shaneo
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Posts posted by shaneo
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Kinetic Isotopic effects (KIE)
Thanks,
Shaneo0 -
ah that last part makes much more sense now! no i didn't realise that i hadn't rearranged eq 1 correctly. My lecturer put the equations down without much explanation. I think i have it up the the point of the last equation in my second image on the description now.
I will post my workings out either tonight or tomorrow.
I think they are correct!
Thanks again,
Shaneo0 -
thanks for your response.. however what i require is an in detail step by step approach to the derivation :/ the mathematics doesn't quite add up in my mind. However i do now know that we take the exponential of the above metioned equation in order to reverse the natural logs in a revised form of the eyring equation.
The equation that i am mentioning is on the bottom of the description0 -
Below is my shot at explaining how to get to the second to last step of the eyring equation from Gibbs laws. However I don't understand two mathematical parts as highlighted and described in the image below.
Could someone please explain how they come about?
Thanks
note:
eq 4 should equal eq 2 (just another way of writing it) (not sure hence the highlighted T in eq 3)
to get to eq 5. i have cancelled the "T" in the equation
to get to eq 6. reversed the minus
to get to eq. 3 is the part that i don't understand.
update: Please see comments below
(not sure how to get from bttom eq in picture to the Eyring equation0 -
Oh it makes sense now. but, you missed out a decimal point "= 10^-10 m^2" on the second line should be "= 1.0^-10"
Thanks for your input!
Shaneo0 -
PSA powerpoint slide.
If the grounded cube has a side length of 10 micrometres then it would have a surface area of 60 micrometres^2. Which is equivalent to 60 x10^-12 metres^2.
To calculate the amount of particles I did 60 x10^-12/ 6x10^-4 (the original size of the MgO cube.) which came out to be 10^6 particles.
For the overall surface area I did the S.A of one particle x by the overall amount of particles so 60x10^-12 x 10^6 = 0.00006 metres^2.
This is the first time I have done this calculation so I don't feel confident in challenging my lecturers note calculations. But, they are different and I cannot see why?
Any suggestions?
ps. if you can't read the photo. it was said that 10^9 crystals or particles would be present in the grounded MgO. 1 particle having a surface area of 6x10^-10 m^20 -
PhLi i gather is not a grignard reagent. otherwise it would be PhMgLi.
I'm wondering if there is a triple bond from Ph to Li because then i will be able to use the Organo metallics with ketone reaction.
Even still, i'm not sure how the valency of a phenyl ring would be able to accommodate for the aforementioned bond?
The starting material is: cyclohexanone with a c=c bond on the 2nd carbon. Which leads me onto a side question, alkene's don't have prefix names and so what is the name of this molecule?Baring in mind ketone groups have a higher priority in IUPAC naming terms than alkenes.
Sorry about my amateur layout.
Shanebtw the C=C bond is not an R group off of one of the Cyclohexanone carbons. It is part of the ring system
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Thanks for the help.
I found out my mistake and I now find it easier to treat the equation in a linear format. i.e. v= 1 ÷ 2 pi . Instead of saying v = 1/2pi.0 -
I don't think I understand your question but this is the starting equation...
v = (1/2π) x √kf / m
does that make more sense?
in brackets is 1 divided by 2 pithe square root encompasses the kf and the m
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I'm trying to use the vibrational frequency equation to calculate the frequency constant, but i can't seem to rearrange it correctly!
Here is the method i used:
v = 1/2π x √kf / m
therefore: v/1/2π = √kf / m
(v/1/2π)2 = kf / m
(v/1/2π)2 x m = kf
Can someone please pick a hole in my rearranging, because i know the actual rearranged equation and this is not it!0
Sterically hindered carbonyl on Enone (Enone + Grignard reaction)
in Organic Chemistry
Posted
Hi all,
I am of the understanding that when a carbonyl group is in a sterically hindered environment on an enone the grignard reagent instead chooses to opt to react with the alkene group on the enone instead. (More specifically cleavage of a hydrogen atom). I have to write out a mechanism for this but i can't get to grips with it.
It would be much appreciated if someone could outline a general mechanism of a reaction of this type with the sterically hindered carbonyl group! or alternatively point me to a webpage or ebook that can! as i have scoured the web to no avail.
Thanks,
shaneo!