Chemistry 221 - Test 3 Review Sheet Answers
(Chap. 10-13)
For each spectrum below indicate (a) type of spectrum, and (b) what the x-axis measures.
a) mass spectrum
a) infrared spectrum
b) m/z (mass/charge)
b) cm–1 or micrometer
In the IR, each peak represents a different functional group (or its bond vibration).
In the MS, each peak represents a different cation fragment (or its mass).
Given the Molecular Formula, the Infrared Spectrum, and the Nuclear Magnetic Resonance Spectrum, what
would you look for to verify the presence of the following functional groups in a molecule.
O
OH
NH2
H
Which of these reagents (1-6) would react with each of these molecules (a-e)?
1) Br2 ; 2) HBr ; 3) H2, Lindlar Cat. ; 4) Mg ; 5) KOH ; 6) PtO 2 , H2
a) Alkane; b) Alkene; c) Alkyne; d) Alcohol; e) Alkyl Halide
What reaction conditions favor the E2 mechanism over the E1 mechanism?
What reaction conditions favor the Sn2 mechanism over the E2 mechanism?
How is the Isotope Effect different from the Element Effect? How are they similar?
Which technique(s) destroy the sample?
MS
IR
NMR
Which technique(s) require very low pressure?
MS
IR
Which technique(s) use magnets?
NMR
S
S
Br
H
MS
IR
NaOH
DMSO
OH
H
Sn1 or Sn2
H
R
Retention or Inversion or Racemization
OH
TosCl
pyridine
NMR
S
+
OTos
H
NaNH2
major
MgBr
MgH2
ether
Retention or Inversion or Racemization
Br
NaI
NaHS
I
SH
* Unsaturation # ≥ 2
* Shows up in the IR as a sharp peak near 2200 cm1-. A second sharp peak will be present
at about 3300 cm1- if the triple bond is at the end of the chain: C≡C–H.
Shows up in the 13C NMR between 65-85ppm.
Shows up in the 1H NMR around 2.7 ppm only if it is at end of chain: C≡C–H.
* Unsaturation # ≥ 4
Shows up in the IR as a sharp peak near 1600 cm1-.
Shows up in the 13C NMR as 2 or more signals between 110-140ppm.
* Shows up in the 1H NMR as one or more peaks near 7.3 ppm. The integration indicates
the number of branches on the chain. (5H is one branch; 4H is 2 branches)
Does not affect the unsaturation number.
OH
* Shows up in the IR as a broad peak near 3500 cm1-.
Does not show up in the 13C NMR.
* Shows up in the 1H NMR as a singlet with integration of 1. Chemical shift varies widely.
Unsaturation # ≥ 1
O
* Shows up in the IR as a strong peak near 1700 cm1-.
H
Shows up in the 13C NMR near 200 ppm. (A doublet if coupling is indicated.)
* Shows up in the 1H NMR around 9.8 ppm.
* Unsaturation # ≥ 2 (Higher # than accounted for by multiple bonds requires a ring.)
Shows up in the IR as a peak near 1650 cm1-. (C=C)
* Shows up in the 13C NMR as 2 signals (2C’s in only one C=C) between 110-140 ppm.
Shows up in the 1H NMR as one signal (C=C–H) near 6.0 ppm.
Does not affect the unsaturation number.
NH2
* Shows up in the IR as a forked peak near 3400 cm1-. For a secondary amine (only one H
on N) the peak would not be forked.
Does not show up in the 13C NMR.
* Shows up in the 1H NMR as a singlet with integration of 2. Chemical shift varies widely.
For a secondary amine there would be a singlet with integration of 1.
* indicates the most easily spotted information for each functional group.
Which reagents react?
a) Alkanes react with 1.
b) Alkenes react with 1, 2, and 6.
c) Alkynes react with 1, 2, 3, and 6.
d) Alcohols react with 2.
e) Alkyl Halides react with 4, and 5.
Can you draw examples of each of these reactions?
Br
Br2
a/1
Br
+
hν
(+ HBr)
Br
Br2
H2
Br
b/1
PtO2
b/4
Br
HBr
Br
+
major
b/2
Br
Br
Br2
Br
Br
or twice.
Br
HBr
2 HBr
c/2 Reacts once
or twice.
H2
Lindlar cat.
c/3 Reacts once
c/4 Reacts twice
HBr
.
H2
H2
PtO2
PtO2
Br
(+ H2O)
d/2
Br
Mg
MgBr
ether
e/4
Br
e/5
Br
2 Br2
c/1 Reacts once
OH
Br
KOH
(+ KBr +H2O)
Br
Br
Sn1 vs. Sn2 vs. E1 vs. E2 are summarized nicely on pages 407-408 of your textbook.
E2 vs. E1: Both favor 3°>2°>1° substrates. The E2 mechanism is influenced by the strength and concentration
of the base (Rate = k [base] [substrate]) while the E1 mechanismis not (Rate = [substrate]). So a high
concentration of a strong base should favor E2 over E1.
