37 how many double bond equivalents does compound b

37. How many double bond equivalents does Compound B contain? D
(a) 0 (b) 1 (c) 2 (d) 3 (e) 4 (f) 5 (g) 6 (h) 7 (i) 8 (j) 9
IR : 3420 (broad), 3305 (sharp), 2147, 1687 cm
-1
Note : the singlet at δ 9.1 ppm disappears on addition of D
2
O and this resonance becomes a
triplet at low temperature.
1
H NMR
(300 MHz) :
DEPT-135 Spectrum
13
C{
1
H} NMR

Chem 213 Final Examination 2014 CRN10488 Page 10 of 17
The NOESY AND COSY spectra of compound C are shown on the next two pages. Make
use of NOESY AND COSY spectra, chemical shift values and multiplicity to decide which
isomer of C is the correct one. Enter the number of the structure you have chosen on the
yellow answer sheet. [12 marks]
ANSWER CHOICES:
H
CH
3
H
H
3
C
OCH
2
CH
3
OCH
3
Cl
H
H
CH
3
H
H
3
C
OCH
3
OCH
2
CH
3
Cl
H
H
3
C
H
H
H
3
C
OCH
2
CH
3
OCH
3
Cl
H
H
3
C
H
H
H
3
C
OCH
3
OCH
2
CH
3
Cl
H
H
CH
3
H
H
3
C
OCH
2
CH
3
OCH
3
H
Cl
H
CH
3
H
H
3
C
OCH
3
OCH
2
CH
3
H
Cl
H
3
C
H
H
H
3
C
OCH
2
CH
3
OCH
3
H
Cl
H
3
C
H
H
H
3
C
OCH
3
OCH
2
CH
3
H
Cl
H
OCH
3
H
H
3
C
OCH
2
CH
3
CH
3
Cl
H
H
OCH
3
H
H
3
C
CH
3
OCH
2
CH
3
Cl
H
H
3
CO
H
H
H
3
C
OCH
2
CH
3
CH
3
Cl
H
H
3
CO
H
H
H
3
C
CH
3
OCH
2
CH
3
Cl
H
H
OCH
3
H
H
3
C
OCH
2
CH
3
CH
3
H
Cl
H
OCH
3
H
H
3
C
CH
3
OCH
2
CH
3
H
Cl
H
OCH
3
CH
3
H
H
CH
3
H
3
CH
2
CO
Cl
H
3
CO
CH
3
OCH
2
CH
3
H
H
CH
3
H
Cl
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16

Chem 213 Final Examination 2014 CRN10488 Page 11 of 17
NOESY SPECTRUM OF C (note COSY correlations have been removed and resonances
are shown as a simplified line representation).

Chem 213 Final Examination 2014 CRN10488 Page 12 of 17
COSY SPECTRUM OF C:

Chem 213 Final Examination 2014 CRN10488 Page 13 of 17
COMPOUND D is a widely used halogenated (containing F, Cl, Br OR I) defoliant found in
Agent Purple. This compound has a molecular ion at m/z = 310 mass units (100% relative
abundance) with additional isotopic peaks at 311 (M+1, 13%), 312 (M+2, 96%), 313 (M+3,
13%), 314 (M+4, 31%), 315 (M+5, 4%), 316 (M+6, 3%) . The fragment containing no
halogens has an exact mass of 205.085 ± 0.004 . Use this information and the following table to
determine the molecular formula of Compound D and enter this on the yellow sheet in the
space provided. [4 pts]
C H N O Exact Mass C H N O Exact Mass
10 9 2 3 205.061 12 17 2 1 205.134
10 11 3 2 205.085 12 19 3 0 205.158
10 13 4 1 205.109 13 17 0 2 205.123
10 15 5 0 205.133 13 19 1 1 205.147
11 9 0 4 205.050 13 21 2 0 205.170
11 11 1 3 205.074 14 9 2 0 205.077
11 13 2 2 205.098 14 21 0 1 205.159
11 15 3 1 205.122 14 23 1 0 205.183
11 17 4 0 205.145 15 9 0 1 205.065
12 13 0 3 205.086 15 11 1 0 205.089
12 15 1 2 205.110 15 25 0 0 205.196
16 13 0 0 205.102
IR: 1735 and 1225 cm
-1
1
H NMR
[500 MHz]:
13
C{
1
H} NMR (multiplicity in the proton-coupled
13
C spectrum indicated): δ 169.6(s), 154.7(s),
131.0(d), 129.8(s), 126.3(s), 122.7(s), 116.8(d), 66.5(t), 64.8(t), 30.8(t), 19.1(t), 13.8(q) .
Selected long range (J = 10 Hz) HMQC: δ 7.4 (s) with 154.7 (s) and 129.8 (s)
δ 5.0 (s) with only 154.7 (s)
δ 4.1 (t) with 169.6 (s) and 19.1 (t)
Enter your structure for Compound D on the yellow answer sheets. [10 pts]

Chem 213 Final Examination 2014 CRN10488 Page 14 of 17
COMPOUND E has the formula C
14
H
18
O
3
and is related to antioxidants and flavour
compounds found in many plants including corn, peanuts and cocoa beans. Use the IR and NMR
data below to determine the structure of E and enter it on the yellow answer. [14 pts]
HMQC:

Chem 213 Final Examination 2014 CRN10488 Page 15 of 17