-step reaction mechanism.

Doable but not entirely easy either.

It does require a pure short chain fatty acid.

aga - 3-9-2016 at 15:13

Bring it on ! (had to google what "short chain fatty acid" is)

I got glacial acetic acid and sodium formate.

Butyric acid may be available if i cap, shake and vomit (i tend to vent frequently anyway).

What other reagents needed ?

blogfast25 - 3-9-2016 at 15:55

 Quote: Originally posted by aga What other reagents needed ?

A short alcohol: propanol or butanol, e.g. Ethanol or methanol work too but a bit boring, really.

> 95 w% H2SO4 (catalyst).

Look up a procedure on orgsyn (or similar) and adapt it.

[Edited on 3-9-2016 by blogfast25]

Darkstar - 3-9-2016 at 21:44

As a reminder, I made a post a while back discussing some of the basic theory behind the mechanism of the Fischer esterification. Maybe re-read that post if you do decide to try one. We can always go back over anything that you still aren't clear on.
aga - 4-9-2016 at 11:39

Checking at the back of all cupboards reveals no butanol or propanol.

Ethanol there is aplenty, also Butane.

Would it be too boring with ethanol ?

Edit:

Maybe boring, but at least i know what ethyl acetate smells like, plus it follows the mechanism as Darkstar kindly posted, and provided a link to above.

[Edited on 4-9-2016 by aga]

blogfast25 - 28-10-2016 at 04:22

I'm considering reviving this thread, by revisiting and refreshing some core concepts of QM/QC, while keeping it Lite.

Concepts under consideration are:

* Complex numbers
* Functions as vectors and as matrices
* The TDSE and TISE
* Quantum operators
* Measurement and mean values of observables

Let me know if there's any appetite for it.

Thanks.

aga - 30-10-2016 at 00:06

Sorry i'm late.

Has class started yet ?

blogfast25 - 30-10-2016 at 16:39

Nope but we can start on the morrow, if it pleases you.

Complex numbers (recap)

blogfast25 - 1-11-2016 at 08:09

Complex numbers (aka imaginary numbers) arise because when we ask "What's the square root of a (a Real number)?", then if the answer is b, then:

$$\sqrt{a}=b \iff b^2=a\:\text{if } (a\in\mathrm{R})$$

As the square of a Real number is always positive, we can't take the square root of a negative Real number.

1. The imaginary number i:

We define:

$$i=\sqrt{-1}$$
Then, with a a Real number:
$$\sqrt{-a^2}=\sqrt{a^2\times -1}=a\sqrt{-1}=ai$$
And:

$$(ai)^2=a^2i^2=a^2\times (\sqrt{-1})^2=a^2\times(-1)=-a^2$$

2. Composite complex numbers:

Most (but not all) complex numbers have a Real and a Complex part, where Re is the Real part and Im is the Complex part:

$$c=Re+Im\times i=x+yi$$

For example:
$$-5+3i$$

We say that c is a member ("an element") of the group of Complex numbers:

$$c\in\mathrm{C}$$

3. Representation in the Complex space:

The Complex number as a vector:

Algebraically:

$$c=r\cos\varphi +ri\sin\varphi$$
$$(r\cos\varphi)^2 +(r\sin\varphi)^2=r^2\cos^2\varphi + r^2\sin^2\varphi$$
And because, always:
$$\cos^2\varphi + \sin^2\varphi=1$$
$$\implies (r\cos\varphi)^2 +(r\sin\varphi)^2=r^2(\cos^2\varphi + \sin^2\varphi)=r^2$$
And because:

$$Re=r\cos\varphi\:\text{and }Im=r\sin\varphi$$
$$\implies r^2=Re^2+Im^2=x^2+y^2$$

The value r:

$$r=\sqrt{x^2+y^2}$$

... is often called the modulus of the complex number.

Exercise:

If:

$$r^2=\frac12$$

And:

$$\varphi=\frac{\pi}{4}$$

Then write the complex number as a + bi.

Next up: complex conjugation > basic operations with complex numbers > ...

[Edited on 1-11-2016 by blogfast25]

aga - 1-11-2016 at 10:03

Gulp.

I forsee disturbing dreams where a guy from Trinidad comes up to me and says :

" i is your imaginary friend "

blogfast25 - 1-11-2016 at 13:29

 Quote: Originally posted by aga Gulp.

We've seen all this above, it's just presented more beautifully.

