有机化学08

This means that the transition state leading to formation of the 2° radical is slightly lower in energy than the transition state leading to formation of the 1° radical. The secondary radical is formed faster! It is important to remember that Ea depends on, but will never exceed Ea H°
1
Primary (1°): There are six primary hydrogens. 有六个伯氢。
H H Cl H H H H H H H H H H Cl
1° 自由基 radical 一个伯氢被取代可得到相应产物。 to the primary product. Abstraction of a primary hydrogen leads
iso-butyl bromide (1°) 异丁基溴
Me Me Me Br Me Me Br
sec-butyl bromide (2°) 仲丁基溴
tert-butyl bromide (3°) 叔丁基溴
Radical Chlorination of Propane (cont.)
Cl Cl Cl
或 Δ or hv
C-H
4
Draw the Reaction-Energy Diagram!
potential energy
1° Ea
Ea
2° Ea
Cl
பைடு நூலகம்
1 kcal/mol
势 能
1° radical + HCl
H° = 3 kcal/mol
2° radical + HCl
reaction coordinate 反应进程
仅有产物是能量不同的：二级自由基比一级自由基的能量低。 In this case, the reactants are the same for both processes (same energy). 这表明形成二级自由基的所经历的过渡态要比形成一级自由基所经历 radical. 的过渡态的能量要稍微低一些。
This means that it is thermodynamically C-H favorable to break a 2° more 1° 2° C–H bond than a 1° C–H bond. But how does this control the kinetic selectivity?
H R C H H
R
R
1°
R C H H
2°
R C H R
3°
2
对于取代基我们也定义一级，二级或三级。 We also refer to substituents as primary, secondary or tertiary.
Me Me Br Me Br
n-butyl bromide (1°) 正丁基溴
自由基的形成是由under kinetic control, so abstraction of the 2° Radical formation is 的，因此失去仲氢进而形成二 hydrogen to form the 2° radical must be faster than abstraction of 级自由基比失去伯氢而形成一级自由基要 。 the 1° hydrogen to form the 1° radical. In other words, the secondary hydrogens are more reactive than the primary hydrogens under these conditions!
H H H H Cl H H H H H H H H Cl
form –103 kcal/mol 形成 H–Cl: break 解离 1° C–H: +98 kcal/mol overall: 总计 ΔH° = –5 kcal/mol
H H H H
1° radical Secondary:
H H H H Cl H H H H Cl
This reaction is under kinetic control, so the product ratio is determined by Ea. Ea How can you predict the value of Ea?
Ea
Analyze the Reaction-Energy Diagram!
Cl
1° (40%)
2° (60%)
但是二级氯代产物更有利。 有六个伯氢和两个仲氢， There are six primary hydrogens and two secondary hydrogens, but the secondary chloride is favored.
我们如何解释该反应的选择性？ this reaction? How can we explain the selectivity of
有两种可能产物！ 为什么？ There are two possible products! Why?
丙烷有两种可在链增长起始步骤中被取代的 can be abstracted in the Propane has two inequivalent types of hydrogens that 氢。 first propagation step. 起始的链增长步骤决定了将形成哪种产物 will be formed (1° 或 2°). The first propagation step determines which product or
Suggested Problems: 41， 4-39,41,44–46,54 ： 4-39， 44-46， 54
E. Radical Chlorination of Propane: Inequivalent Hydrogens
Cl Cl Cl Δ 或 hv or Cl
40% (1°)
60% (2°)
3
2. Relative Reactivity
2°
H H H H
Δ or hv 或
Cl Cl
H H Cl H H
Cl H H
1°
H H H H
1°
H H H H
H H H H
1° chloride (40%) 氯
氯 2° chloride (60%)
Calculating Relative Reactivity: primary: 40% = 6.67% per H 6 H's secondary: 60% = 30.0% per H 2 H's
form –103 kcal/mol 形成 H–Cl: break 解离 2° C–H: +95 kcal/mol overall: 总计 ΔH° = –8 kcal/mol
H H H H
H H H H
2° radical
C-H C-H 2° C–H bonds are weaker than 1° C–H bonds.
H°!
The only energy difference is in the products: the 2° radical is lower in energy than the 1°
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F. Bromination of Propane: Higher Selectivity
Br Br Br
H H H H
1° radical Secondary:
H H H H Br H H H H Br
form H–Br: –88 kcal/mol break 2° C–H: +95 kcal/mol overall: ΔH° = +7 kcal/mol
(1° Hydrogens (1°, 2°, 3°) 1. Inequivalent 2° 3° )
伯碳 (1°)： Carbon: carbon that has only one other carbon attached to it 仅与有另一个碳原子与其相连 Primary (1°)
Secondary 仲碳 (2°)： (2°) Carbon: carbon that has two other carbons attached to it 有另外两个碳原子与其相连 Tertiary (3°) 叔碳 (3°)： Carbon: carbon that has three other carbons attached to it 有另外三个碳原子与其相连
有两个仲氢。 Secondary (2°): There are two secondary hydrogens.
H H Me Me H Cl Me Me H Cl
2° radical 自由基 Abstraction of a secondary hydrogen leads to the secondary product. 一个仲氢被取代可得到相应产物。
secondary primary
=
30.0 6.67
= 4.5
The secondary hydrogens are 4.5 二级氢比一级氢活泼4.5。 times more reactive than the primary hydrogens.
为什么？
Why?
We can answer this using thermodynamics, kinetics, and reaction-energy diagrams! Primary: