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aga
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| Quote: | | Step 1 - The nucleophilic oxygen of the hydroxyl group donates a lone pair to the electrophilic hydrogen on hydronium. |
Am i being Thick or does 'Lone Pair' mean TWO electrons ?
Is ONE electron being donated or TWO ?
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blogfast25
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Quote: Originally posted by aga  | | Quote: | | Step 1 - The nucleophilic oxygen of the hydroxyl group donates a lone pair to the electrophilic hydrogen on hydronium. |
Am i being Thick or does 'Lone Pair' mean TWO electrons ?
Is ONE electron being donated or TWO ? |
I've drawn the two electron pairs (MOs) on the hydroxyl group:
It's the one marked green '1' that now attacks the proton on the hydronium (oxonium) and forms a bond with it. So it's not a question of 'donating ONE
or TWO electrons': the lone, non-bonding MO 1 now becomes a bonding MO, bonding that second proton to the O atom.
Personally I would describe that step more easily as the protonation of that hydroxyl group by a
H<sub>3</sub>O<sup>+</sup> cation (hydronium = oxonium): it's really a classic acid/base reaction:
R-OH(aq) + H<sub>3</sub>O<sup>+</sup>(aq) < === > R-OH<sub>2</sub><sup>+</sup>(aq) +
H<sub>2</sub>O(l)
But in terms of movement of MOs, what I said above is what happens.
I prepared 2-chloropropane by that method but using ZnCl<sub>2</sub>(anh.) as a catalyst, not so long ago:
http://www.sciencemadness.org/talk/viewthread.php?tid=30180#...
[Edited on 10-11-2015 by blogfast25]
[Edited on 10-11-2015 by blogfast25]
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aga
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Sorry, still not Clear.
It's the term 'Lone Pair' that i'm struggling with.
Does that mean 'An orbital with between 0 and 2 electrons in it'
or does it mean 'An orbital with two electrons in it'
or does that mean 'An orbital with 0-2 electrons that is currently not involved in bonding' ?
The confusion arises from previous descriptions of electrons being swapped, and now we're into bonding orbitals where the electrons in it (or not) are
somewhat abstracted.
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blogfast25
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| Quote: Originally posted by aga | Sorry, still not Clear.
It's the term 'Lone Pair' that i'm struggling with.
Does that mean 'An orbital with between 0 and 2 electrons in it'
or does it mean 'An orbital with two electrons in it'
or does that mean 'An orbital with 0-2 electrons that is currently not involved in bonding' ?
The confusion arises from previous descriptions of electrons being swapped, and now we're into bonding orbitals where the electrons in it (or not) are
somewhat abstracted. |
In the context of Step 1:
An orbital with two electrons in it, that is currently not involved in bonding
But also in the context of Step 3:
The iodide ion has 4 lone electron pairs and uses one of them to bond with the carbonium ion with: one of the carbonium MOs is empty. The iodide acts
as the Lewis base, the carbonium ion as the Lewis acid.
[Edited on 10-11-2015 by blogfast25]
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aga
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Thanks for that.
So 'Lone Pair' is a misnomer and should be interpreted as 'electron orbital' irrespective of it's apparent electron configuration, due to
hybridisation ?
Sorry to labour this point, however i feel it is rather important to get the idea, seeing as it appears to be very important.
Edit:
If you would not mind, please explain it in terms of your recent illustrated example (the one with the OH, red arrow, H3O) etc)
[Edited on 10-11-2015 by aga]
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blogfast25
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| Quote: Originally posted by aga | Thanks for that.
1) So 'Lone Pair' is a misnomer and should be interpreted as 'electron orbital' irrespective of it's apparent electron configuration, due to
hybridisation ?
2) If you would not mind, please explain it in terms of your recent illustrated example (the one with the OH, red arrow, H3O) etc)
|
1) Yes! 'non-bonding molecular orbital' is probably the most accurate term, actually.
2) I prefer the other example:
The structure of the iodide anion is an sp<sup>3</sup> hybridised tetrahedron, all four MOs are full (2 electrons, Pauli conformant).
