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!

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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. (Hx = H in question)

1 peak     = 0 adjacent H atoms on C (CHxO, CHxNH, CHxC=O);

2 peaks= 1 adjacent H atoms on C (CHxCH)

3 peaks= 2 adjacent H atoms on C (CHxCH2)

4 peaks= 3 adjacent H atoms on C (CHxCH3)

5 peaks= 4 adjacent H atoms on C (CH2CHxCH2) or (CHCHxCH3)

6 peaks= 5 adjacent H atoms on C (CH3CHxCH2)

7 peaks= 6 adjacent H atoms on C (CH3CHxCH3)


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.


b)       Find how many H the compound would have if it were saturated (saturated H)(remember a saturated hydrocarbon is CnH2n+2).

C7H14 ----> C7H16

c)   Units of unsaturation    =              Saturated H - Actual H


                                        =              16 -14


                                        =              1 (ring or double bond)