Chapter 29 – Review

29.1 Chromatography Basics

  1. What is the purpose of chromatography? Check answer[1]
  2. Describe the purpose of the stationary phase and the mobile phase. Check answer[2]
  3. Why do substances travel at different rates? Check answer[3]

29.2 Thin Layer (TLC) and Paper Chromatography (PC)

  1. Try your own paper chromatography using household items.  Stationary phase – paper towel or coffee filter, Mobile phase – water or rubbing alcohol, Mixture – water soluble markers, water soluble wet paint, food colouring.  Describe the results and how you might adjust the experiment to get different results.
  2. Calculate the Rf values for the following separation: Solvent travelled 5.8 cm, Compound A travelled 2.1 cm, Compound B travelled 2.3 cm, Compound C travelled 4.0 cm, and Compound D travelled 5.6 cm.  What conclusions can you make about Compounds A through D? Check answer[4]
  3. Describe some of the factors that will influence a component’s Rf value.

29.3 Chromatographic Columns

  1. Research some uses of ion exchange chromatographic (IEC) columns and size exclusion chromatographic (SEC) columns.
  2. What do all types of chromatographic columns have in common? Check answer[5]
  3. What are some benefits of chromatographic columns over paper chromatography or thin-layer chromatography? Check answer[6]

29.4 Chromatography Technology

  1. Explain the principles of gas chromatography (GC).
  2. Explain the principles of high-performance liquid chromatography (HPLC).
  3. Consider what a scientist may need to consider when choosing a chromatography method.  Complete this chart.  Research may be needed.
Chart 1: What to consider when choosing a chromatography method. Columns are left blank for you to fill in.
Chromatography Method PC/TLC IEC GC HPLC
Can the method be used to physically separate sample into different containers?
Can the method be used to check purity of sample?
Can the method be used in combination with another method to determine compound identity?
What type of samples are required? Consider states, amount of sample, polarities, charges.
Is the method inexpensive or expensive to run?
Is the method readily available or does it require specialized lab equipment?
How long does the method take to get a result? (quick or slow)

Links to Enhanced Learning

For more questions about Chromatography, try Separation of solutions and mixtures chromatography (practice) [New tab]

29.5 Spectroscopy Basics

  1. What is the purpose of spectroscopy? Check answer[7]
  2. What is a spectrum? Check answer[8]

29.6 Infrared (IR) Spectroscopy

  1. What functional groups give the following signals in an IR spectrum?
    1. 1700 cm-1
    2. 1550 cm-1
    3. 1700 cm-1 and 2510-3000 cm-1  Check answer[9]
  2. How can you distinguish the following pairs of compounds through IR analysis?
    1. CH3OH (Methanol) and CH3CH2OCH2CH3 (Diethylether)
    2. cyclopentane and 1-pentene.
    3. 1-propanol and 2-propanol. Check answer[10]
  3. The following spectrum is for the accompanying compound, benzoic acid. What are the peaks that you can identify in the spectrum? Check answer[11]
    The IR spectrum for benzoic acid
    (Credit: “Benzoic acid – compound 673″ © National Institute of Advanced Industrial Science and Technology via SDBSWeb , accessed 11/21/2023.)
  4. Using the video IR spectra practice | Spectroscopy | Organic chemistry | Khan Academy – YouTube, predict which molecule has the shown IR spectrum.  Spectrum 1 – stop the video at 0:16. Determine which of the three molecules is the correct one.  Watch the video for an explanation of the answer. Spectrum 2 – starts at 2:00, stop at 2:05. Determine which of the three molecules is the correct one.  Watch the video for an explanation of the answer. Spectrum 3 – starts at 3:34, stop at 3:42. Determine which of the three molecules is the correct one.  Watch the video for an explanation of the answer.
  5. Where might the following compounds have IR absorptions? Check answer[12]
    a)

    A chemical structure of 1-cyclohexen-1-ylmethanol
    (credit: Organic Chemistry (OpenStax),  CC BY-NC-SA 4.0)

    b)

    A chemical structure of a methyl ester with a five-carbon chain off the carbonyl. There is a terminal alkyne and a methyl on the beta C.
    (credit: Organic Chemistry (OpenStax),  CC BY-NC-SA 4.0)
  6. The IR spectrum of an unknown compound is shown. What functional groups does the compound contain? Check answer[13]
  7. Where might the following compounds have IR absorptions? Check answer[14]

