2120 CHM Assignment 6 ANSWERS 1. Associate each of the following IR spectra with one of the following compounds and justify your answer. a. Propanoic acid: look for a carbonyl stretch and a broad OH stretch b. 2-Pentanol: look for an OH peak (broad). No carbonyl peak, no sp2 CHs. c. Benzyl alcohol; Look for an OH peak and sp2 CHs. No carbonyl peak. d. Acetophenone: Look for a carbonyl peak and sp2 CHs. No OH peak. Note: the key functional groups must be identified. Identifying the peaks on the spectrum directly is an straightforward way to do this.
sp3 CHs OH
sp3 CHs OH sp2 CHs
Assignment 6 NMR, IR ANSWERS
sp2 CHs
C=O
OH sp3 CHs
C=O
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Assignment 6 NMR, IR ANSWERS 2. Associate each of the following IR spectra with one of the following compounds and justify your answer. a. 2-Propyn-1-ol: Look for an alkyne peak CC plus an OH peak (broad) and a CH. b. 1-Pentyne: Look for an alkyne peak CC plus CH. No other key groups. c. 4-Methylpentanenitrile: Look for a CN. No OH, C=O or CH. d. p-Acetylbenzonitrile: Look for C=O, sp2 CHs and CN. No CH. Note: the key functional groups must be identified. Identifying the peaks on the spectrum directly is an straightforward way to do this.
CN
sp3 CHs
CN C=O
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Assignment 6 NMR, IR ANSWERS
CC sp3 CHs
OH CC-H
CC
sp3 CHs CC-H
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Assignment 6 NMR, IR ANSWERS 3. Give the number of peaks expected for the indicated protons in each of the following structures: Note: s = singlet, d = doublet, t = triplet, q = quartet
a) H3 C 2 (d) d) H 3C 1 (s) H H 2 (d) O CH3 1 (s) H 3C H2 C O H 1 (s) Note: usually a broad singlet C H2 2 (d) e) O CH3 3 (t) b) H 3C 3 (t) 5 (quintet) 4 (q) f) H2 N CO2 H C H2 O Cl H 2 (d) c) H O CH 3
H CH3 4 (q) 2 (d)
4. How many signals would be expected in the following molecules?
a) H3 C CH3 CH C H2 5 signals d) H 3C HB HA O 4 signals HB HA CH3 H 3C e) H2 C H C H2 O CH3 b) H3 C O CH3 c) Cl H O CH 3 H 3 signals f) O H 3C C H2 2 signals H2 C O CH3
1 signal
O
3 signals
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Assignment 6 NMR, IR ANSWERS 5. Explain the following differences in chemical shift.
Me a) Me O H H H H Me O Me The "H" at 7.7 ppm is deshielded by resonance. Another way to put this is that resonance with the carbonyl group removes electron density from the proton.
= 7.1 = 7.7 ppm
= 6.7 ppm b) H O H O The "H" at 6.7 ppm is strongly deshielded by resonance. The proton at 6.1 ppm is deshielded only by proximity to the electron-withdrawing carbonyl group. This inductive effect is less powerf ul than the resonance ef fect.
H = 6.1 ppm
H
c)
O O
H2 C
= 4.41 ppm C H2 O CH 3
O O
H2 C
C H2
O
CH 3
= 3.6 ppm The oxygen of the ester is even more electronegative than the oxygen of the ether because resonance in the ester is strongly drawing electron density to the carbonyl oxygen. Electron density is consequently pulled more strongly away from the H's at 4.41 ppm. d) Me N = 5.6 ppm = 4.2 ppm Me N
H H
H H
Resonance with the nitrogen pushes electron density onto the H at 4.2 ppm, effectively "shielding" it with electrons.
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Assignment 6 NMR, IR ANSWERS 6. Determine the structure of the following unknowns using the table to fill in your answers. a.
Signal
~4.3 ~2.0 ~1.3
Integration
1H 3H 6H
Multiplicity
m (multiplet) or septet s d
Comments
CH next to O (based on ) CH3 next to C=O 2 x CH3 (identical) next to CH (i.e. an isopropyl group)
A B C
DU = [5(2) + 2 10]/2 = 1 Integration: (6 mm + 18 + 34)/10H = 5.8 mm/H
The pieces: CH3 CH H3 C O ? ? O CH 3
The molecule: CH3 CH H3 C O CH 3 O C 5H 10 O2
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Assignment 6 NMR, IR ANSWERS b.
A
B C
D
E
Signal
~7.8
Integration
2H (CH2 or 2 x CH)
Multiplicity
d
Comments
Not all the 2Hs can be CH2 or there wouldnt be enough carbons accounted for in the molecule (there are 10 carbons in the molecule); These Hs are in the aromatic regionthey cant be CH2s only CH group exist on aromatic rings. Each CH is next to another CH
H X H H H Y
A
plane of symmetry The red H has one neighbor (n = 1) and is therefore a doublet.
