jane1234's chemistry questions!!

[nacho]

Interesting,
i would have said
A) 3
b) 4

and, yea, those are supposed to be H atoms. being the lazy conventional chemist i am, i didn't write them out

What is the book answer xD? Hi Five Vea.


Sorry..but the suspense is killing me.

no book answer, i made it up
um, actually im pretty lost myself, i was hoping to receive to get the same answer as you guys lol..


edit: some of you may find this diagram helpful, credits to luken93


this is how i did it:

[luken93]

http://www.nmrdb.org/cheminfo/servlet/org.cheminfo.hook.appli.HookServlet?imageID=1003527262802998733.png

It's a pretty bad molecule to choose, however it's got 4 with a multiplet (may as well say 4.5)

[luken93]

Honestly guys, don't worry too much about it. They are never going to give you anything this hard, as multiplicity isn't studied at VCE level. They will choose something that is usually user friendly.

[Water]

Luken, what about the Carbon Environment?

[nacho]

Honestly guys, don't worry too much about it. They are never going to give you anything this hard, as multiplicity isn't studied at VCE level. They will choose something that is usually user friendly.

well, at least we've seen the hardest now! (clap)

[Andiio]

Honestly guys, don't worry too much about it. They are never going to give you anything this hard, as multiplicity isn't studied at VCE level. They will choose something that is usually user friendly.

Yeah multiplicity/multiplets are shifty as. Have you done them in Uni Chem?

[RobM8]

Luken, what about the Carbon Environment?

Anyone know the correct answer?

[luken93]

Luken, what about the Carbon Environment?

Can't find one, sorry

Honestly guys, don't worry too much about it. They are never going to give you anything this hard, as multiplicity isn't studied at VCE level. They will choose something that is usually user friendly.

Yeah multiplicity/multiplets are shifty as. Have you done them in Uni Chem?

Yeah, only really pops up in Benzene (when the environments are almost exactly the same)

[nacho]

Luken, what about the Carbon Environment?

Can't find one, sorry

Honestly guys, don't worry too much about it. They are never going to give you anything this hard, as multiplicity isn't studied at VCE level. They will choose something that is usually user friendly.

Yeah multiplicity/multiplets are shifty as. Have you done them in Uni Chem?

Yeah, only really pops up in Benzene (when the environments are almost exactly the same)

just enquiring, for the multiplicity question, would one round that up to 5 environments? Or is it still 4, because technically, it's such an insignficant difference between the amount of deshielding? (as stated by relevance in accounting)

[luken93]

No, it'd be 4 - the multiplet is still a group of peaks in the end.

DON'T WORRY ABOUT IT GUYS, EVERYONE CHILL!

[RobM8]

I can't chill with you turning caps lock on buds.

[nacho]

Yeah I am stupid, it is 4 1H environments.

My reasoning was:

number of protons/1H's --> none-3-2-none-2-2-3-none

So environments are ---> none32, 32none, none22, 223, 23none (23none is the same as none32) so 4?

Do the non-immediate neighbour atoms matter in 1H NMR?

Any flaw in the way I worked it out?

there may be, but i don't even know what your working out means

[RobM8]

Hahaha, don't worry about it. It is hard to do this on the computer.

[Graphite]

a) 13C , 5 peaks
b) 5 peaks.

that is the right answer. this molecule is not symmetrical

[Pixon]

It's 5 and 5. I saw this last night when Water responded...I didn't think there would be so much discussion... :/
It's not symmetrical...so all the carbons/proton environments are being deshielded in a different way. Don't over think it with this 4.5 business.  Make sure you look at the whole molecule, not just the adjacent protons.

And sorry to say so, but I honestly don't think that is the hardest thing you will get in VCE. Consider Aspirin which is commonly examined since so many students are familiar with it...it's probably a good idea to give that one a go...at least at low-res 1H NMR.

I don't know why multiplicity even came up since splitting was never mentioned....but basically there's only the one proton environment based off the carbon in between the methylene and methyl group which has "strange" splitting. It is first split into 4 due to the 3 hydrogens on the methyl, and these 4 peaks are split into three due to the hydrogens on methylene. Thus... 4x3=12 peaks for that proton environment...I may just be a VCE student but this is what I've been taught anyways...could be wrong...

P.S Some schools do teach multiplicity...and Luken, what kind of multiplicity are you seeing in a benzene ring? I'm very curious to know what this Uni Chem is teaching you...