29.7 Mass Spectrometry (MS)

  1. What are three things that can be determined using MS? Check answer[15]
  2. Assume that you have two unlabeled samples, one of methylcyclohexane and the other of ethylcyclopentane. How could you use mass spectrometry to tell them apart? The mass spectra of both are shown. Check answer[16]
    The mass spectra of two unlabeled samples A and B.
    (credit: Organic Chemistry (OpenStax),  CC BY-NC-SA 4.0)
  3. The sex hormone testosterone contains only C, H, and O and has a mass of 288.2089 amu, as determined by high-resolution mass spectrometry. What is the likely molecular formula of testosterone? Check answer[17]

29.8 Nuclear Magnetic Resonance (NMR)

  1. What is a chemical shift? Check answer[18]
  2. What effect does the magnetic field have on atoms? Check answer[19]

29.9 1H NMR Spectroscopy

  1. How many different sets of protons do the following molecules contain?
    4 chemical structures: a) pyridoxine (vitamin B6); b) DEET (insect repellent); c) isocitrate (a citric acid cycle intermediate) and d) paroxetine (an antianxiety drug, trade name Paxil).
    (Credit: Map: Organic Chemistry (Wade), Complete and Semesters I and II, CC BY-NC-SA 4.0).
  2. How many non-equivalent hydrogens are in the following molecules? How many different signals will you see in a H1 NMR spectrum? Check answer[20]
      1. CH3CH2CH2Br
      2. CH3OCH2C(CH3)3
      3. Ethyl Benzene
      4. 2-methyl-1-hexene
  3. Methyl 2,2-dimethylpropanoate (CH3)2CCO2CH3 has two peaks in its 1H NMR spectrum. What are their approximate chemical shifts? Check answer[21]
  4. Each of the following compounds has a single 1H NMR peak. Approximately where would you expect each compound to absorb? Check answer[22]
    a)

    A chemical structure of cyclohexane.
    (credit: Organic Chemistry (OpenStax),  CC BY-NC-SA 4.0)

    b)

    A chemical structure of acetone.
    (credit: Organic Chemistry (OpenStax),  CC BY-NC-SA 4.0)

    c)

    A chemical structure of benzene.
    (credit: Organic Chemistry (OpenStax),  CC BY-NC-SA 4.0)

    d)

    The structure of dichloromethane whose condensed structural formula reads, C H 2 C l 2.
    (credit: Organic Chemistry (OpenStax),  CC BY-NC-SA 4.0)

    e)

    The structure of ethane-1, 2-dial has a 2-carbon chain. Each carbon is single-bonded to a hydrogen atom and double-bonded to an oxygen atom.
    (credit: Organic Chemistry (OpenStax),  CC BY-NC-SA 4.0)

    f)

    The structure of N,N-dimethylmethanamine in which nitrogen atom is bonded to three methyl groups.
    (credit: Organic Chemistry (OpenStax),  CC BY-NC-SA 4.0)
  5. How many peaks would you expect in the 1H NMR spectrum of 1,4-dimethylbenzene (para-xylene, or p-xylene)? What ratio of peak areas would you expect on integration of the spectrum? Check answer[23]
    The structure of para-xylene (1,4-dimethylbenzene).
    (credit: Organic Chemistry (OpenStax),  CC BY-NC-SA 4.0)
  6. Predict the splitting patterns you would expect for each proton in the following molecules: Check answer[24]
6 chemical structures used to predict the splitting
(credit: Organic Chemistry (OpenStax),  CC BY-NC-SA 4.0)
    1. Match spectra 7-12 below to their corresponding structures G-L.
      6 chemical structures labelled G through L used to match spectra 7-12
      (Credit: Map: Organic Chemistry (Wade), Complete and Semesters I and II, CC BY-NC-SA 4.0).
Spectrum 7 table
δ splitting integration
9.96 d 1
5.88 d 1
2.17 s 3
1.98 s 3
Spectrum 8 table
δ splitting integration
9.36 s 1
6.55 q 1
2.26 q 2
1.99 d 3
0.96 t 3
Spectrum 9 table
δ splitting integration
9.57 s 1
6.30 s 1
6.00 s 1
1.84 s 3
Spectrum 10 table
δ splitting integration
9.83 t 1
2.27 d 2
1.07 s 9
Spectrum 11 table
δ splitting integration
9.75 t 1
2.30 dd 2
2.21 m 1
0.98 d 6
Spectrum 12 table
δ splitting integration
8.08 s 1
4.13 t 2
1.70 m 2
0.96 t 3

