課程名稱︰有機化學一
課程性質︰系上必帶
課程教師︰汪根叢
開課系所︰化學系
考試時間︰2006/10/17
是否需發放獎勵金：是
（如未明確表示，則不予發放）
試題 :
1.Please compare the relative proton acidity(pKa) of the following compounds
and give your reasons.(10%)
H H H O H O
| ┌─┐ | ┌─┐ | ∥ ┌─┐ | ∥ ┌─┐
(a) H-C-│H │ H-C-O-│H │ H-C-C-O-│H │ H-C-C-O-│H │
| └─┘ | └─┘ | └─┘ | └─┘
H H H Cl
(b) 3HC-CH3 2HC＝CH2 HC≡CH
2.(1)Give the IUPAC systematic names of the following compounds.(10%)
∣
╲╱ Cl
∣ ∣ ∣ ∣ ∣ ∣ ∣
(a) ╱∣╲╱∣╲ (b) ╲╱╲╱╲╱╲╱╲ (c) ╱╲╱╲╱╲╱
∣ ∣ ∣ ∣
Cl ╱╲ ╱ Br
∣
╱
∣ ▕(指向外的粗線)
(d) ╱╲ (e) ╱╲
∣ ∣ ∣ ∣
╲╱╲ ╲╱╲(這裡是向內的虛線)
∣ Cl ▽
╱∣╲
(2)Accirding to the following IUPAC names, draw the corresponding structures.
(10%)
(a)2,3-dimetheyl-5-(2-methylbutyl)decane
(b)6-(1,2-dimethylpropyl)-4-propyldecane
(c)4-bromo-2chloro-1-methylcyclohexane
(d)cis-1-ethyl-3-isopropylcyclopentane
(e)cyclopentylcyclohexane
3.(1)Give you reason for which of the following diaxial-substituted cyclohexane
has the highest energy.(4%)
CH3 CH3
∣ ∣
∕╲╱ ∕╲╱ ∕╲╱
╱╲∕ ╱╲∕ ╱╲∕
∣ ∣ ∣ ∣
Cl Cl CH3 Cl
(2)Draw the most stable conformation of the following molecules.(6%)
CH3 CH3
▕ Cl ▕
╱╲╱(虛線) ╱╲
(a) ∣ ∣ ﴠ (b) ∣ ∣
╲╱ (虛線)╱╲╱╲(虛線)
╱∣╲ CH3 CH3
4.(1)Which of the following conformers of isobutyl chloride is the most stable,
why?(4%)
H CH3 H H CH3 Cl
╲︵︵︵╱ ╲︵︵︵╱
(a) ( ∣ ) (b) ( ∣ )
( ╱╲ ) ( ╱╲ )
H︶︶︶CH3 H︶︶︶CH3
∣ ∣
Cl H
(2)Draw the most stable and unstable eclipsed conformers of isobutyl
chloride.(6%)
5.For chemical reactions shown below, the starting W firstly transforms to
intermediates A and B, then subsquently from products C and D. Based on
Curtin-Hammett principle explain how you can control the product formation
ratio of C to D.(15%)
W
Ea2 ∣ Ea1
<---- ↓ → Ea3
C → 〔 A ← B 〕→ D given: Ea3 > Ea2 > Ea1
6.Given your reasons for the following observations of bond dissociation
energy.(15%)
(a) 3HC-X X = F Cl Br I (3%)
138 103 87 71
(b) 3HC-X X = F OH NH2 (3%)
136 119 108
R = CH3CH3 (CH3)2CH (CH3)3C
(c) R-H 101 98.5 96.5 (6%)
(d) R-Cl 84 84 85 (3%)
7.Based on Hammond postulate, give your reason for the selectivity of the
following halogenations.(20%)
F
∣
CH3 CH2 CH3
∣ F2 ∣ ∣
3HC——H ——→ 3HC——H + 3HC——F
∣ hνor △ ∣ ∣
CH3 CH3 CH3
9 ： 1
CH3 CH3
∣ Br2 ∣
3HC——H ——→ 3HC——Br only product
∣ hνor △ ∣
CH3 CH3
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