# CH ENGR 104C Lecture Notes - Lecture 20: Infrared Spectroscopy, Unified Atomic Mass Unit, Mass Spectrum

For unlimited access to Class Notes, a Class+ subscription is required.

SOLVING COMBINED SPECTROSCOPY PROBLEMS:

Lecture Supplement: page 50-53 in Thinkbook

CFQ’s and PP’s: page 216 – 241 in Thinkbook

Introduction:

The structure of an unknown molecule can be determined using three methods, which

each reveal different aspects of the whole:

o Mass Spectrometry indicates formula (by varying the magnetic field to

manipulate the flight path of ions and plotting the number of ions recorded)

o Infrared (IR) Spectroscopy indicates presence/absence of functional groups

(using the principle that similar functional groups absorb photons of similar

energies)

o Nuclear Magnetic Resonance (NMR) Spectroscopy indicates how atoms are

attached to give the “structural skeleton” (by using differences in nuclear spin flip

energies)

Procedure:

Note: On the majority of practice problems, the given information will not be

presented in a way that is easy to read; it will just be listed. So to help solve the

problem, it’s always a good idea to organize the data in a table or some other form

that makes it easy to understand.

1. Mass Spectrometry: The first set of data that should be analyzed is the Mass

Spectrometry. This process will yield the formula for the molecule.

Molecular Ion:

m/z: Relative Intensity:

M mass of M 100% *M should always be set to 100%

M+1 mass of M+1 relative intensity of M+1

M+2 mass of M+2 relative intensity of M+2

o *M should always be set to 100% relative intensity. If it is anything other

than 100%, you must set it to 100% and scale the M+1 and M+2 peaks by the

same factor.

For example: (Example is from page 50 of the Thinkbook)

Molecular Ion: m/z: Relative Intensity:

M 120 48% 48 * 2.083 100%

M+1 121 4.85% 4.85 * 2.083 10.104%

M+2 122 <0.1% 0.1 * 2.083 <0.208%

To make the M peak equal to 100% it must be multiplied by a factor of

(100/48) = 2.083. You must then multiply M+1 and M+2 by this factor as

well.

o Next, you should analyze the data from the mass spectrum:

• M: gives information about the presence of Nitrogen using the Nitrogen

Rule:

If m/z for M is even, then there is an even number of N (0, 2, 4 …)

If m/z for M is odd, then there is an odd number of N (1, 3, 5 …)

• M+1: indicates the number of Carbons in the molecule:

# of Carbons = Intensity of M+1 peak

1.1

*It is divided by 1.1 because 1.1% of Carbon isotopes contribute to the

M+1 peak.

*If dividing these numbers gives a remainder of less than 0.3, round down.

If you get a remainder of greater than 0.7, round up. If you get a

remainder between 0.4 and 0.7, you should continue the problem

considering both cases (for example: if you get 10.5, then consider the

case of 10 Carbons and 11 Carbons).

• M+2: indicates the presence of Br, Cl, and S.

If M+2 = ~50%, then there is a Br in the molecule.

If M+2 = ~33%, then there is a Cl in the molecule.

If M+2 = ~4%, then there is a S in the molecule.

Note: there could be more than one of these atoms in molecule. For

example, M+2 = 66% would indicate that there are two Cl atoms.

However, these cases are rare in the problems given in the Thinkbook,

OWLS, and old exams.

o Finally, you must use this information to generate possible formulas for the

molecule.

• Begin by noting the m/z for the M peak. Using this number, you will

subtract the mass of any atoms you know for certain will be in the formula

The number of Carbons determined from the M+1 peak multiplied by

the mass of Carbon-12 (12 amu) tells you the mass of Carbon that can

be subtracted from the m/z.

If you found from the m/z for M that there are an odd number of

Nitrogens, you know for certain that there is at least one N so you can

subtract its mass (14 amu) from the m/z. If even, you cannot subtract

anything in this stage because there may be zero N.

If M+2 indicated the presence of Br, S, or Cl, you can subtract the

mass of the atom found. (For example, you can subtract 35 amu if

M+2 = 33% indicating Cl).

• Next, use the Hydrogen Rule to calculate the maximum number of

Hydrogen in the formula:

Max. # of H = 2c + n + 2,

c = # of Carbon and

n = # of Nitrogen

• Using the number of amu left (and taking into consideration the maximum

number of Hydrogens), construct possible formulas using C, H, O, N (and

Br, S, or Cl if you determined that they were present in previous steps).

You should not consider other elements like Iodine or Fluorine unless

indicated in the problem.

• For example: (Example is from page 218 of the Thinkbook).

Molecular Ion

m/z Relative Intensity Conclusions:

M 132 100% MW = 132

Even number of N

M+1 1 33 6.89% 6.89 ÷ 1.1 = 6.26

Six Carbons

M+2 134 1.42% No S, Cl, Br

M – (6 Carbons) = 132 – (6 x 12) = 60 amu left for O, N, H

Maximum # of H = 2c + n + 2 = (2)(6) + 0 + 2 = 14 H’s

Oxygen Nitrogen

60 – O – N = H Formula

Notes

0 0 60 – 0 – 0 = 60 C6H60 Violates H-rule

1 0 60 – 16 – 0 = 44 C6H44O Violates H-rule

2 0 60 – 32 – 0 = 28 C6H28O2Violates H-rule

3 0 60 – 48 – 0 = 12 C6H12O3Reasonable

0 2 60 – 0 – 28 = 32 C6H32N2Violate H-rule

0 4 60 – 0 – 56 = 4 C6H4N4Reasonable

1 2 60 – 16 – 28 = 16 C6H16N2O Reasonable

2 2 60 – 32 – 28 = 0 C6N2O2Reasonable

2. Proton NMR Spectroscopy: The next set of data that should be analyzed is the

NMR data.

(It is possible to analyze the IR before analyzing NMR. However, analyzing

NMR first gives the advantage of knowing the number of Hydrogens in the

molecule, which will help indicate the correct formula from the list of

possibilities. Having the correct formula will tell the number of DBE’s, a useful

tool in analyzing the IR spectrum).

• Information will be given about chemical shifts, splitting patterns, and

integrals. Arrange this data in a table (like shown below) to make it easy to

read. Using this data, you will learn how many sets of equivalent protons are

in the molecule and how many neighbors they have. This information will

give you structural information about the molecule.

For example: (Example is from page 218 of the Thinkbook).

Chemical Shift Splitting Integration #H Implication

4.0 ppm triplet 1.0 2 H CH2CH2

3.9 ppm singlet 1.5 3 H CH3

3.0 ppm singlet 1.0 2 H CH2

1.3 ppm sextet 1.0 2 H CH3CH2CH3

0.9 ppm triplet 1.5 3 H CH2CH3

6.0 12 H

• The integration does not tell you how many of each type of proton you

have. Instead, it tells you the ratio of equivalent protons in the molecule.

If the integration does not add up to the number of Hydrogens in your