So if earlier you were only physically present or were really daydreaming about girls with really big t*ts, now's your chance!

aga - 1-11-2016 at 13:45

If i've seen all THAT before, call me a cobblerwonky and slap my bodkin with a thrice folded cornichon !

Alternatively:-

Ah yes, of course.

I must take some time to consider such dazzlingly beautiful representations of awesome Mathematical Purity.

(buys enough time, hopefully, to run off and find the Mafs book, then frantically revise)

blogfast25 - 1-11-2016 at 15:29

 Quote: Originally posted by aga to run off and find the Mafs book, then frantically revise)

Get over yourself! You know enough to revise this quite quickly!

Chop, chop...

j_sum1 - 1-11-2016 at 15:57

I had to look up what a cornichon is and I have no idea why it would be thrice-folded.

The mathematics here is rather beautiful and elegant. But the initial concepts of complex numbers involve the intersection of numerous different fields of mathematics as well as some counterintuitive ideas and some rather unfortunate terminology. Compounding this is the fact that computation with complex numbers manually quickly becomes its own special headache. (I distinctly remember asking my maths teacher the value of 2i and it taking him a long time to figure it out.)

The combined effect of this is that it easy to feel overwhelmed. But in reality it is not that daunting.

aga - 1-11-2016 at 16:11

Cornichons are just Gherkins.

Overwhelmed, a bit, over-imbibed, a lot.

Partake !

On the morrow i'll get some new car insurance.

After that, well, who knows ?

blogfast25 - 2-11-2016 at 09:50

 Quote: Originally posted by j_sum1 (I distinctly remember asking my maths teacher the value of 2i and it taking him a long time to figure it out.)

Even more "fun" are things like:

$$\sqrt [4]{-6}$$

But we're not going there. "Lite" here means addition, substraction, multiplication and division only!

[Edited on 2-11-2016 by blogfast25]

aga - 2-11-2016 at 11:52

The car insurance got complicated, a bit like where i'm heading on this first (presumably easy) question.

So far i got some pythagoras going on to give :-

$$y = \sqrt { \frac {1}{2} } \sin{ \frac {\pi}{4} }$$

Is this going in remotely the right direction, or is it an overthunk, drunk and sunk direction ?

[Edited on 2-11-2016 by aga]

blogfast25 - 2-11-2016 at 12:22

 Quote: Originally posted by aga The car insurance got complicated, a bit like where i'm heading on this first (presumably easy) question. So far i got some pythagoras going on to give :- $$y = \sqrt { \frac {1}{2} } \sin{ \frac {\pi}{4} }$$ Is this going in remotely the right direction, or is it an overthunk, drunk and sunk direction ? [Edited on 2-11-2016 by aga]

No, it is in fact correct but it's not the full solution.

aga - 2-11-2016 at 12:43

Phew ! I was once again worried that i'd gone all around the houses for no reason.

So,

$$y = \sqrt { \frac {1}{2} } \sin{ \Big ( \frac {\pi}{4} \Big ) }$$

$$x = \sqrt { \frac {1}{2} } \sin{ \Big ( 90 - \frac {\pi}{4} \Big ) }$$

The y part is the Imaginary bit, so, tentatively ...

$$\sqrt { \frac {1}{2} } \sin{ \Big ( 90 - \frac {\pi}{4} \Big ) } + i \sqrt { \frac {1}{2} } \sin{ \Big ( \frac {\pi}{4} \Big ) }$$

The LaTeX is coming back, slowly ...

blogfast25 - 2-11-2016 at 15:24

 Quote: Originally posted by aga $$\sqrt { \frac {1}{2} } \sin{ \Big ( 90 - \frac {\pi}{4} \Big ) } + i \sqrt { \frac {1}{2} } \sin{ \Big ( \frac {\pi}{4} \Big ) }$$ The LaTeX is coming back, slowly ...

... is basically correct but slightly strangely expressed. But, don't mix (360) degrees wit radians. Instead of:

$$90-\frac{\pi}{4}\:\text{write}\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$$
But that transformation wasn't even necessary:

$$c=r\cos\varphi+ri\sin\varphi=\sqrt{\frac12}\cos\frac{\pi}{4}+i\sqrt{\frac12}\sin\frac{\pi}{4}$$

And with:

$$\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$$
$$\implies c=\frac12+\frac12 i$$

Quick backcheck:
$$r^2=x^2+y^2=\Big(\frac12\Big)^2+\Big(\frac12\Big)^2=\frac14+\frac14=\frac12$$

Thanks for playing!

