Jump to content

shaneo

Members

11

0 Neutral

• Rank
Quark

Profile Information

• Favorite Area of Science
Chemistry
1. Sterically hindered carbonyl on Enone (Enone + Grignard reaction)

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!
2. Why do SN2 reactions not show secondary KIE?

Kinetic Isotopic effects (KIE) Thanks, Shaneo
3. Eyring Equation Derivation

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, Shaneo
4. Eyring Equation Derivation

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 description
5. Eyring Equation Derivation

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 equation
6. Calculation of Surface Area

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! Shaneo
7. Calculation of Surface Area

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^2
8. 1.) PhLi 2.) H3O+ reagents to produce... ?

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. Shane btw the C=C bond is not an R group off of one of the Cyclohexanone carbons. It is part of the ring system
9. Rearranging equations with square roots...

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.
10. Rearranging equations with square roots...

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 pi the square root encompasses the kf and the m
11. Rearranging equations with square roots...

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!
×
×
• Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.