stonecold's chem questions :)

[fady_22]

But it doesn't say that it is pulsed for the simple spectrophotometer.

[vexx]

But it doesn't say that it is pulsed for the simple spectrophotometer.

I don't think it is, but regardless, i think you are right.

I only recalled what i said from memory where in AAS it specifically says the light is chopped whereas in UV-Vis it just has a light source shining a beam of light whereby my assumption that the light source was fairly simple was probably wrong.

[stonecold]

we'll find out soon enough.

and get rid of that 12.

[stonecold]

Okay, so I have a few questions.  Been doing the Neap Exam guide, and there are a few things which I need clarified.

First one is, how does a wavenumber of 10⁵m^-1 = 1000 cm^-1?

Doesn't cm mean x 10^-2

And the other is damn NMR.  I have attached the spectrum below.
I get how the peak heights work , no worries.  It basically tells you how many hydrogens there are in each category of environments.
Also number of peaks = number of environments.

What confuses the shit out of me is this:
Peak Z makes sense, as it represents the CH₃ group, which is adjacent to an environment with 2 hydrogens, so it has 3 fine peaks.
Peak Y makes sense, as it represents the CH₂ group, which is adjacent to environments with a total of 4 hydrogens, so it has 5 fine peaks.
Peak X WTF!  It represents the COH group, which is adjacent to an environment with 2 hydrogens, but it only has 1 fine peak...

I'm so confused!  Help much appreciated.  

[fady_22]

For the wavenumber question:
First change the wavenumber into wavelength, i.e. take the reciprocal. You will get . Now multiply this by to get a wavelength of . Now take the reciprocal to get 1000 cm-1.

[vexx]

Stonecold- i just did that question today strangely:)

Peak Z makes sense, as it represents the CH3 group, which is adjacent to an environment with 2 hydrogens, so it has 3 fine peaks.
Peak Y makes sense, as it represents the CH2 group, which is adjacent to environments with a total of 4 hydrogens, so it has 5 fine peaks.
Peak X WTF!  It represents the COH group, which is adjacent to an environment with 2 hydrogens, but it only has 1 fine peak...

Peak X at ~ 9-10ppm represents a Hydrogen environment that is connected to a carbon that double bonds to an oxygen. It is purely talking about the hydrogen environments as it is a Proton NMR, not the group as a whole. The structure of propanal you can see has a hydrogen attached to a carbon which is connected also to a double bond oxygen and another carbon, which means NO other hydrogens. so the hydrogen in this environment, using the n+1 rule has no adjacent hydrogens and hence is a single peak.

so basically the proton NMR has proton environments, and not functional group environments - look at the hydrogen atom alone. that answer your question?

[fady_22]

Stonecold- i just did that question today strangely:)

Peak Z makes sense, as it represents the CH3 group, which is adjacent to an environment with 2 hydrogens, so it has 3 fine peaks.
Peak Y makes sense, as it represents the CH2 group, which is adjacent to environments with a total of 4 hydrogens, so it has 5 fine peaks.
Peak X WTF!  It represents the COH group, which is adjacent to an environment with 2 hydrogens, but it only has 1 fine peak...

Peak X at ~ 9-10ppm represents a Hydrogen environment that is connected to a carbon that double bonds to an oxygen. It is purely talking about the hydrogen environments as it is a Proton NMR, not the group as a whole. The structure of propanal you can see has a hydrogen attached to a carbon which is connected also to a double bond oxygen and another carbon, which means NO other hydrogens. so the hydrogen in this environment, using the n+1 rule has no adjacent hydrogens and hence is a single peak.

so basically the proton NMR has proton environments, and not functional group environments - look at the hydrogen atom alone. that answer your question?

But peak splitting occurs when there are hydrogens bonded to the adjacent atom to the atom the hydrogen is bonded to. So technically, that peak should be split into three, as there are two hydrogens bonded.

I think its got to do with the fact that its a aldehyde... it think that hydrogens in this functional group do not show peak splitting (like hydrogens in hydroxyl groups don't show splitting). 

[stonecold]

thanks all.  i understand the wavenumber stuff now fady.

there is some reason why certain hydrogens don't show peak splitting.  i wish i could find the basic rules somewhere.  textbook is far too vague.  I want to know which groups won't show peak splitting.

maybe some university chem pplz will know haha...

other than that, gotta say, l love chem.  it is soo easy and interesting.  bio is also quite awesome.
everything else go and die in a hole!  especially umat and regular english!

[vexx]

Stonecold- i just did that question today strangely:)

Peak Z makes sense, as it represents the CH3 group, which is adjacent to an environment with 2 hydrogens, so it has 3 fine peaks.
Peak Y makes sense, as it represents the CH2 group, which is adjacent to environments with a total of 4 hydrogens, so it has 5 fine peaks.
Peak X WTF!  It represents the COH group, which is adjacent to an environment with 2 hydrogens, but it only has 1 fine peak...

Peak X at ~ 9-10ppm represents a Hydrogen environment that is connected to a carbon that double bonds to an oxygen. It is purely talking about the hydrogen environments as it is a Proton NMR, not the group as a whole. The structure of propanal you can see has a hydrogen attached to a carbon which is connected also to a double bond oxygen and another carbon, which means NO other hydrogens. so the hydrogen in this environment, using the n+1 rule has no adjacent hydrogens and hence is a single peak.

so basically the proton NMR has proton environments, and not functional group environments - look at the hydrogen atom alone. that answer your question?

