Lecture 9
Aromatics: Long-Range Coupling
H’s on aromatic rings may couple with non-neighboring protons due to
long-range coupling. You will see this in lab! Why? Nuclei “communicate”
via bonding electrons - p electrons that are in resonance will allow
non-neighboring H’s to “communicate” and couple/split. This leads to
complex splitting.
Nuclear Magnetic Resonance
Use: To assist in the elucidation of a molecule’s structure
Information Gained:
• Different chemical environments of nuclei being analyzed (1H nuclei):
chemical shift
• The number of nuclei with different chemical environments: number of
signals in spectrum
• The numbers of protons with the same chemical environment:
integration
• Determine how many protons are bonded to the same carbon:
integration
• Determine the number of protons that are adjacent to one another:
splitting patterns
• Determine which protons are adjacent to one another: coupling
constants
Integration
• Area underneath signal; NMR machine will give integrals
• First, gives the relative ratio of different types of protons in compound
• Second, allows determination of actual ratio of different types of protons
1. Measure the length of the integral with a ruler
2. Establish a relative ratio of protons (divide each length by
the lowest number)
Coupling Constants (J)
Protons that split each other’s peaks will have the same coupling constant
or J value.
Problem: LG 15.2
Draw the structure of a compound that fits each molecular formula and
has a 1H NMR spectrum showing a single peak (a singlet). (Hint: Consider
HDI).
Single peak must mean equivalent Hs! (Only one peak and no splitting.)
(a) C2H6O
(c) C4H6
(f) C3H6O
(g) C4H9Br
Problem: LG 21.29 (a-d)
Steps to solve problem:
1. Calculate HDI
2. Draw out possibilities of structure
3. Determine how many signals would be seen for each possibility & their
splitting patterns
4. Work with the NMR data:
(a) Count # of signals = the # of different types of H’s
(b) Look at the splitting of each signal
(c) Consider the ppm values for each signal - use correlation chart
(d) Look at the integration values
LG 21.29(a)
C4H10O
1.28 ppm (s, 9H)
4.5 ppm (s, 1H)
LG 21.29(b)
C3H7Br
1.71 ppm (d, 6H)
4.32 ppm (m, 1H)
LG 21.29(c)
C4H9Cl
1.04 ppm (d, 6H)
1.95 ppm (m, 1H)
3.35 ppm (d, 2H)
LG 21.29(d)
C8H10
1.25 ppm (t, 3H)
2.68 ppm (q, 2H)
7.23 ppm (m, 5H)
LG 21.31
The 1H NMR spectrum of a compound C3H3Cl5 shows peaks at 4.5 ppm (t, 1H)
and 6.0 ppm (d, 2H). What is the compound’s structure?
IR Correlation Chart
Correlation of Bond Stretching and IR Absorption (See also Correlation Chart & Table in Lab
Guide)
Wavenumber Range (cm-1)
Type of Bond
Group
Family of Compounds
Single Bonds
—C—H
Alkanes
2850-3300
=C—H
Alkenes, aromatics
3000-3100
ºC—H
Alkynes
3300-3320
O—H
Alcohols
3200-3600
N—H
Amines
3300-3500
C—O
Ethers, Esters, Alcohols
Carboxylic Acids
1330-1000
C=C
Alkenes, aromatics
1600-1680
C=O
Carbonyls
1680-1750
Aldehydes, ketones
1710-1750
Carboxylic acids
1700-1725
Esters, amides
1680-1750
C=N
Imines
1500-1650
CºC
Alkynes
2100-2200
CºN
Nitriles
2200-2300
Double Bonds
Triple Bonds
1H
NMR Correlation Chart
Antioxidants & Chocolate
Antioxidants:
Protect against cardiovascular disease, cancer and cataracts
Thought to slow the effects of aging
Chocolate:
High levels of antioxidants - complex mixtures of phenolic
compounds
By weight, has higher concentration of antioxidants than red
wine or green tea
20x higher concentration of antioxidants than tomatoes
Dark chocolate has more than 2x the level of
antioxidants as milk chocolate.
Side note: The main fatty acid in chocolate, stearic
acid, does not appear to raise blood cholesterol
levels the way other saturated fatty acids do.