CHEMISTRY 2500: Organic Chemistry I
MIDTERM-1
Friday, January 30, 2015
Instructions:
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This exam paper has 5 pages, double sided (including the cover page and data sheet) and consists of 8 questions.
The exam is worth a total of 58 marks. Most of these marks are for explanation/showing your work rather than for reaching the
correct answer. Explain all of your answers fully using diagrams where appropriate (a picture really is worth a thousand words!).
Marks will be deducted for poorly drawn structures.
Remember to write your name on every page of the exam.
No calculators allowed. No other electronic devices can be present with you during the exam unless authorized by the instructor.
You may use a molecular model kit.
There is a 2-hour time limit.
If your work is not legible, it will be given a mark of zero.
Read the questions carefully. Good luck.
Confidentiality Agreement:
I agree not to discuss (or in any other way divulge) the contents of this exam until they have all been marked
and returned. I understand that, if I were to break this agreement, I would be choosing to commit academic
misconduct, a serious offense that will be punished. The minimum punishment would be a mark of 0 on this
exam and removal of the “overwrite midterm mark with final exam mark” option for my grade in this course;
the maximum punishment would include expulsion from this university.
Signature: _______________________________
Course: CHEM 2500 (Organic Chemistry I)
Semester: Spring 2015
The University of Lethbridge
NAME(Print):
Date: _________________________
ID #:
Question Breakdown
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
/4
/2
/4
/10
/9
/8
/10
/11
Total
/58
1.
[4 marks]
(a) The following molecule has been named incorrectly, however, the structure of the molecule can still be
deduced.
2-chloro-4-nitro-5-hydroxy-1-isopropylcyclohex-1-ene
OH
NO 2
Cl
(b) Based on the structure above, give the correct name of this molecule according to IUPAC rules.
4-chloro-3-isopropyl-6-nitrocyclohex-3-en-1-ol
(note that the 1 is not necessary for the alcohol)
2. Name the following molecule according to IUPAC rules. Note, stereochemistry does NOT have to be
assigned.
[2 marks]
Cl
F
3.
(a)
F
2-chloro-1,5-difluoro-4-methylpentane
[4 marks]
Draw R-2-methoxypropanoic acid using line-bond notation.
O
OH
O
(b)
Draw the enantiomer of R-2-methoxypropanioc acid.
O
OH
O
4. Using the appropriate letter or letters, indicate the relationship(s) between the following pairs of molecules.
If there is more than one relationship, provide all the letters that apply. No explanation is necessary.
[10 marks]
A = stereoisomers
E = enantiomers
B = constitutional isomers
F = the same molecule
C = conformers
G = none of the above
D = diastereomers
relationship(s)
H
H
F
H
and
F
Cl
Cl
Cl
F
H
H
and
Cl
A,E
H
HO
H
H
OH
H
and
O
Br
CO2Me
and
H
A,D
B
O
and
A,E
Br
CO2Me
5.
[10 marks]
(a) Given the formula CH3COCHN2, use the implied connectivity to draw its fully expanded Lewis structure.
Add formal charges where appropriate.
O
N
H
C
H
N
C
C
H
H
(b) Based on your structure in part (a), draw two other resonance structures (you should have a total of 3
structures, the original and two new resonance structures). Use curly arrows to show the electron movement
converting one structure into another.
O
H
O
N
H
C
N
C
C
H
H
O
N
H
H
A
C
N
C
C
H
H
B
N
H
H
C
N
C
C
H
H
C
(c) Rank these resonance structures in terms of their contribution to the character of this species and indicate
degenerate structures, if any. Explain your reasoning.
C
>
A
>
B
All structures obey the octet rule and all structures have formal charges. The only
difference between the resonance structures is where the formal negative charge is
placed. C is best because the negative charge is on the more electronegative oxygen atom.
A is better than B because the negative charge is on N which is more electronegative than
C.
6. Using only the formula C4H7O2Cl, give an example that exemplifies each of the following. All structures
must be drawn in proper line-bond format. Note, your answers may contain more than one functional
group.
