Download Fall 2009 Exam 2 Key for CH341/CH540: Organic Chemistry - Prof. Bruce Hathaway and more Exams Organic Chemistry in PDF only on Docsity!
CH341/CH540, Exam 2 Key, Fall 2009
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Name
This exam consists of several parts of unequal difficulty. Look over the whole exam right now, and start on something you know well. If you don’t know something, skip it, and come back to it later. Use your time wisely.
Keep Your Eyes On Your Own Paper!
Grading Summary
Page 2. _______________ (16) Page 5. _______________ (10)
Page 3. _______________ (30) Page 6. _______________ (13)
Page 4. _______________ (25) Page 7. _______________ (6 +)
Total ___________________
A Partial Periodic Table of the Elements
1A IIA IIIB IVB VB VIB VII B VIII IB IIB IIIA IVA VA VIA VII A VIII A 1 1.01^ H
2 He 4. 3 6.94^ Li
4 9.01^ Be
5 10.8^ B
6 12.0^ C
7 14.0^ N
8 16.0^ O
9 19.0^ F
10 Ne 20. 11 Na 23.
12 Mg 24.
13 27.0^ Al
14 28.1^ Si
15 31.0^ P
16 32.1^ S
17 35.5^ Cl
18 Ar 40. 19 39.1^ K
20 Ca 40.
21 22 23 24 25 26 27 28 29 30 31 Ga 69.
32 72.6^ Ge
33 As 74.
34 79.0^ Se
35 79.9^ Br
36 Kr 83. 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 Te 128
53 127^ I
54
- (12) Give IUPAC names for the following structures. Use R, S, E, or Z, when needed.
a. H 3 C
H 3 C
CH 3
(E)-4-ethylnon-4-ene
b.
H Br
CH 3
(R)-2-bromo-1-cyclopropyloctane
c.
CH 3
CH 3
CH 3
1- tert -butylcycloocta-1,3,6-triene
- (4) Why do cyclopentane and cyclopentene have similar boiling points? Why are they essentially insoluble in water?
Both have about the same size and shape, so they have the same London forces attracting other molecules of themselves, so they have similar boiling points. They contain only C and H, so all of the bonds are non-polar, so they are not attracted to the polar water.
- (25) Supply the needed reagents and experimental conditions to accomplish each of the following transformations. More than one step may be required in some cases.
a.
CH 3 OH CH^3
H 2 SO 4 , heat
b.
CH 3
CH 3
OH
H 2 O, H 2 SO 4 (cat)
c.
H 3 C CH 3 CH 3
H 3 C CH 3
CH 3
Br HBr
d.
H 3 C H 3 C
C
CH 2
CH 3
Br H 3 C H 3 C
C
CH 2
CH 3
O CH
CH 3
H O CH^ CH^3
CH 3
CH 3
e.
- Br 2
- 2 KOH, heat
(10) Multiple Choice: Choose the best answer to each question.
C
What type of alkyl halides react best with weak nucleophiles to do substitution reactions? A. Primary B. Secondary C. Tertiary D. Aromatic
A
What type of alkyl halides react best with strong nucleophiles to do substitution reactions? A. Primary B. Secondary C. Tertiary D. Aromatic
D
Which of the halide ions is the best nucleophile in water? A. Fluoride B. Chloride C. Bromide D. Iodide
C
Which carbocation below is the least stable? A. B. C. D. CH 3 CH 3 CH 2 CH 3
D
_________
Which alkene below is the most stable? A. B. C. D. C C
H 3 C CH 2
H 3 C H H
C
C
H 3 C CH 3
H 3 C H
C
C
H 3 C CH 2
H 3 C H CH 3
C
C
H 3 C CH 3
H 3 C CH 3
7(6) a. What does Markovnikov’s rule say? Why does it work?
The H of the HX ends up bonded to the carbon of the C=C that is bonded to the most H’s, and the X ends up bonded to the C of the C=C bonded to the least H’s.
It works because adding H-X this way forms the more stable carbocation intermediate.
b. Draw the generic structure for the polymer formed from the following alkene.
C CH 2
Br
H 3 C
C C
H
H
CH 3
Br ( )n
Bonus: ( 5) When the following reaction is run, only the indicated stereoisomer of the product is obtained. Explain why, using words, as well as a mechanism.
H 2 C
C
C
CH 3
Br (^) H
H
HBr (^) H 3 C C
C
CH 3
Br (^) H
H Br
H 2 C
C
C
CH 3
Br (^) H
Br H H
H 3 C
C
C
CH 3
Br+ H
Br H
H 3 C
C
C
CH 3
Br (^) H
H Br
When H-Br is attacked by the C=C. the adjacent Br can form a bond with the developing carbocation, to form a bromonium ion. Bromide attacks from below to make the dibromocompound with the indicated stereochemistry.
