Awwh I wish you went deeper into fragmentation because I guess there is a lot of possibilities for different functional groups, and that's what I seem to have the most trouble with. But thanks anyways for these videos!!
Fragmentation rate is based on the strength of the bond correct? The stability of the ion only determines if it lasts long enough for you to detect it as such? Therefore the rate of fragmentation would also depend on the strength of the bond only while your detection rate would depend on the stability and rate?
@93Meercat actually i had the same question,but he answered it within the vid , that as (carbon-13) has a 1.1% , which is a very low so we don't have to worry about it , and that it has to be only 1 type available,its not like a probability that there would be carbon-12 or carbon-13 in the same compound of (CH2CL2/CH2BR2) as he mentioned in his vid of MS part 3 =)
@93Meercat@93Meercat actually i had the same question,but he answered it within the vid , that as (carbon-13) has a 1.1% , which is a very low so we don't have to worry about it , and that it has to be only 1 type available,its not like a probability that there would be carbon-12 or carbon-13 in the same compound of (CH2CL2/CH2BR2) as he mentioned in his vid of MS part 3 =)
@93Meercat In the real world you would but you dont have to make guesses as to what the rate is. The rate is determined experimentally and recorded. It's called calibration, for each component the fragmentation pattern is determined and logged, then plugged into fourier transforms to determine composition and concentration because you get too much overlap doing a complex analysis, that is, a sample with a broad range of components.
wow..I have an exam on spectroscopy tomorrow and I wasnt feeling good abt it..after viewing all 5 of your videos on mass spec..I cannot wait to sit my exam. You really are a good teacher..ill go on ur website now. Thanks alot!!
This is an incredibly well explained video, i'm doing AS chemistry, and this helped me immensly, so thanks man :) I'd love if you did an AS and A2 section I'm real stuck on some stuff, and your teaching helps me understand things so much better, I guess, just thanks for putting these up :)
Awwh I wish you went deeper into fragmentation because I guess there is a lot of possibilities for different functional groups, and that's what I seem to have the most trouble with. But thanks anyways for these videos!!
babygurll7 1 month ago
i really enjoy watching your videos while learning and thanks for making our life much easier :)
am414able 4 months ago
Grateful! It is fantastic!
chillwinternight 4 months ago
You make Organic chemistry so easy~ Thank you Thank you Thank you!!
wishpuppy 4 months ago
who the fuck disliked this? i want to kill those shits
boringgrass 4 months ago
Much appreciated, freelanceteach! This was very helpful. :)
Liser728 5 months ago
amazing :)
lol now i spend more time ENJOYING what im learning for a change than complaining that i have to study!
Ms12213 6 months ago 2
Comment removed
dmravi13 7 months ago
Thank you!!
dmravi13 7 months ago
Fragmentation rate is based on the strength of the bond correct? The stability of the ion only determines if it lasts long enough for you to detect it as such? Therefore the rate of fragmentation would also depend on the strength of the bond only while your detection rate would depend on the stability and rate?
msingletary1984 7 months ago
Thank you, This was very helpful :D
However, I do have one question: Why don't we take into account the isotope of carbon (Carbon-13) when working out the peaks of CH2Cl2 / CH2Br2?
Thank you once again! This has really helped my understanding!!
93Meercat 10 months ago
@93Meercat actually i had the same question,but he answered it within the vid , that as (carbon-13) has a 1.1% , which is a very low so we don't have to worry about it , and that it has to be only 1 type available,its not like a probability that there would be carbon-12 or carbon-13 in the same compound of (CH2CL2/CH2BR2) as he mentioned in his vid of MS part 3 =)
princessmaram 10 months ago
@93Meercat @93Meercat actually i had the same question,but he answered it within the vid , that as (carbon-13) has a 1.1% , which is a very low so we don't have to worry about it , and that it has to be only 1 type available,its not like a probability that there would be carbon-12 or carbon-13 in the same compound of (CH2CL2/CH2BR2) as he mentioned in his vid of MS part 3 =)
princessmaram 10 months ago
@93Meercat In the real world you would but you dont have to make guesses as to what the rate is. The rate is determined experimentally and recorded. It's called calibration, for each component the fragmentation pattern is determined and logged, then plugged into fourier transforms to determine composition and concentration because you get too much overlap doing a complex analysis, that is, a sample with a broad range of components.
msingletary1984 7 months ago
an actual life saver. Come and teach in England!
kano20100 1 year ago
5:05 is the ultimate satisfaction of a knowledgeable philosopher... exactly what @sodr2 said. Thank you sooooo much for these videos!!!!
terminaterjohn 1 year ago
5:05 = the pose of a great thinker... thanks a lot for these videos
sodr2 1 year ago
wow..I have an exam on spectroscopy tomorrow and I wasnt feeling good abt it..after viewing all 5 of your videos on mass spec..I cannot wait to sit my exam. You really are a good teacher..ill go on ur website now. Thanks alot!!
ennybubbles 1 year ago
Thank you so much. :) The videos have helped loads! Please do keep it up!
miatsanglp 1 year ago
explained very well. Great vid for review.
chobostrata 1 year ago
extremely helpful video!!!!! thank you so much i can finally do my hw =D
kiliwia13 2 years ago
thank u very much .. may God bless u..
mad1207 2 years ago 11
thank you very much for the information about mass spectra interpretation you ROCK
crambus 2 years ago 3
these videos are amazing!
pagan8tor 2 years ago 3
This is an incredibly well explained video, i'm doing AS chemistry, and this helped me immensly, so thanks man :) I'd love if you did an AS and A2 section I'm real stuck on some stuff, and your teaching helps me understand things so much better, I guess, just thanks for putting these up :)
xx
azerion66 2 years ago 5
THANK YOU!!!!!!!!!! u rock dude XD
blackhappiness2 2 years ago 16