[8 marks]
**The following represent one possible answer**
(a) a carboxylic acid
Cl
O
OH
(b) an acid chloride
OH
O
Cl
(c) an ester
O
O
Cl
(d) a ketone
O
OH
Cl
(e) a Z alkene
Cl
OH
HO
(h) a trans alkene
OH
Cl
H
H
(f) an aldehyde
OH
OH
O
Cl
H
(g) a cyclic ether
O
Cl
HO
7. For each of the molecules below, assign the stereochemical configuration(s) as E, Z, R or S.
For full marks, you must show the priority numbers you used to assign each configuration and it must be
clear what part of the molecule is being described as E, Z, R or S.
[10 marks]
(a)
1
Cl
S
2
2
H
3
H
4
4
Cl
1
3
S
(b)
higher priority groups on same side therefore ‘Z’
two identical groups on one side therefore can’t
assign E or Z.
(c) Hint: The 3-dimentionality of this molecule is implied.
4
H
2
3
S
O
1
(d)
higher priority groups on opposite sides therefore ‘E’
O
OCH3
OCH3
8. Draw all isomers (constitutional and stereo) for C5H11Cl. There are more than 1 and less than 20. All
molecules are neutral and every atom has a formal charge of zero. Only structures in line-bond notation will
be graded (no expanded or condensed structures).
[11 marks]
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Some Useful Data
Principal Functional Group Priority List
Carboxylic acid
Sulfonic acid
Ester
Acid chloride
Amide
Nitrile
Aldehyde
Ketone
Alcohol
Thiol
Amine
1
CHEM 1000 Standard Periodic Table
18
1.0079
4.0026
H
He
2
13
14
15
16
17
6.941
9.0122
10.811
12.011
14.0067
15.9994
18.9984
Li
Be
B
C
N
O
F
Ne
3
22.9898
4
24.3050
5
26.9815
6
28.0855
7
30.9738
8
32.066
9
35.4527
10
39.948
1
2
20.1797
Na
Mg
11
39.0983
12
40.078
3
4
5
6
7
8
9
10
11
12
44.9559
47.88
50.9415
51.9961
54.9380
55.847
58.9332
58.693
63.546
65.39
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
19
85.4678
20
87.62
21
88.9059
22
91.224
23
92.9064
24
95.94
26
101.07
27
102.906
28
106.42
29
107.868
30
112.411
31
114.82
32
118.710
33
121.757
34
127.60
35
126.905
36
131.29
Rb
Sr
37
132.905
38
137.327
Cs
Ba
55
(223)
56
226.025
Fr
87
Ra
Y
39
La-Lu
Ac-Lr
88
P
S
Cl
Ar
15
74.9216
16
78.96
17
79.904
18
83.80
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
41
180.948
42
183.85
43
186.207
44
190.2
45
192.22
46
195.08
47
196.967
48
200.59
49
204.383
50
207.19
51
208.980
52
(210)
53
(210)
54
(222)
Hf
Ta
W
Re
Os
Ir
Pt
Au
72
(261)
73
(262)
74
(263)
75
(262)
76
(265)
77
(266)
78
(281)
79
(283)
Rf
Db
Sg
105
106
138.906
140.115
140.908
144.24
La
Ce
Pr
Nd
57
227.028
58
232.038
59
231.036
60
238.029
Ac
Si
14
72.61
40
178.49
104
89
25
(98)
Al
13
69.723
Th
90
Pa
91
U
92
Bh
107
Hs
Mt
Dt
Hg
Tl
Pb
Bi
Po
At
80
81
82
83
84
85
174.967
Rg
108
109
110
111
(145)
150.36
151.965
157.25
158.925
162.50
164.930
167.26
168.934
173.04
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
61
237.048
62
(240)
63
(243)
64
(247)
65
(247)
66
(251)
67
(252)
68
(257)
69
(258)
70
(259)
71
(260)
Np
93
Pu
94
Am
95
Cm
96
Bk
97
Cf
98
Es
99
Fm
100
Md
101
No
102
Developed by Prof. R. T. Boeré
Lr
103
Rn
86