http://www.chem.hmc.edu/chem/nmr/page7.htm
Problem 1
C3H5BrO2
MW = 152
1 H (s) broad > 10 probably COOH
1 double bond/ring
Cl
OH
H3C
d
or
H
q
O
H2
C
Cl
C
H2
t
OH
t
O
Problem 3
C9H12
MW = 120 C9H20 – C9H12 = 4 rings/double bonds – think aromatic if protons
show up at 7.2-7.8
Possible?
With an aromatic system, you can determine the number of substitutions
by counting the number of aromatic H’s = 5 so it is mono-substituted.
What about the side chain? Subtract C6H5 from the formula of the
starting material: C9H12 - C6H5 = C3H7
Two possible side chains:
1 carbon with two methyl groups branching off if it: 1 m @ 2.5 (1 H) , 1
large d @ 1.8 (6 H)
In a linear system, the splitting pattern would be: 3 signals: 1 t @ 2.8 (2
H), 1 m @ 1.8 (2 H), 1 t @ 1.1 (3 H)
What do you see?
Problem 7
C6H10 C6H14- C6H10 = H4/2 = 2 rings or double bonds
With only three signals, this molecule is HIGHLY symmetrical.
What does this mean? The peak at 5.7 is a single peak and the only electronegative
thing that could be on that carbon is a double bond, hence the shift. As you more
away from the C=C, the shift decreases. Why aren’t the shifts at 2 ppm/1.6 ppm
sharper? It could be a transition that they undergo which tends to flatten out the
multiplets. Possible structures?
My guess? Cyclohexene.
Problem 14
C8H6O2 C8H18 - C8H6 = H12/2 = 6 rings or double bonds
Also highly symmetrical with only 2 signals
The easy part of this is the sharp peak at 10. Only a few FG’s show up at
10 ppm. Also, 4 H around 8 ppm would tell me an aromatic group with
something pulling the electrons out of the ring, deshielding the ring and
letting the H signal be shifted.
What do you think?
My guess? A aromatic o or p dialdehyde.
Problem 27
C8H19N – No rings or double bonds.
5 signals with an H exchange. Also pretty symmetrical!
a 4H t @ 2.5 is probably attached to the N, which contains the H @ 1.1
The next 2 4H m @ 1.4 and 1.3 respectfully are 2 and 3 carbons away
from N and the 6H t @ 0.9 are two methyl groups at the end of the
chain. Dibutyl amine!
Problem 29
C3H7OCl no rings or double bonds
Not too tough! The 3 H singlet has to be a methyl group attached to an
O – we have some sort of ether:
C3H7OCl - CH3O = C2H4Cl
Cl will bond like H so it must be on one of the two possible positions: on
the C attached to the O or the adjacent carbon. Since we have 4 H in
very close shift, the Cl must be on the adjacent carbon.
Problem 30
C6H6O2 - C6H14 = H8/2 = 4 rings or double bonds.
Also symmetrical with only three signals. Aromatic? I don’t think so.
6.6 – 6.8 ppm is too low for aromatic H stretches but not for alkene
stretches which are close to the aromatic region.
The splitting pattern that
shows up here is a very common type of pattern that you see for alkenes.
The sharp coupling is known as J coupling and it occurs between the H’s
on two ends of a double bond. The s 2H @ 9.8 is an aldehyde proton
stretch. O=C-C=C-C=C-C=O
Problem 33
C6H4Cl2 This is an aromatic spectra with two symmetrical “alkene” type
stretches. At first blush, it is p-dichlorobenzene.
Problem 46 C5H12O2 no rings or double bonds Also lots of symmetry!
This tells me there are 2 sets of methyl groups, two of them with O as
ethers (shift to 3.2-3.6 ppm) and two @ 1.35 ppm.
C
C-C-OC
OC
Problem 50
C3H7OCl No rings or double bonds
4 signals? 4 different H environments so this would be a linear alcohol
(broad H @ 1.5) with a Cl on the terminal C Cl – C-C-C-O-H
t m t b.singlet