The electronic structure of the central, charge carrying C atom in the carbonium cation is:
* sp<sup>3</sup> hybridisation
* two of the sp<sup>3</sup> bond to two carbons
* one of the sp<sup>3</sup> bonds to one hydrogen
* one of the sp<sup>3</sup> is EMPTY!
Iodine then 'donates' one of its full sp<sup>3</sup> orbitals to the empty C sp<sup>3</sup> MO and a C-I covalent bond is
formed. Et voila, everybody's happy.
aga too?
[Edited on 11-11-2015 by blogfast25]
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blogfast25
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And here's a question for Darkstar and/or Annaandherdad.
I found it on a physics forum I'm a member of and got stuck trying to solve it.
Here's the question:
Another member part-answered the question as follows:
Because I'm not firm on my legs with Dirac notation, I wanted to calculate the energy expectation value using:
I first calculated the normalisation constant (the wave function is zero everywhere but in -a/2 < 0 < +a/2) and found it to be
10/√(61a<sup>5</sup> .
But then it dawned on me we don't know V(x), so we don't know the Hamiltonian operator?
I'm stuck.
[Edited on 11-11-2015 by blogfast25]
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aga
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Yes ! Thanks for the detailed clarification.
Notes to self:-
1. Do not just count electrons.
2. Imagine the electron density 'clouds' to get a feel for where the charges are.
3. Lone Pair = unbonded orbital.
4. Use Electronegativity tables !
http://www.sciencemadness.org/talk/files.php?pid=421505&...
5. Electrophile wants more electrons
Nucleophile has too many.
N.B. RELATIVELY too many/few
6. Resonance is the dislocation of the bonding orbitals giving
more than 1 possible stable bonding configuration to a molecule
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blogfast25
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In a nutshell, aga, in a nutshell...
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annaandherdad
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Blogfest, I got your message. The problem is not very well posed. The problem is probably trying to say that the potential is 0 for -a/2 < x
< +a/2, and infinity outside that range. The energy eigenfunctions are
u_n(x) = sqrt(2/a) sin( 2n pi x/a)
and the energy eigenvalues are
E_n = (2/m) (n pi hbar /a)^2
However, the first question, what is the energy value you get on the first measurement, cannot be answered. The wave function is a linear
combination of all of the energy eigenstates, so there are probabilities for measuring the different energy eigenvalues. You can't say which one will
be measured, only the probabilities.
For the second question, the number of times you get the ground state is approximately the probability of being in the ground state times 1000.
The probability of being in the ground state is an integral that is a bit of a pain to do.
Any other SF Bay chemists?
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Darkstar
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That's what I thought, too.
| Quote: Originally posted by aga | OK. Exhausted my Butane lighter testing that out.
Can i bill the B&D Education Store for the replacement butane ? |
We give you a full scholarship to B&D University and you have the audacity to ask US to pay YOU for something that trivial? Well, I never!
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blogfast25
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| Quote: Originally posted by annaandherdad | Blogfest, I got your message. The problem is not very well posed. The problem is probably trying to say that the potential is 0 for -a/2 < x
< +a/2, and infinity outside that range. The energy eigenfunctions are
u_n(x) = sqrt(2/a) sin( 2n pi x/a)
and the energy eigenvalues are
E_n = (2/m) (n pi hbar /a)^2
However, the first question, what is the energy value you get on the first measurement, cannot be answered. The wave function is a linear
combination of all of the energy eigenstates, so there are probabilities for measuring the different energy eigenvalues. You can't say which one will
be measured, only the probabilities.
For the second question, the number of times you get the ground state is approximately the probability of being in the ground state times 1000.
The probability of being in the ground state is an integral that is a bit of a pain to do. |
So there's no way to determine the amplitudes in the superposition? Fourier transform, anyone?
Can the superposition be be normalised? After all, the particle is bound...
Thanks!
Moderators: when will this site adopt MathJax (Latex rendering of math formulas) like below:
It's truly bizarre that on a forum about science in the 21<sup>st</sup> Century we still have to render the language of science with sub
and sup tags...