Next installment tomorrow.

[Edited on 2-11-2016 by blogfast25]

aga - 2-11-2016 at 15:32

Mind: As boggled as a thing that can be boggled, but with extra boggle-ness.

Why was the expression 'strange' and where does 'c' some into it ?

Edit:

I recall that Raidans exist, but am not aware of what they are, apart from being an alternate way of expressing Degrees.

I would Google, but it's bed time, and Google does not always return the Best results.

[Edited on 2-11-2016 by aga]

j_sum1 - 2-11-2016 at 16:32

Quote: Originally posted by blogfast25
 Quote: Originally posted by j_sum1 (I distinctly remember asking my maths teacher the value of 2i and it taking him a long time to figure it out.)

Even more "fun" are things like:

$$\sqrt [4]{-6}$$

But we're not going there. "Lite" here means addition, substraction, multiplication and division only!

[Edited on 2-11-2016 by blogfast25]

Division of complex numbers can be its own special kind of torture too. Fourth root of neg 6 is easy by comparison IMO. And yes, fun.

blogfast25 - 2-11-2016 at 19:05

 Quote: Originally posted by aga Why was the expression 'strange' and where does 'c' some into it ? Edit: The notion of Radians did not enter my head, at all. I recall that Raidans exist, but am not aware of what they are, apart from being an alternate way of expressing Degrees. Gradians as well. Same story.

c is just representation of a complex number.

Radians are NOT an "alternate way of expressing Degrees", rather it's the other way around. "Degrees" are a crappy non-math way of expressing angles: good for compasses and maps, rubbish for math.

In math a full circle is:

$$2\pi$$

So a quarter circle is a right angle or "90 degrees" or:

$$\frac{\pi}{2}$$
and half of that is "45 degrees" or:

$$\frac{\pi}{4}$$

Even "radians" is a misnomer: in the "2 pi system" angles are dimensionless numbers. They don't need a "name".

You invoked the famous trigonometric identity:

$$\sin\theta=\cos\big(\frac{\pi}{2}-\theta\big)$$

That was fine but not really needed.

Manana!

aga - 3-11-2016 at 14:33

I really did not know what they meant.

Hasta mañana, Maestro.

(alt+164 for the n + tilde: changes it from 'n' to 'ny', hence 'manyana')

Maroboduus - 3-11-2016 at 16:59

Gradians are 400ths of a full circle.
They were an option on various calculators in the 1980s.

At the time I heard they were used by some engineers.

Your memory is not faulty here.

blogfast25 - 3-11-2016 at 17:07

 Quote: Originally posted by Maroboduus Gradians are 400ths of a full circle. They were an option on various calculators in the 1980s. At the time I heard they were used by some engineers. Your memory is not faulty here.

G*d yeah, I vaguely remember that now (not that I want to!)

Complex numbers: complex conjugation

blogfast25 - 4-11-2016 at 12:32

With c a complex number, c* is said to be its complex conjugate, if and only if:

$$c=a+bi \implies c^*=a-bi$$
$$c=a-bi \implies c^*=a+bi$$

The product of a complex number with its own complex conjugate is the modulus of the complex number (and thus a Real number):

$$c^*\times c=(a-bi)(a+bi)=a^2+abi-abi+(bi)^2=a^2-b^2i^2=a^2+b^2\in\mathrm{R}$$

If a number is Real, conjugation has no effect:

$$c\in\:\mathrm{R}\implies c=c^*\implies c^*c=c^2$$

Flash back/flash forward to quantum physics:

Higher up in this thread we saw that the probability distribution P(x) in QP is given by, with psi the wave function:
$$P(x)=\psi^*(x)\psi(x) \in\:\mathrm{R}$$

Probabilities are of course always Real numbers:

$$0\leq P \leq 1$$

If psi is a Real function:

$$\psi\in\mathrm{R}\implies \psi^*=\psi\implies P(x)=\psi(x)^2$$

The term:

$$\psi^*(x)\psi(x)$$

... is often referred to as the Norm.

Wave functions for which:

$$\int_{all\:\mathrm{ space}}\psi^*\psi dV=1$$

... are said to be Normalised.

Next up: complex numbers basic operations > vectors >...

[Edited on 5-11-2016 by blogfast25]

aga - 4-11-2016 at 14:21

To be sure, what does the sigma-like symbol mean ?
$$c\in\:\mathrm{R}$$
I guessed at 'in the realm of' but then it all went odd.

