kaplan doesn't give a lot for ochem on their normal retail books, but for the people in their class holy crap their science section is very thorough for ochem. Compare how much content is in general chem in the normal version, that is how much ochem there is for the people that is in their class. They have stuff on NMR and stuff which seriously couldn't possibley be on the PCAT! I don't know what it is but I know a lot of friends that taken Ochem2(i only taken ochem1) and they say that stuff is the hardest stuff ever lol and take forever hah.
It is hard to imagine there will be a long NMR problem on the PCAT. But if there is such a problem, I'm sure it would be one that recalls basic knowledge in using NMR to predict the structure of an organic compound.
A typical NMR problem I learned from organic2 would look like the following:
Given C4 H10 0 is reactant, Predict the structure of the product
C4 H10 O -----> ??
Givens:
After analysis, here are the product's IR and NMR data:
IR
2950 cm-1
3500 cm-1
H NMR
Shift------Split------Integration
3.4 ppm---- t-----------2H
2.4 ppm-----t-----------1H
1.7 ppm-----m----------1H
0.9 ppm-----d-----------6H
First thing to do is to find Degree of Unsaturation or IHD
DOU or IHD = (((2*(#C) - #H)+2)/2 + #N/2
DOU or IHD = number of double bonds and/or # of rings
(e.g. benzene ring has IHD = 4) since there are 3 double bonds and 1 ring
Shift tells you what type of structure Hydrogens are next to? (e.g. H next to benzene ring would be around 7.2ppm) so a chart must be supplied
Split = N + 1 where N is the total number of hydrogens on neighboring Carbons
Integration is total number of Hydrogen in the product compound
To solve the problem I wrote above.
IHD or DOU = zero (so I know there is no double bonds nor any Rings)
Interpret IR data:
2950 cm-1 tells me there are sp3 C-H bonds
3500 cm-1 tells me there is an O-H bond
Integration shows product compound will have a total of 10 hydrogens
Interpret Split (usually the hardest part of NMR)
Look at the Final compound to understand the following:
1.7 ppm m(multiplit) 1H shows C-H bond at (a)
since there are a total of 6 hydrogens on the two neighboring carbons connected to C-H at (a) => N=6 6+1 = 7 (7 is considered a multiplit)
0.9 ppm d (doublet) 6H shows 2 sets of CH3 (a total of 6 hydrogens here) at (b)
since 2 sets of CH3 is connected to one neighboring carbon connected to only 1 Hydrogen => N=1
1 +1 = 2 (doublet)
3.4 ppm t(triplet) 2H shows the 2 hydrogens at (c)
since its Carbon has a neighboring Carbon with 1 hydrogen and another neighboring Oxygen with 1 hydrogen: N = 2 (added those 2 hydrogens)
2 + 1 = 3 (triplet)
2.4 ppm t(triplet) 1H shows 1 hydrogen at (d)
This hydrogen is attach to oxygen and the oxygen has 1 neighboring carbon with 2 hydrogens => N = 2
2 + 1 = 3 (triplet)
Hence,
Final compound looks like this:
-------(a)--(c)
--------H---H
-(b)----l----l
H3C---C---C---H (c)
--------l----l
------H3C-OH
-------(b)-(d)
IUPAC name for the product: 2-methyl Propanol
(If I made a mistake; please correct me)
This is a basic NMR problem. Based on what I heard from a chemistry major who wants to go to med school, there are specific undergraduate Chemistry classes dedicated to studying NMR and using NMR to interpret compound larger than the one I solve above.
This may sound crazy to some people (it did to my friends) but I think this stuff is really interesting; the fact, we can use IR and NMR technology to accurately predict an unknown compound and its structure.
Oh and i have to add..... This NMR problem won't be on the PCAT at least not to this extend or magnitude.(hopefully)