[Edited on 12-11-2015 by blogfast25]
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annaandherdad
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| Quote: Originally posted by blogfast25 |
So there's no way to determine the amplitudes in the superposition? Fourier transform, anyone?
Can the superposition be be normalised? After all, the particle is bound...
Thanks!
|
Yes, the amplitudes of the superposition can be determiined, and their squares are the probabilities, but that doesn't tell you what energy value you
will get on a measurement, only the probabilities of the different outcomes. You would only be able to predict exactly what energy would be
measured if the system were in an energy eigenstate, which this one is not.
And yes, the superposition can be normalized.
I recall someone making the comment that the 1000 measurements had to be on independent systems, not repeated measurements on a single system, but I
can't find the comment now. Anyway, that is correct, if you measure energy once, you leave the system in an energy eigenstate corresponding to the
value of energy measured. Successive measurements of energy on the same system will yield the same value.
The probabilities given by quantum theory refer to measurements made on independent, identically prepared systems.
The wave function given (an inverted parabola) is qualitatively very similar to the ground state wave function (a sine function), so most of the
probability will be in the ground state.
I think this problem requires too much algebra for the educational value, even it were properly stated.
Any other SF Bay chemists?
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blogfast25
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@AAHD:
Agreed with all your points. Thanks.
This is the thread:
http://physics.stackexchange.com/questions/217939/infinite-p...
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aga
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| Quote: Originally posted by Darkstar | | For extra credit, explain why the two molecules in the top box are the exact same molecule, while the two molecules in the bottom box are not. What
have I changed by adding a phenyl ring? |
The molecules in the top box are the same thing rotated 180 degrees around the Y axis.
The ones in the bottom box are mirror images of each other.
I was about to triumphantly blurt 'steroisomer !' then had the sense to look that up.
The mirror-image thing makes them Enantiomers.
Edit:
If it's right, take the extra credits in liu of the butane
[Edited on 12-11-2015 by aga]
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Darkstar
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| Quote: Originally posted by aga | The molecules in the top box are the same thing rotated 180 degrees around the Y axis.
The ones in the bottom box are mirror images of each other.
I was about to triumphantly blurt 'steroisomer !' then had the sense to look that up.
The mirror-image thing makes them Enantiomers. |
Correct! They are mirror images of one another, called enantiomers. This is actually a very important concept, particularly in fields like medicine.
Changing the spatial orientation of just a single atom or functional group in a biologically-active compound can sometimes drastically alter its
effects. Welcome to the world of chirality.
| Quote: | | If it's right, take the extra credits in liu of the butane
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Well, I didn't want to bring this up, but I'm afraid you have forced my hand. There's a small matter of a room in one of the freshman dormitories that
you completely trashed last weekend while drinking and doing unauthorized experiments with the undergrads. I've already received over two dozen
complaints regarding broken glassware in the sink, chemical spills/splatters/spatters/puddles/sprays all over the walls, floors and counter tops,
stains, holes in the carpet from what looks like a large acid spill, complaints of foul and obnoxious odors, hazardous waste that was improperly
disposed of (halogenated compounds, toxic heavy metals etc), empty beer cans all over the place...
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blogfast25
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I heard it was experiments ON the undergrads. 
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aga
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| Quote: Originally posted by Darkstar |
Well, I didn't want to bring this up, but I'm afraid you have forced my hand. There's a small matter of a room in one of the freshman dormitories that
you completely trashed last weekend while drinking and doing unauthorized experiments with the undergrads. I've already received over two dozen
complaints regarding broken glassware in the sink, chemical spills/splatters/spatters/puddles/sprays all over the walls, floors and counter tops,
stains, holes in the carpet from what looks like a large acid spill, complaints of foul and obnoxious odors, hazardous waste that was improperly
disposed of (halogenated compounds, toxic heavy metals etc), empty beer cans all over the place...
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As per the instructions of my Binary Logic Lecturer i will answer the allegations thusly :-
1110x1111111110x11 defintely 1
I take issue with the 'obnoxious odours' allegation.