Oscilllator - 4-11-2016 at 15:25

 Quote: Originally posted by aga To be sure, what does the sigma-like symbol mean ? $$c\in\:\mathrm{R}$$ I guessed at 'in the realm of' but then it all went odd.

It technically means "in the set of" and R means "Real numbers" . Real numbers are just normal numbers that can be positive, negative and have as many decimal points as you like e.g. -1, 1.3, 3.14159...

So c∈R means c is a sensible, non-imaginary number. There are other letters you can use like N for "natural number", which is all the whole positive numbers and Z for integer, which can have negative numbers too.

aga - 4-11-2016 at 15:39

If even Numbers have limits, they cannot be a linear sequence, which, in turn, makes even Numbers meaningless

The status of Zero has always bothered me.

Ponderment required.

blogfast25 - 4-11-2016 at 16:44

 Quote: Originally posted by aga If even Numbers have limits, they cannot be a linear sequence, which, in turn, makes even Numbers meaningless The status of Zero has always bothered me. Ponderment required.

Zero took a long time to get accepted: the other numbers didn't want to know it! But accepting it was a true revolution!

The first part of your ponderment is cryptic. Care to explain?

aga - 6-11-2016 at 14:51

 Quote: Originally posted by blogfast25 The first part of your ponderment is cryptic. Care to explain?

Sorry, i missed the question earlier.

It was some random drunken notion more than likely.

Something about Numbers spanning an infinity, yet allowing an infinity between each number.

0 to 1 (or -1) being an infinity, equal to, er, infinity, making 0 the only True number.

Every other number is infinitely distant from 0, and infinitely distant from the next number, making them all basically the same number.

j_sum1 - 6-11-2016 at 16:13

Quote: Originally posted by aga
 Quote: Originally posted by blogfast25 The first part of your ponderment is cryptic. Care to explain?

Sorry, i missed the question earlier.

It was some random drunken notion more than likely.

Something about Numbers spanning an infinity, yet allowing an infinity between each number.

0 to 1 (or -1) being an infinity, equal to, er, infinity, making 0 the only True number.

Every other number is infinitely distant from 0, and infinitely distant from the next number, making them all basically the same number.

Don't contemplate infinity for too long. Especially when drunk. You know that Cantor went mad don't you? He was trying to resolve the very questions you have asked and eventually went nuts.

Not only are there an infinite number of numbers between any two given numbers (uncountably infinite actually), there are an infinite number of infinities; each one infinitely bigger than the one that preceded it.

Complex numbers basic operations:

blogfast25 - 7-11-2016 at 05:58

$$(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i$$

2. Multiplication:

$$(a+bi)\times(c+di)=ac+adi+bci+bdi^2=ac+(ad+bc)i-bd=(ac-bd)+(ad+bc)i$$

3. Division:

Division is carried out by multiplying denominator and numerator by:

$$\frac{c^*}{c^*}=1$$

$$\implies\frac{a+bi}{c+di}=\frac{a+bi}{c+di}\times\frac{c-di}{c-di}\tag{1}$$
$$=\frac{(a+bi)(c-di)}{c^2+d^2}=\frac{ac-adi+bci-bdi^2}{c^2+d^2}$$
$$=\frac{(ac+bd)+(-ad+bc)i}{c^2+d^2}=\frac{ac+bd}{c^2+d^2}+\frac{-ad+bc}{c^2+d^2}i$$

4. Exponential complex numbers:

Euler's formula:

$$e^{ai}=\cos a+i\sin a$$

Complex conjugation:

$$(e^{ai})^*=e^{-ai}$$

So:

$$e^{-ai}=\cos a-i\sin a$$

Exercise:

Using the 'trick' of (1), determine:

a)

$$\frac1c=\frac{1}{2-3i}$$

b)

$$\frac1i$$

Next up: vectors > functions as vectors > ...

[Edited on 7-11-2016 by blogfast25]

aga - 16-11-2016 at 11:43

$$\frac1c=\frac{1}{2-3i}$$
So
$$c = 2-3i$$
and
$$(2-3i) - (c - 0i) = 0$$
Let a = 2, b=3, c=c, d=0, then with rule (1)
$$(2-c) + (3+0)i = 0 = 2-c+3i$$
Which basically winds up back at the start !?!?

Confused.

aga - 17-11-2016 at 12:29

To be 100% honest, i do not understand the first queston, nor how (1) applies.
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