Organic Chemistry With Biological Applications International Edition 2nd Edition by John E. McMurry – Test Bank
Chapter 11 – Structure Determination: Nuclear Magnetic Resonance Spectroscopy
1. Nuclear magnetic resonance spectroscopy provides information about a molecule’s:
a. conjugated pi electron system
b. size and formula.
c. carbon-hydrogen framework.
d. functional groups.
2. Which of the following would not produce nuclear magnetic resonance?
3. Explain why all protons in a molecule do not absorb rf energy at the same frequency.
ANS: All nuclei in molecules are surrounded by electron clouds. When a uniform external magnetic field is applied to a molecule, the circulating electron clouds set up tiny local magnetic fields of their own. These local magnetic fields act in opposition to the applied field, so that the effective field actually felt by a nucleus is a bit smaller than the applied field.
Beffective = Bapplied – Blocal
This effect is termed shielding. Each nucleus is shielded to a slightly different extent, so each unique kind of proton in a molecule resonates at a slightly different frequency and gives rise to a unique NMR signal.
The following questions pertain to the display of NMR spectra. Match a term to each description below. Place the letter of the term in the blank to the left of the description.
b. high-field or upfield side
e. low-field or downfield side
f. chemical shift
4. When looking at an NMR spectrum the right-hand part of the chart is the .
5. The exact place on the spectrum at which a nucleus absorbs is called its .
6. The calibration standard for and NMR is:
7. The NMR spectrum are calibrated using an arbitrary scale that is divided into units.
8. The vertical axis of spectrum displays the of the signal.
For each of the compounds below tell how many signals you would expect the molecule to have in its normal, broadband decoupled NMR spectra.
20. What is the ratio of peak areas upon integration of the spectrum for A, B, and C respectively?
21. a.) Treatment of tert-butyl alcohol with hydrogen chloride yields a mixture of tert-butyl chloride and 2- methylpropene. After chromatographic separation, how would you use 1HNMR to help you decide which was which?
ANS: 2-Methylpropene has two kinds of hydrogens. It has a vinylic absorption (4.5–6.5 *) representing two hydrogens and an unsplit signal (1.0–1.5 *) due to the six equivalent methyl hydrogens.
tert-Butyl chloride has only one kind of hydrogen, which results in one unsplit signal.
b.) How would the 13CNMR for the two compounds differ?
ANS: 2-Methylpropene would have three signals with the signals due to the alkene carbons being
futher downfield than any of the signals in tert-butyl chloride. tert-Butyl bromide would have
only two signals.