29.10 13C NMR Spectroscopy

  1. At what approximate positions would you expect ethyl acrylate, H2C═CHCO2CH2CH3, to show 13C NMR absorptions? Check answer[25]
  2. Classify the resonances in the 13C NMR spectrum of methyl propanoate, CH3CH2CO2CH3. Check answer[26]
    The 13 C N M R spectrum of methyl propanoate shows peaks at 0 (T M S), 9, 29, 52, and 175.
    (credit: Organic Chemistry (OpenStax), CC BY-NC-SA 4.0)
  3. Predict the number of carbon resonance lines you would expect in the 13C NMR spectra of the following compounds: Check answer[27] a) Methylcyclopentane b) 1-Methylcyclohexene c) 1,2-Dimethylbenzene d) 2-Methyl-2-butene e)
    Chemical structure of 2-methyl-3-pentanone.
    (credit: Organic Chemistry (OpenStax), CC BY-NC-SA 4.0)

    f)

    Chemical structure of 2,3-dimethyl-2-pentene.
    (credit: Organic Chemistry (OpenStax), CC BY-NC-SA 4.0)
  4. This figure shows 13C NMR spectrum for three related molecules: p-nitrophenol, o-nitrophenol, and m-nitrophenol. Identify some 13C NMR differences between these isomers.  Try to explain the differences using the molecule structure.
13C NMR spectra for three nitrophenols
(Credit: Original data used to construct these spectra is © National Institute of Advanced Industrial Science and Technology via SDBSWeb. The spectra were recorded on a 15 MHz instrument (with respect to 13C, or 60 MHz with respect to 1H). Credit: Instrumental Analysis, CC BY-NC-SA 4.0)
  1. 13C-NMR data is given for the molecules shown below. Complete the peak assignment column of each NMR data table.
    Table A
    an oxygen separated by a 3 carbon chain and a methanone group
    δ carbon fragment carbon #
    161.12 CH
    65.54 CH2
    21.98 CH2
    10.31 CH3
    Table B
    a 5 carbon chain. A double done is at carbon 2. At carbon 3 is a carbon doubled bonded oxygen branch
    δ carbon fragment carbon #
    194.72 C
    149.10 C
    146.33 CH
    16.93 CH2
    14.47 CH3
    12.93 CH3
    Table CA complex structure containing two ethers, two ketones and branched alkyl groups
    δ carbon fragment carbon #
    171.76 C
    60.87 CH2
    58.36 C
    24.66 CH2
    14.14 CH3
    8.35 CH3
    Table DA complex structure containing an ester chain attached to a phenol
    δ carbon fragment carbon #
    173.45 C
    155.01 C
    130.34 CH
    125.34 C
    115.56 CH
    52.27 CH3
    40.27 CH2
    Table EA complex amine structure. The nitrogen has 3 branches: a phenyl group, and two ethyl groups
    δ carbon fragment carbon #
    147.79 C
    129.18 CH
    115.36 CH
    111.89 CH
    44.29 CH2
    12.57 CH3

Image Credit: Organic Chemistry with a Biological Emphasis Volume I, (page 296-297) CC BY-NC-SA 4.0

  1. Combining NMR and MS. You obtain the following data for an unknown sample. Deduce its structure.

Combustion Analysis: C (69.7%), H (11.7%)
1H-NMR:

NMR of an unknown sample
(Credit: Organic Chemistry with a Biological Emphasis Volume I, (page 301) CC BY-NC-SA 4.0)

13C-NMR:

NMR of an unknown sample
(Credit: Organic Chemistry with a Biological Emphasis Volume I, (page 301) CC BY-NC-SA 4.0)

Mass Spectrometry:

MS of an unknown sample
(Credit: Organic Chemistry with a Biological Emphasis Volume I, (page 300) CC BY-NC-SA 4.0)

Links to Enhanced Learning

29.11 Visible and Ultra-Violet Spectroscopy (UV-Vis)

  1. What is a conjugated double bond? Check answer[28]
  2. Identify which of the following dienes are isolated, conjugated, or cumulated. Check answer[29]
    3 structures: a) a 5 carbon chain with a double bond at C2 and C4; b) a 5 carbon chain with a double bond at C2 and C3; c) a 5 carbon chain with a double bond at C1 and C4
    (Credit: Map: Organic Chemistry (Wade), Complete and Semesters I and II, CC BY-NC-SA 4.0).
  3. A solution of analyte, with molar absorptivity of 676 cm-1 M-1, is placed in a sample cell that has a pathlength of 1.00 cm. At a wavelength of 490 nm, the solution’s absorbance is 0.228. What is the analyte’s concentration? Check answer[30]
  4. A reaction shown below. While the two starting materials are only slightly coloured, the product is an intense orange-red. Account for this observation.
    A chemical reaction where the two starting materials are only slightly coloured and produces a product that is an intense orange-red
    (Credit: Organic Chemistry with a Biological Emphasis Volume I, (page 232) CC BY-NC-SA 4.0)
  5. Which would be more useful in distinguishing the two compounds shown below. IR or UV spectroscopy? Explain.  Check answer[31]
    Two structures representing a 9 carbon ketone with multiple double bonds. On the left shows 2-nonanone with double bonds at C3, C5 and C7. On the right shows 2-nonanone with double bonds at C4 and C7.
    (Credit: Organic Chemistry with a Biological Emphasis Volume I, (page 232) CC BY-NC-SA 4.0)
  6. Which analytical technique – IR, UV, or MS – could best be used to distinguish between the two compounds below? Explain. Check answer[32]
Two structures. On the left is 5-nonanone. On the right is 4-nonanone.
(Credit: Organic Chemistry with a Biological Emphasis Volume I, (page 232) CC BY-NC-SA 4.0)

Attribution & References

Except where otherwise noted, this material (including the images in solutions) has been adapted by Samantha Sullivan Sauer from:

29.1 to section 29.5 review questions

  • written by Samantha Sullivan Sauer, shared under CC BY-NC 4.0

29.6 review questions

29.7

28.8

  • Q1-2 written by Samantha Sullivan Sauer, shared under CC BY-NC 4.0

29.9

29.10

29.11

  1. Chromatography separates a mixture into it's dissolved components. ↵
  2. The stationary phase (typically a solid) does not move and holds onto the mixture and it's components. The mobile phase moves (typically a gas or liquid) and drags the mixture's components with it at varying rates. ↵
  3. The rate of travel depends on the components affinity for the stationary phase over the mobile phase. More affinity to the stationary phase means the component will travel slower. ↵
  4. Compound A Rf = 0.36; Compound B Rf = 0.40; Compound C Rf = 0.69; Compound D Rf = 0.97.  Compound D has high affinity to the mobile phase and low affinity for the stationary phase.  Compounds A and B have similar affinities for the stationary phase. ↵
  5. All columns require a stationary phase and a mobile phase of different properties to separate the applied mixture. ↵
  6. Columns allow for collection of the mixture's components whereas TLC/PC are only visual and can't capture the components. PC/TLC are typically much quicker and cheaper to perform. ↵
  7. To determine the structure of a compound. ↵
  8. The pattern in which matter absorbs or emits radiation. ↵
  9. A) carbonyl group in aldehydes, ketones, carboxylic acids, amides, esters B) aromatics, amines, nitro, C) carboxylic acids ↵
  10. 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 an 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. ↵
  11. 1680 cm-1 for carbonyl group in carboxylic acid, 2820 cm-1 for OH in carboxylic acid, and 2925 cm-1 for CH in aromatics. ↵
  12. Strategy: Identify the functional groups in each molecule, and then check an IR table of values. (a) Absorptions: 3400 to 3650 cm–1 (O–H), 3020 to 3100 cm–1 (=C–H), 1640 to 1680 cm–1 (C═C). This molecule has an alcohol O–H group and an alkene double bond. (b) Absorptions: 3300 cm–1 (≡C–H), 2100 to 2260 cm–1 (C≡C), 1735 cm–1 (C═O). This molecule has a terminal alkyne triple bond and a saturated ester carbonyl group. ↵
  13. Strategy: All IR spectra have many absorptions, but those useful for identifying specific functional groups are usually found in the region from 1500 cm–1 to 3300 cm–1. Pay particular attention to the carbonyl region (1670 to 1780 cm–1), the aromatic region (1660 to 2000 cm–1), the triple-bond region (2000 to 2500 cm–1), and the C–H region (2500 to 3500 cm–1). Solution: The spectrum shows an intense absorption at 1725 cm–1 due to a carbonyl group (perhaps an aldehyde, –CHO), a series of weak absorptions from 1800 to 2000 cm–1 characteristic of aromatic compounds, and a C–H absorption near 3030 cm–1, also characteristic of aromatic compounds. In fact, the compound is phenylacetaldehyde.
    Chemical structure structure of phenylacetaldehyde. ↵
  14. (a) 1715, 1640, 1250 cm−1 (b) 1730, 2100, 3300 cm−1 (c) 1720, 2500–3100, 3400–3650 cm−1 ↵
  15. mass, formula and structure of compound ↵
  16. Strategy: Look at the possible structures and decide on how they differ. Then think about how any of these differences in structure might give rise to differences in mass spectra. Methyl cyclohexane, for instance, has a –CH3 group, and ethylcyclopentane has a –CH2CH3 group, which should affect the fragmentation patterns. Solution: Both mass spectra show molecular ions at M+ = 98, corresponding to C7H14, but they differ in their fragmentation patterns. Sample A has its base peak at m/z = 69, corresponding to the loss of a CH2CH3 group (29 mass units), but B has a rather small peak at m/z = 69. Sample B shows a base peak at m/z = 83, corresponding to the loss of a CH3 group (15 mass units), but sample A has only a small peak at m/z = 83. We can therefore be reasonably certain that A is ethylcyclopentane and B is methylcyclohexane. ↵
  17. C19H28O2 ↵
  18. The position on the plot at which the nuclei absorbs. ↵
  19. It aligns the nuclear spins of the atoms. ↵
  20. A. 3; B. 3; C. 5; D. 7 ↵
  21. Strategy: Identify the types of hydrogens in the molecule, and note whether each is alkyl, vinylic, or next to an electronegative atom. Then predict where each absorbs.  Solution: The –OCH3 protons absorb around 3.5 to 4.0 δ because they are on carbon bonded to oxygen. The (CH3)3C– protons absorb near 1.0 δ because they are typical alkane-like protons. ↵
  22. (a) 1.43 δ (b) 2.17 δ (c) 7.37 δ (d) 5.30 δ (e) 9.70 δ (f) 2.12 δ ↵
  23. Two peaks; 3 : 2 ratio ↵
  24. (a)  −CHBr2, quartet;  −CH3, doublet (b) CH3O−, singlet;  −OCH2 −, triplet;  −CH2Br, triplet (c) ClCH2− , triplet;  −CH2−, quintet (d) CH3− , triplet;  −CH2− , quartet;   −CH− , septet; (CH3)2, doublet (e) CH3−, triplet;  −CH2−, quartet;   −CH−, septet; (CH3)2, doublet (f) =CH, triplet,  −CH2−, doublet, aromatic C−H, two multiplets ↵
  25. Strategy: Identify the distinct carbons in the molecule, and note whether each is alkyl, vinylic, aromatic, or in a carbonyl group. Then predict where each absorbs. Solution: Ethyl acrylate has five chemically distinct carbons: two different C=C, one C=O, one O–C, and one alkyl C. The likely absorptions are:
    The structure of ethyl acrylate. Delta values of carbon atoms are C 2 of ethyl: 15, C 1 of ethyl: 60, C 1, 180, and C double bond C: 130.The actual absorptions are at 14.1, 60.5, 128.5, 130.3, and 166.0 δ. ↵
  26. −CH3, 9.3 δ;  −CH2− , 27.6 δ; C=O, 174.6 δ;− OCH3, 51.4 δ ↵
  27. (a) 4 (b) 7 (c) 4 (d) 5 (e) 5 (f) 7 ↵
  28. conjugated double bonds are alternating carbon-carbon double bonds separated by carbon-carbon single bonds. ↵
  29. a) conjugated, b) cumulated, c) isolated ↵
  30. 3.37 x 10-4 M ↵
  31. UV because the molecule on the left has conjugation whereas the molecule on the right does not. Their IR spectra would be very similar as their functional groups are similar. ↵
  32. MS because both molecules have the same functional groups and neither have conjugation. The MS spectra will show different fragments due to the location of the carbonyl group. ↵

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Organic and Biochemistry Supplement to Enhanced Introductory College Chemistry Copyright © 2024 by Gregory Anderson; Caryn Fahey; Adrienne Richards; Samantha Sullivan Sauer; David Wegman; and Jen Booth is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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