They were merely natural odours that are much better Out than In.
The positioning of the Nose at such time as they become Out is a matter for the Smeller, not the Producer.
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aga
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Almost forgot the squigly diagram for QM question 8
Edit :
Seeing as we can't send images via U2U it might be easier to do the DarkStar thing and post the Answers in the thread, where images can be included.
OK with you bloggers ?
[Edited on 12-11-2015 by aga]
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blogfast25
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Fur aga:
The normalisation procedure for the problem higher up:
Note that A has two roots, one positive, one negative, which represent the wave function's parity (remember those fellows gerate and
ungerate?)
Now the wave function has been normalised we are certain that the probability density distribution P(x) is entirely correct. The wave function is now
ready to extract information from.
| Quote: Originally posted by aga |
Seeing as we can't send images via U2U it might be easier to do the DarkStar thing and post the Answers in the thread, where images can be included.
OK with you bloggers ?
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As discussed, yes.
[Edited on 13-11-2015 by blogfast25]
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aga
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Ah feck.
i just saw that i got the H in the OH- attacking the 3-Carbon.
Should be the O ripping in there.
Oh the woes of MS Paint when applied to Organic Chemistry.
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aga
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Oh dear.
Looks like i scared the Professors away.
OK. I'll fix the dorm and sort out the polymerised undergrads.
It's just the chair legs and gravity holding them together.
Someone said 'stick and ball model' and i just got carried away.
They're only holding hands in mutual support - not to keep the structure intact or signify pi-bonds anymore.
Should all heal nicely for the Christmas break once the chairs are, er, extracted and returned to the canteen.
[Edited on 13-11-2015 by aga]
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blogfast25
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Hydrolysis of 1-chloropropane by sodium hydroxide:
NaOH fully dissociates in water into sodium and hydroxide ions.
The nucleophilic (Lewis base) hydroxide anion attacks the polarised carbon atom. A new bond is formed, between the OH and the C atom.
The C-Cl MO folds back onto the Cl atom, forming a chloride ion as leaving group.
[Edited on 14-11-2015 by blogfast25]
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Darkstar
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As far as nomenclature is concerned, the carbon that chlorine is bonded to is technically the 1-carbon. There are way too many nomenclature rules to
attempt to explain them all in a brief post, but basically you number carbon chains based on what's attached to them, not left-to-right like you read.
In 1-chloropropane (notice that it's not called 3-chloropropane!), you start on the carbon with the chloro group and count from there. On the other
hand, if the chloro group were attached to the middle carbon, then you could start counting on either end since it wouldn't make a difference (either
way, the carbon that the chloro group ends up on is the 2-carbon); however, you would NOT start counting on the second carbon because you always want
the longest carbon chain as the parent chain. Carbon chains are named using a prefixparentsuffix structure (i.e. 1-chloropropane or 1-chloropropan-2-ol), where the parent chain is considered
the "main" chain and everything else just "additions" to it.
In larger, more complex molecules with multiple functional groups, the order of precedence determines what gets prefixed and what gets suffixed. In
other words, certain groups take priority over others. The most important group on a molecule gets suffixed at the end by itself, and all other groups
get prefixed in alphabetical order in front of the parent chain. An exception to this are groups like chloro, bromo, iodo, fluoro and alkyl groups
(methyl, ethyl, propyl, butyl etc), which never get suffixed regardless. In some cases, this priority will also determine which carbon you start
counting on as well. Also, the prefix name for a functional group is usually different than the suffix name for it (i.e. when prefixed, hydroxyl
groups are "hydroxy," but when suffixed, they are "ol").
But don't worry about trying to make sense of all of this right now. I will do a more in-depth lesson on IUPAC nomenclature in the future. Consider
this extracurricular reading for the time being. Below are some examples I made for your viewing pleasure. Oh, and by the way, that last one is a
sample of the kind of stuff I will be asking you to name on the homework questions!
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aga
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Is that Fly Spray or Slug-knackerer ?
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