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CLAS – Chem 109 B - Ch. 10
Chapter 10 Practice
1.
Br
2.
Br
3.
Br
4.
Cl
5.
Br
Br
+
6.
Br
7.
Cl
8.
Cl
9.
OCH 3
10.
Br
11.
OTs
12.
OCH 3
13.
14.
15.
16.
+
CH3CH2CH2CH2CH2CH2OH
17.
18.
+
19.
20.
O
OH
21.
O
22.
no reaction
23.
O
H
24.
O
25.
O
H
26.
O
27.
OH
CH3I +
28.
+ HO
I
29.
OH
+
Br
30.
Br
HO
31.
O
+ CH CH I
3
2
H
32.
OH
OCH 3
33.
OH
OCH 3
34.
HO
OCH 3
H3CO
OH
35.
36.
1. RCO3H
2. HO-, H2O
OH
OH
OH
+
OH
37.
1. OsO4
2. H2O2, H2O
OH
OH
38.
+
N
I
-
39.
N
+
40.
+ N(CH 3)3
41.
N
42.
+ N(CH3)3
43.
H3C
N
H
CH3
N
44. Rank the following alcohols in order of increasing rate of dehydration in the presence of sulfuric acid.
C<A<B<D
45. Which amine will form the following product when undergoing 2 successive Hoffman degradations?
H
N
46. Indicate how the following compound could be synthesized using the given starting material and any necessary
reagents.
1. (CH3)3N
2. Ag2O, H2O
3. Δ
47. What is the major product of rearrangement for the following arene oxide?
OH
D
48. What is the major product of rearrangement for the following arene oxides? Which arene oxide is more likely to be
carcinogenic?
OH
OH
NO 2
OCH 3
49. Explain why elimination of 2-bromopentane produces 2-pentene whereas elimination of the quaternary ammonium
ion produces 1-pentene.
OH
When a compound with a good leaving group such as a bromide (very weak base) undergoes E2 reactions the
transition state is alkene-like. Alkenes are more stable when the sp2 carbons are more highly substituted. Therefore the
alkene that is more substituted will form faster. When a compound with a bad leaving group such as an amine (good
base) undergoes E2 reactions the transition state is carbanion-like. Carbanions are more stable when they’re least
substituted. Therefore the alkene that is least substituted will form faster.