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12.E: Structure Determination: Mass Spectrometry and Infrared Spectroscopy (Exercises)

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    61717
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    These are the homework exercises to accompany the Textmap for McMurry's Organic Chemistry textbook.

    12.2 Interpreting Mass Spectra

    12.2 Exercises

    Questions

    Q12.2.1

    Caffeine has a mass of 194.19 amu, determined by mass spectrometry, and contains C, N, H, O. What is a molecular formula for this molecule?

    Q12.2.2

    The following are the spectra for 2-methyl-2-hexene and 2-heptene, which spectra belongs to the correct molecule. Explain.

    A:

    B:

    Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 2 December 2016)

    Solutions

    S12.2.1

    C8H10N4O2

    C = 12 × 8 = 96

    N = 14 × 4 = 56

    H = 1 × 10 = 10

    O = 2 × 16 = 32

    96+56+10+32 = 194 g/mol

    S12.2.2

    The (A) spectrum is 2-methyl-2-hexene and the (B) spectrum is 2-heptene. Looking at (A) the peak at 68 m/z is the fractioned molecule with just the tri-substituted alkene present. While (B) has a strong peak around the 56 m/z, which in this case is the di-substituted alkene left behind from the linear heptene.

    12.3 Mass Spectrometry of Some Common Functional Groups

    12.3 Exercises

    Questions

    Q12.3.1

    What are the masses of all the components in the following fragmentations?

    Solutions

    S12.3.1

    12.5 Spectroscopy and the Electromagnetic Spectrum

    12.5 Exercises

    Questions

    Q12.5.1

    Which of the following frequencies/wavelengths are higher energy

    A. λ = 2.0x10-6 m or λ = 3.0x10-9 m

    B. υ = 3.0x109 Hz or υ = 3.0x10-6 Hz

    Q12.5.2

    Calculate the energies for the following;

    A. Gamma Ray λ = 4.0x10-11 m

    B. X-Ray λ = 4.0x10-9 m

    C. UV light υ = 5.0x1015 Hz

    D. Infrared Radiation λ = 3.0x10-5 m

    E. Microwave Radiation υ = 3.0x1011 Hz

    Solutions

    S12.5.1

    A. λ = 3.0x10-9 m

    B. υ = 3.0x109 Hz

    S12.5.2

    A. 4.965x10-15 J

    B. 4.965x10-17 J

    C. 3.31x10-18 J

    D. 6.62x10-21 J

    E. 1.99x10-22 J

    12.7 Interpreting Infrared Spectra

    12.7 Exercises

    Questions

    Q12.7.1

    What functional groups give the following signals in an IR spectrum?

    A) 1700 cm-1

    B) 1550 cm-1

    C) 1700 cm-1 and 2510-3000 cm-1

    Q12.7.2

    How can you distinguish the following pairs of compounds through IR analysis?

    A) CH3OH (Methanol) and CH3CH2OCH2CH3 (Diethylether)

    B) Cyclopentane and 1-pentene.

    C)

    12.7(2).png

    Solutions

    S12.7.1

    S12.7.2

    A) A OH peak will be present around 3300 cm-1 for methanol and will be absent in the ether.

    B) 1-pentene will have a alkene peak around 1650 cm-1 for the C=C and there will be another peak around 3100 cm-1 for the sp2 C-H group on the alkene

    C) Cannot distinguish these two isomers. They both have the same functional groups and therefore would have the same peaks on an IR spectra.

    12.8 Infrared Spectra of Some Common Functional Groups

    12.8 Exercises

    Questions

    Q12.8.1

    The following spectra is for the accompanying compound. What are the peaks that you can I identify in the spectrum?

    Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 2 December 2016)

    Q12.8.2

    What absorptions would the following compounds have in an IR spectra?

    Solutions

    S12.8.1

    Frequency (cm-1) Functional Group

    3200 C≡C-H

    2900-3000 C-C-H, C=C-H

    2100 C≡C

    1610 C=C

    (There is also an aromatic undertone region between 2000-1600 which describes the substitution on the phenyl ring.)

    S12.8.2

    A)

    Frequency (cm-1) Functional Group

    2900-3000 C-C-H, C=C-H

    1710 C=O

    1610 C=C

    1100 C-O

     

    B)

    Frequency (cm-1) Functional Group

    3200 C≡C-H

    2900-3000 C-C-H, C=C-H

    2100 C≡C

    1710 C=O

     

    C)

    Frequency (cm-1) Functional Group

    3300 (broad) O-H

    2900-3000 C-C-H, C=C-H

    2000-1800 Aromatic Overtones

    1710 C=O

    1610 C=C