NMR Pattern Identification Problems Fall 2015 KEY

For Questions 1-3, select the structure from the following choices that best matches the NMR spectrum. The idea is to eliminate possible structures based mainly on the observed coupling patterns and intensities, although chemical shift information is helpful too.

Question 1

A

The indicates a H with 6 equivalent H next door and when you see this and a

large 6H doublet, you can be sure you have an isolated isopropyl group CH(CH3)2. This eliminates all but A, B or C. Of these three, the correct answer must be A because it is the only isomer that can have two doublets in the aromatic region.

Question 2

J

In this case, the pattern indicates 5 near neighbours and, along with a triplet for a methyl group, is consistent with a -CH2CH2CH3. The additional presence of a pentet indicates a CH2 with four near neighbours and along with the remaining upfield triplet, suggests we have a -CH2CH2CH2CH3 chain. This eliminates everything except J. Note the aromatic pattern should be a doublet (ortho) and two triplets (meta and para) for a mono-substituted benzene ring, although it is common for these resonances to overlap.

Question 3

E

Two overlapping triplets and two strongly suggests there are two ethyl groups narrowing our choices down to D, E or F. The presence of a singlet, doublet and triplet pattern in the aromatic region is only consistent with meta-substitution so the correct answer is E. ______

For Questions 4-6, select the structure from the following choices that best matches the NMR spectrum.

ABC D E

N N N N N

H N N N N N

FGHIJ

Question 4

J

There is only one possible structure that can have two inequivalent triplets and two singlets and that is J. The singlets would be in 9:6 ratio which looks reasonable from the spectrum. Question 5

E

The downfield resonances are in the alkene region and clearly indicate there are two different alkene H. This eliminates all except C, D, E and G. The absence of an isopropyl group (no septet and doublet) eliminates C, D and G leaving only E as a possibility. Note E should have two singlets in 3:1 ratio for the t-Bu and methyl groups and the alkene H should be approximately dt (large coupling to trans alkene H and smaller coupling to adjacent CH2) and dq (large coupling to trans alkene H and smaller coupling to adjacent CH3) as observed.

Question 6

B

The lone small resonance downfield suggests this compound only has one alkene H which eliminates all except A and B. A can be eliminated as there is no isopropyl groups (septet and doublet) so the correct answer is B. This fits as it should have a and triplet of an ethyl group (there are two but they are equivalent) as well as

two singlets for the alkene CH3 groups (which are different). The alkene H is coupled to a neighbouring CH2 making it a triplet and that CH2 is just coupled to the single alkene H making it a doublet.

______

For Questions 7-10, select the structure from the following choices that best matches the NMR spectrum. In this case, use the additional IR data to help distinguish between possibilities.

ABCD

O O O O O O O O

O O O O O O O O O O

EFGHI

Question 7 IR at 1706 cm-1

B

A band at 1706 cm-1 could be a C-conjugated ester or a non-conjugated ketone which leaves only B, E and G as possibilities. E would be expected to have the septet and doublet pattern of an isopropyl group and G would have three singlets of integration 2:2:3 so by the process of elimination it must be B. B should have two triplets (2:3 ratio)

and a sextet (-CH2CH2CH3) characteristic of an n-propyl chain.

Question 8 IR band at 1768 cm-1

F

A band at 1765 cm-1 is consistent with an O-conjugated ester which eliminates all but C and F. The septet and doublet pattern confirms an isopropyl group so the answer must be F.

Question 9 IR band at 1722 cm-1

G

The IR is probably close enough that this could be either a C-conjugated ester or a non- conjugated ketone, although the latter might be a better fit. That leaves just C, E and G as possibilities but we can quickly settle on G since it is the only one of the three that will have any singlets in the 1H NMR.

Question 10 IR band at 1688 cm-1

H

You don’t even need the NMR to identify H as the only possibility because of these choices, only the conjugated ketone H would have an C=O stretch below 1700 cm-1.

The NMR matches too as it has an CH2 singlet downfield due to the adjacent oxygen and the carbonyl plus the characteristic triplet and quartet of an ethyl group.

END