But peak splitting occurs when there are hydrogens bonded to the adjacent atom to the atom the hydrogen is bonded to. So technically, that peak should be split into three, as there are two hydrogens bonded.

I think its got to do with the fact that its a aldehyde... it think that hydrogens in this functional group do not show peak splitting (like hydrogens in hydroxyl groups don't show splitting).  

hahaha sorry, my mind is actually not working at all lately, and the make matters worse i would have realised what i had said after i wrote that if i had proof read or looked at the question, (as i replied with the molecule and what i wrote from memory). arghh i got everything on this right here so i really only said that peak X was CHO because Y and Z were the other groups (CH2/3) quite obviously as you pointed out above.
we haven't covered that bizarre aldehyde exception in first year for the splitting, so maybe someone in higher years knows?

[stonecold]

yeah, i got it right as well, because I was able to work out the other 2 groups like you did vexx.

i wonder if we are expected to know this stuff for the exam...

[vexx]

yeah, i got it right as well, because I was able to work out the other 2 groups like you did vexx.

i wonder if we are expected to know this stuff for the exam...

i would be quite certain that if we ever got a question to do with these exceptions it would be like that question were you can work it out anyway by elimination! i heard that the NEAP questions are very good in terms of replicating the VCAA questions if not harder, so hopefully we have no need to worry:)

[stonecold]

^i really like the NEAP questions.  SO much better than the textbook.  I actually quite enjoy doing them.  I wish I could just do chem and bio for the next 6 weeks, but that just cant happen.

Also, I wonder if for NMR, do you think they will get us to use the chemical shift to identify groups?
They'll probably give us the molecule already, so it won't be too hard, but I haven't really seen any questions to practise it.

And also, just me, or is infrared spec really easy?

[vexx]

^i really like the NEAP questions.  SO much better than the textbook.  I actually quite enjoy doing them.  I wish I could just do chem and bio for the next 6 weeks, but that just cant happen.

Also, I wonder if for NMR, do you think they will get us to use the chemical shift to identify groups?
They'll probably give us the molecule already, so it won't be too hard, but I haven't really seen any questions to practise it.

And also, just me, or is infrared spec really easy?

im not sure what they could ask, but they could ask anything as long as it is relevant so a question like that in NMR could be asked.
yeah infrared is pretty easy, there isn't really much to it. not too sure about mass spec at this point as i've barely going through it / done any questions, but it seems slightly more complicated.

and i would love to just do 6 weeks of chem haha. BUT you can't complain! you have like one class a day, i have classes at 8.30 and finish anywhere between 3:15pm and 6.45pm (have umep right after class and finishes at 6.45, and englang is right after a class and finishes at 5.30) heheh

[stonecold]

ahh i know.  i have it relatively easy compared to everyone else here.  i  shouldn't whinge...
i go to school practically one day a week and that's it.  i somehow still magage to waste a lot of time

anyhow, been progressing through the NEAP book, and just noticed something on chromatography.

basically, peak area on a chromatogram = concentration/relative amounts present.  not peak height, which is sweet, as i knew that.

what i don't get later on is, they give you two different molecules, and because of the properties, you have to be able to determine which one has the longer retention time, and given the concentrations of each, draw the chromatogram.  easy, but they actually show the peak heights.

e.g.  for 700mg ethanol and 200mg propanol, propanol has larger retention time, and ethanol has largest peak area.
on the graph in the answer though, they actually draw the peaks up to 200mg and 700mg.  Isn't this technically incorrect, as it is peak area and not peak height that indicates relative concentration?

Would you be expected to show this in an exam, because i just showed that one peak was larger than the other (relative sizes of course) and labelled the peaks and the axes.

Question 1.4.8 and 1.4.9 in the NEAP guide for anyone who has it...

[vexx]

ahh i know.  i have it relatively easy compared to everyone else here.  i  shouldn't whinge...
i go to school practically one day a week and that's it.  i somehow still magage to waste a lot of time

anyhow, been progressing through the NEAP book, and just noticed something on chromatography.

basically, peak area on a chromatogram = concentration/relative amounts present.  not peak height, which is sweet, as i knew that.

what i don't get later on is, they give you two different molecules, and because of the properties, you have to be able to determine which one has the longer retention time, and given the concentrations of each, draw the chromatogram.  easy, but they actually show the peak heights.

e.g.  for 700mg ethanol and 200mg propanol, propanol has larger retention time, and ethanol has largest peak area.
on the graph in the answer though, they actually draw the peaks up to 200mg and 700mg.  Isn't this technically incorrect, as it is peak area and not peak height that indicates relative concentration?

Would you be expected to show this in an exam, because i just showed that one peak was larger than the other (relative sizes of course) and labelled the peaks and the axes.

Question 1.4.8 and 1.4.9 in the NEAP guide for anyone who has it...

haha what! i just did that about 10 seconds ago, we are actually doing it at the same times.

yeahh i think i trust neap with this because i got that one right by thinking:
- ethanol has a smaller mass therefore it will come out of the column quicker, having a lower retention time, then proponal which is a heavier molecule (higher retention time) so you know that the left peak is eth and the right is prop.
- the peak height is apparently the relative amount, with the peak area being the relative concentration, so since we can only assume the latter is the same. and we determine the peak height using a ration of eth:prop 7:2 and then drawing the peaks high enough so they reach these relative numbers for each.

and voila, i think that is it.