Week 10 - NMR Assignment

Week 10

• Tu and Th Morning Labs start at 9:30 AM.

• Laboratory notebook check
- Bring your completed Lab notebook
- Make sure you have a table of contents
- See p. 5 of reader for details on grading of lab notebook

• NMR Lab

This Week's Experiment: This assignment will be completed in the Science Learning Center (4th Floor of Young Hall)

THIS ASSIGNMENT WILL NOT BE GRADED. However, I recommend that you attempt these problems. These problems will help you acquire the problem solving skills required to identify the structure of an unknown compound given only the molecular formula, IR, Carbon NMR and DEPT NMR. I recommend that you work in groups of 2 or 3.

First, go to the WebSpectra NMR page (at www.chem.ucla.edu/~webspectra/).

Once at the WebSpectra page, do the following:

1. Identify the structure of Compound #21 (Beginner level) using Carbon NMR and DEPT NMR. Note: for the carbon NMR, the signals at 77.4, 77.1 and 76.8 are due to the NMR solvent (CDCl₃) and should be ignored. [On the final exam I will clearly identify the solvent signals].

Suggested plan of attack:

a) First, determine the unsaturation number for C8H10.

Thus, unsat # = [ 8 x (4 - 2) + 10 x (1 - 2) + 2 ] ÷ 2

= [ 16 - 10 + 2 ] ÷ 2 = 4

when the unsaturation number is 4 or greater, one immediate possibility to be consider is "is there an aromatic ring present". Benzene, for example, has an unsaturation number of 4 (i.e., there are three double bonds and a ring closure).

b) Try to account for the degrees of unsaturation.

Is the compound aromatic? is an alkene present? carbonyl group? IR spectroscopy is an excellent way to determine which functional groups are present. Carbon NMR can also be used to identify functional groups.

c) Are all of the degrees of unsaturation accounted for?

If the answer is no to this question, then a good possibility is that your compound has a ring closure.

d) Does the compound have symmetry?

For example, the carbon NMR for compound #21 has only 6 signals. Since this compound has 8 carbons, there has to be some symmetry present that explains the equivalency of two carbons.

e) Using DEPT NMR, identify the carbon units present.

For example, Compound #21 has C, CH, CH, CH, CH₂ and CH₃ units only. Also note if the units are sp2 carbons (between 220-100 ppm) or sp3 carbons (between100-0 ppm). Thus, {C, CH, CH, CH} are part of the aromatic ring. {CH₂ and CH₃} are the saturated chain.

DEPT NMR Logic Table: (KNOW THIS TABLE FOR THE FINAL EXAM!)

	C	CH	CH2	CH3
DEPT 135	0	1	-1	1
DEPT 90	0	1	0	0
DEPT 45	0	1	1	1

e) Attach the units in order that best fits the spectral data.

Just start proposing possible structures! Don't try to solve the complete structure in your head.

f) Finally, once you have a proposed structure, then go back and make sure that is fits the spectral data given.

2. Identify the structure of Compound #22 (Beginner level) using IR, Carbon NMR and DEPT NMR.

3. Identify the structure of Compound #15 (Intermediate 1 level) using IR, Carbon NMR and DEPT NMR.

4. Identify the structure of Compound #12 (Intermediate 2 level) using IR, Carbon NMR and DEPT NMR.

5. Identify the structure of Compound #13 (Intermediate 2 level) using IR, Carbon NMR and DEPT NMR.