B C D E
~7.2 ~2.7
2H (CH2 or 2 x CH) 2H (CH2 or 2 x CH) 3H 3H
d q
See explanation above next CH2 to CH3 (b/c the peak is a quartet); not directly next to O, could be next to C=O (based on ) CH3 next to C=O (based on ) or on phenyl ring (which acts like an electron-withdrawing group) CH3 next to CH2
~2.6 ~1.2
s t
DU = [10(2) + 2 -12]/2 = 5 HINT: DU of 4 = phenyl ring In this molecule, there is probably a phenyl ring (DU = 4) + one ring or one double bond Integration: (11 mm + 11 + 11 + 17 + 17)/12Hs = 5.6 mm/H
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Assignment 6 NMR, IR ANSWERS
H The pieces: D ? CH3 ? O C ? A X H H B H ? Y C H2 C E CH 3
? = carbonyl or phenyl
The molecule: O C H H B H H OR C H2 CH3 H 3C C H2 A O C H H B H H CH3
H3 C A
Note: Carbonyl group should be placed at "X". We can tell this because H A is the most deshielded of the aromatic protons (by resonance with the carbonyl group).
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Assignment 6 NMR, IR ANSWERS c.
B D
E
A C
Signal
~3.4 ~3.3 ~1.8 ~1.3 ~0.9
Integration
2H 3H 1H 4H 6H
Multiplicity
d s m m t
Comments
CH2 (probably not 2 x CH, but keep this in mind, just in case), next to the CH, next to O (based on ) CH3 isolated, next to O (based on ) CH multiplet (hard to say what the neighbors are) 2 x CH2 (identicalsymmetry exists in the molecule); hard to say what the neighbors are 2 x CH3, identical in chemical environment, each next to CH2 (also identical)
A B C D E
DU = [ 7(2) + 2 16 ]/2 = 0
no rings or double bonds
Integration: (12 mm + 17 + 6 + 23 + 34) / 16 Hs = 5.8 mm/H
T he pieces: ? ? O C H2 A Note: the formula contains only one oxygen E C D H2 ? CH 3 x 2 C 7 H16O CH C ? H3 C B B H 3C O A C H2 CH C H2 D O ? The molecule: D H2 C C CH3 E E CH3
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Assignment 6 NMR, IR ANSWERS
7.
Determine the most likely structure of a compound, with the molecular formula C9H12, which gave a 1H NMR spectrum consisting of: a doublet at 1.25 a septet at 2.90 and a multiplet at 7.25
8.
A compound with the molecular formula C10H13Cl gave the following 1H NMR spectrum: singlet, 1.6 singlet, 3.1 multiplet, 7.2 (5H) The most likely structure for the compound is:
Cl Cl Cl
]
I
II
Cl Cl
III
IV
V
9. Determine the likely structure for a compound A (C6H10O), which is found to decolorize bromine in carbon tetrachloride. Its spectral data is as follows: H NMR triplet, 1.0 singlet, 2.4 singlet, 1.4 singlet, 3.4 quartet, 1.6
OH
1
IR 2200 cm (sharp) 3300 cm-1 (sharp) 3500 cm-1 (broad)
-1
OH
OH
I
II
III
OH
OH
IV
V
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Assignment 6 NMR, IR ANSWERS 10. Briefly explain how you might distinguish between the following substances by comparing their 1H-NMR spectra:
O O O O
I II The main difference would likely be in the chemical shift of the methine proton of the isopropyl group. The methine proton in I is likely to be more deshielded, and produce a signal (septet) at about 3.5 ppm, while the analogous proton in II is likely to be found, also as a septet, at about 2.3 ppm. O O O O
I
~3.5 ppm
II ~2.3 ppm
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Assignment 6 NMR, IR ANSWERS 11. An unknown compound has the formula C6H12O. Elucidate the structure of the molecule by scrutinizing its IR, 1H NMR and 13C NMR spectra, shown below.
C =O
B A C D
E
Signal
~2.4 ~2.1 ~1.6 ~1.3 ~0.9
Integration
2H 3H 2H 2H 3H
Multiplicity
Triplet (t) Singlet (s) Multiplet (m) Multiplet (m) Triplet (t)
Comments
CH2 next to CH2 (based on multiplicity) and next to C=O (based on ) CH3 next to C=O (based on ) CH2 CH2 CH3 next to CH2 (based on multiplicity)
A B C D E
DU = [ 6(2) + 2 12]/2 = 1 C=O (based on IR) Integration: (9 mm + 13 + 10 + 10 + 14) / 12 Hs = 4.7 mm/H
The molecule: C O C H2 C E C H2 D CH3
H 3C B
C H2 A
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Assignment 6 NMR, IR ANSWERS An unknown compound, I, has the formula C3H7NO2. Elucidate the structure of I by scrutinizing its IR, 1H NMR and 13C NMR spectra, shown below.
N OT a C=O
C A B
Signal
~4.4 ~2.0 ~1.0
Integration
2H 2H 3H
Multiplicity
Triplet (t) M t
Comments
CH2, next to CH2 (based on multiplicity), next to O or NO2 (based on ) CH2 CH3, next to CH2
A B C
DU = [ 3(2) + 2 +1 7 ] /2 = 1
NOT a C=O based on IR maybe NO2?
Integration: (17 mm + 17 + 26 ) / 7Hs = 8.6 mm/H The molecule:
B H2 C O N
H3 C C
C H2 A
O
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