Post by: methodsboy on April 02, 2009, 06:10:08 pm
i've got a chem SAC tommorrow and i dnt know how to do this question :-[
could someone please help - Question 12 and 13
[IMG]http://img12.imageshack.us/img12/2094/massspectometry.jpg[/img]
Thanks
Post by: TrueTears on April 02, 2009, 06:24:54 pm
if it is, then I shall continue, if not, then my method must be wrong.
Post by: methodsboy on April 02, 2009, 06:27:10 pm
Post by: TrueTears on April 02, 2009, 06:28:54 pm
Post by: d0minicz on April 02, 2009, 06:33:45 pm
and the highest m/e is 60
so i guess you play around ensuring that you have a COOH and the carbons and hydogens are proportional etc :S
then work your way from there
sorry if isnt any help lols :S
Post by: TrueTears on April 02, 2009, 06:37:43 pm
Post by: d0minicz on April 02, 2009, 06:41:51 pm
Post by: Mao on April 02, 2009, 06:51:05 pm
For Mass Spec, IR and NMR in VCE, you assume it is an organic molecule,
Parent peak (heaviest and abundant) - 60
Base peak (most abundant) - 43
Second most abundant peak - 45
From 60 to 43, a fragment of weight of 17 has been lost, this is -OH
From 60 to 45, a fragment of 15 has been lost, this is CH3
It is an acid, most likely carboxylic acid.
Checking Mr of CH3COOH, bingo.
Ethanoic acid, peak at 60 is CH3COOH+, peak at 45 is COOH+, peak at 43 is CH3C(=O)+, peak at 15 is CH3+
Post by: TrueTears on April 02, 2009, 06:51:59 pm
Q11)oh i see, gotta note that from now on.
For Mass Spec, IR and NMR in VCE, you assume it is an organic molecule,
Parent peak (heaviest and abundant) - 60
Base peak (most abundant) - 43
Second most abundant peak - 45
From 60 to 43, a fragment of weight of 17 has been lost, this is -OH
From 60 to 45, a fragment of 15 has been lost, this is CH3
It is an acid, most likely carboxylic acid.
Checking Mr of CH3COOH, bingo.
Ethanoic acid, peak at 60 is CH3COOH+, peak at 45 is COOH+, peak at 43 is CH3C(=O)+, peak at 15 is CH3+
Post by: Mao on April 02, 2009, 06:57:22 pm
There are two carbon environments. Since there are three carbons in the molecule (propanol), it must be symmetrical about the central carbon (otherwise there will be three carbon enviroments).
Hence the molecule must be 2-propanol.
Remembering that atoms closer to electronegative atoms have higher chemical shifts:
on the carbon graph, the small peak on the left is the central carbon connected to the hydroxyl group, the tall peak on the right is the two outer carbons
on the proton graph, the sextuplet on the left is the single H connected to the central carbon (there are 6 H in its neighbourhood), the singlet is the H in OH (shielded by the O), and the doublets account for the 6 hydrogens on the outer carbons, as both of the CH3 group's neighbour is the H on the central carbon
Post by: methodsboy on April 02, 2009, 09:09:58 pm
13)You Rock!
There are two carbon environments. Since there are three carbons in the molecule (propanol), it must be symmetrical about the central carbon (otherwise there will be three carbon enviroments).
Hence the molecule must be 2-propanol.
Remembering that atoms closer to electronegative atoms have higher chemical shifts:
on the carbon graph, the small peak on the left is the central carbon connected to the hydroxyl group, the tall peak on the right is the two outer carbons
on the proton graph, the sextuplet on the left is the single H connected to the central carbon (there are 6 H in its neighbourhood), the singlet is the H in OH (shielded by the O), and the doublets account for the 6 hydrogens on the outer carbons, as both of the CH3 group's neighbour is the H on the central carbon