Proton NMR Exercises with Structure Builder

This activity was designed by Dr. Richard Musgrave with programming by Mike Carnahan, William Haun and Alan Shapiro at St. Petursburg College, PO Box 13489, St. Petersburg, FL, 33733 USA

The following exercises require you to look at a ¹H NMR and build the structure. This is a summary of the required steps. There is also a good help page available within the application. Make sure you have the latest version of Flash Player installed. Some guidance for solving NMR structures given a molecular formula and spectrum can be found at the bottom of this page.

1. Calculate the number of Double Bond Equivalents (DBE). This is also called Units of Unsaturation and it defines the number of rings and/or double bonds based on the molecular formula (see below).

2. Using the integration of the peaks, pick the pieces you can see based on the spectrum. Watch the Molecular Formula and DBE count down as you pick the pieces!

3. If the Molecular Formula and DBE have not counted down to zero, select the other groups that make up the compound. For example, if there is one C, one O and one DBE left over, there is probably a C=O group.

4. After everything has counted down to zero, you will be allowed to put the structure together! Drag the fragments into the field and put them together using multiplicity and chemical shift information from the spectrum. There is one tip that is VERY important here. All fragments are rotatable by right clicking, but in order to ensure your answer matches the correct answer, rotate METHYL (CH₃) GROUPS only!

5. Keep trying until you get the right structure, but most of all: HAVE FUN. You can find the answers here if you get stuck!

Number 1	Number 2	Number 3	Number 4
Number 5	Number 6	Number 7	Number 8
Number 9	Number 10	Number 11	Number 12
Number 13	Number 14	Number 15	Number 16
Number 17	Number 18	Number 19	Number 20
Number 21	Number 22	Number 23	Number 24
Number 25	Number 26	Number 27	Number 28
Number 29	Number 30	Number 31	Number 32
Number 33	Number 34	Number 35	Number 36
Number 37	Number 38	Number 39	Number 40
Number 41	Number 42	Number 43	Number 44
Number 45	Number 46	Number 47	Number 48
Number 49	Number 50

¹H NMR

a) Correlation chart:

One peak for each DIFFERENT H.

b) Integration: Tells how many H there are of a given type.

For peaks < 5 ppm the following usually applies:

3H = CH₃                                                                                             9H = 3 x CH₃

2H = CH₂; NH₂ (NH₂ single peak)                                      6H = 2 x CH₃  OR 3 x CH₂

1H = CH; NH; OH (NH, OH single peak)                                          4H = 2 x CH₂

For peaks 7-8 ppm (aromatic) the following usually applies. (There may be two or more peaks).

4H = aromatic ring with 4H attached

5H = aromatic ring with 5H attached.

c) Multiplicity: Tells how many H are on adjacent Carbon atoms.

n peaks = n-1 adjacent H atoms on C atoms. (H_x = H in question)

1 peak     = 0 adjacent H atoms on C (CH_x—O, CH_x—NH, CH_x—C=O);

2 peaks= 1 adjacent H atoms on C (CH_x—CH)

3 peaks= 2 adjacent H atoms on C (CH_x—CH₂)

4 peaks= 3 adjacent H atoms on C (CH_x—CH₃)

5 peaks= 4 adjacent H atoms on C (CH₂—CH_x—CH₂) or (CH—CH_x—CH₃)

6 peaks= 5 adjacent H atoms on C (CH₃—CH_x—CH₂)

7 peaks= 6 adjacent H atoms on C (CH₃—CH_x—CH₃)

DOUBLE BOND EQUIVALENTS (# of rings and/or double bonds) may be obtained from the molecular formula in the following way:  (Example: C₇H₁₃NF₂O₃)                         

a)       For the purpose of this calculation, find effective number of H.

i)         For each halogen, add 1 H.

ii)       For each nitrogen, subtract 1H.

iii)      Oxygen has no effect on this calculation.

So C₇H₁₃NF₂O₃ becomes C₇H₁₄ (JUST FOR THIS CALCULATION).

b)       Find how many H the compound would have if it were saturated (saturated H)(remember a saturated hydrocarbon is C_nH_2n+2).

C₇H₁₄ ----> C₇H₁₆

c)   Units of unsaturation    =              Saturated H - Actual H

                                                                           2

                                        =              16 -14

                                                             2

                                        =              1 (ring or double bond)