Sn2 vs. E2: The E2 mechanism favors 3°>2°>1° substrates while the Sn2 mechanism favors 1°>2°>3°
substrates. A more basic “nucleophile” (like OH–) will favor the E2 mechanism while a less basic
nucleophile (H2 O) will favor the Sn2 mechanism.
Interpreting Spectra:
Figure 8.51; Problem 8.46
C8H10O2
Unsaturation # = 4 (benzene confirmed in the NMR)
IR:
and
–OH
NMR:
C=C (benzene)
4 signals (benzene counted as one signal)
integration: 4 (a disubstituted benzene), 2 (a CH2), 3 (a CH3 ), and 1 (an OH). This accounts for
the 4 unsat.; the eight C; the ten H; and the OH. One O remains from the formula (–O–).
We have with 3 “middle” pieces (benzene, CH2 , and -O-) and 2 “end” pieces (OH and CH 3).
Possible Stuctures:
O
O
HO
OH
HO
reasonable
O
wrong coupling
O
OH
O
reasonable
OH
HO
wrong chemical shift
wrong chemical shift
O
wrong coupling
With additional information it would be possible to distinguish between the two reasonable answers.
Figure 8.52; Problem 8.46
C10H12O2
Unsaturation # = 5 (benzene confirmed in the NMR)
IR:
and
C=O
C=C (benzene)
NMR:
5 signals (benzene counted as one signal)
integration: 1 (a C-H way downfield indicating aldehyde); 4 (a disubstituted benzene),
2 (a CH2 ), 2 (a CH2), and 3 (a CH 3). This accounts for the 4 unsat.; the ten C; the
twelve H; and C=O. One O remains from the formula (–O–).
We have with 4 “middle” pieces (benzene, two CH2 , and -O-) and 2 “end” pieces (O=C-H and
CH3 ). To reduce the number of possible structures (about 12) consider the coupling. One
of the CH2 is a triplet and the CH3 is a triplet, so they are both connected to a CH 2 . There
is only one other CH2 (a multiplet), so both are connected to it.
Now we have with 2 “middle” pieces (benzene and -O-) and 2 “end” pieces (O=C-H and
CH2 CH2 CH3).
Possible Stuctures:
O
H
O
O
H
O
The two circled carbons would have very different chemical shifts. Which is correct?
Figure 8.53; Problem 8.46
C6H11O2 Cl
IR:
Unsaturation # = 1
C=O
NMR:
4 signals
integration: 1 (a C-H); 2 (a CH2 ), 2 (a CH2), and 6 (two CH3 ). This accounts for the 1 unsat.
(IR: C=O); the 6 C (5 with H above and the C=O); the eleven H; and C=O. One Cl (–Cl)
and one O remain from the formula (–O–).
We have with 5 “middle” pieces (C=O, CH, two CH2 , and -O-) and 3 “end” pieces (Cl and
two CH3 ). To reduce the number of possible structures consider the coupling. Both
of the CH2 are triplets and must be connected to each other (–CH 2CH2–). The two CH 3
are a doublet, so they are connected to the CH.
Now we have with 3 “middle” pieces (–CH2 CH2–, C=O, and -O-) and 2 “end” pieces (Cl and
CH3 CHCH3 ).
O
O
O
Use of more
extensive IR tables
O
Cl
O
Cl
O
Cl indicates that the
correct structure is
on the right.
wrong coupling
reasonable
reasonable
Cl
O
Cl
O
wrong chemical shift
O
Cl
O
wrong chemical shift
O
O
wrong coupling
Figure 8.54; Problem 8.46
C6H10O2
IR:
Unsaturation # = 2 (benzene confirmed in the NMR)
C=O and no C=C (Structure must have either two C=O or one C=O and one ring.)
NMR:
2 signals (suggests a symmetric structure)
integration: 2 (a CH2), 3 (a CH3 ). These numbers give the correct proportions but the actual
numbers are double. So the actual integration is 4 (two CH2) and 6 (two CH3 ).
This integration accounts for the 4 of the six C; and the ten H. If there are two C=O it
would account for the remaining two C; two O; and Unsat. # = 2. While this is the most
likely explanation of the spectra, let’s also consider the possibility that there is one C=O;
one –O–; one C without H; and one ring. This would also fit the formula, the
unsaturation #, and the integration.
These two possibilities would be easily distinguished with a 13C NMR spectrum which
would show whether there was only one signal for C without H (for two identical C=O)
or two different signals for C without H (a C=O and a different C no H).
Possible Stuctures:
with 2 C=O
O
wrong # signals
with C=O, –O–, C (no H), and ring
There are many possible structures with these
pieces. None of the possibilities has the
symmetry needed to have only two signals.
Some examples are shown below.
O
O
O
wrong # signals
O
O
O
O
O
reasonable
O
O
O
O
O
O
wrong coupling
O