Chapter 13

Question 1

Nuclear Magnetic Resonance (NMR) Spectroscpoy

Answer 1

a stectroscopic technique that gives information about the number and types of atoms in a molecule

Question 2

nuclear spin states

Answer 2

• an electron has a spin quantum number of 1/2 (can be +1/2 or -1/2)
• a moving charge has an associated magnetic field
• nuclei need multiple spin states to be NMR active, which include those with an odd mass and /or atomic number

Question 3

orientation of nuclear spin in an applied magnetic field

Answer 3

• placing nuclei in a strong magnetic field aligns their molecular spin with (+1/2 = lower energy) and against (-1/2 =higher energy) the field
• the difference in energy between these states is on the order of radio waves and is of the magnet
• stronger magnets have greater sensitivity and require higher radio frequency

Question 4

nuclear magnetic resonance

Answer 4

• irradiating the nuclei with a radio frequency equal to the difference in energy between the spin states excites the spin from +1/2 to -1/2
• upon return to the ground state, the nuclei emit radio waves equal in energy to that absorbed, which is measured and recorded as signal
• a nucleus is in resonance when absorption occurs
• an applied magnetic field causes electrons to circulate, and the orientation of this field dictates the direction of circulation

Question 5

deshielding

Answer 5

electron density circulation adds to the applied magnetic field, thus requiring higher energy to cause resonance

Question 6

chemical shift

Answer 6

the difference in resonance frequencies caused by shielding/deshielding of nuclei

Question 7

NMR Spectrometer

Answer 7

-NMR samples are prepared by dissolving the compound of interest in a deuterated solvent, to avoid H resonances from the solvent that would saturate the signal

Question 8

equivalent hydrogens

Answer 8

hydrogens that have the same chemical environment, and thus show up as a single signal in H-NMR

Question 9

rules to determine equivalency

Answer 9

1. bonded to the same sp3 hybridized carbon that can rotate freely
2. a plane or point of symmetry in a molecule exists and hydrogens equidistant from the plane or point are the same

Question 10

signal area

Answer 10

• the area under the signal is proportional to the number of equivalent hydrogens giving rise to that signal
• integrating the peaks in a H-NMR spectrum tells you the relative number of hydrogens represented by each peak

Question 11

chemical shift

Answer 11

• incredibly valuable in determining structure since the “electronic environment” around the H influences its chemical shift
• influenced by electronegativity, hybridization of nearby atoms, and magnetic induction within an adjacent pi bond

Question 12

electronegativity of nearby atoms

Answer 12

-EN atoms pull electron density away from atoms bonded to it, thus deshielding them, and causing a downfield shift in resonance

Question 13

hybridization of adjacent atoms

Answer 13

-more s character for sp2 and sp hybridized carbon than sp3, making it more EN

Question 14

diamagnetic effects from pi-Bonds

Answer 14

-shielding vs. E.N=> “deshielding”

Question 15

ring current

Answer 15

circulation of pi-electrons on an aromatic ring under an applied magnetic field
-this current induces a strong magnetic field that opposes the applied field in the middle of the ring, but reinforces it outside the ring

Question 16

signal splitting

Answer 16

spin-spin coupling with adjacent nuclei split NMR signals depending on the extent of coupling and the number of adjacent nuclei

Question 17

(n+1) rule

Answer 17

if a hydrogen has n hydrogens nonequivalent to it, but equivalent among themselves on the same or adjacent atoms, its H-NMR signal is split into (n=1) peaks
-for molecules that can freely rotate, sp3 hybridized, the rule simplifies to m+n+1 peaks

Question 18

vicinal coupling

Answer 18

spin-spin coupling involving the H atoms on two C atoms that are bonded to each other

Question 19

germinal coupling

Answer 19

spin-spin coupling that occurs between nonequivalent H atoms bonded to the same C atom (often due to restricted bond rotation, such as on a terminal double bond

Question 20

13C-NMR

Answer 20

• 13C-NMR active, and 13C has a natural abundance of 1%
• (n+1) rule applies, but splitting is not observed due to the nearly negligible probability of two 13C isotopes being adjacent to each other in a molecule
• 1H will split 13C signals resulting in complex splitting patterns, which is why H signals are decoupled and no splitting is observed, resulting in relative clear spectra