本篇論文藉由計算化學方法密度泛函理論（Density Functional Theory，DFT），計算化學反應中具有選擇性的還原消去(Reductive Elimination)反應、以及金屬配位路徑的選擇性，探討決定反應的速率決定步驟（Rate Determining Step，RDS），動力學所需之活化能、產物熱力學之穩定性；由計算所得之結構與能量，討論選擇性的決定因素。本篇論文內容分成二個主題:

第一個主題-銅催化試劑: C-CF2H鍵之選擇性生成：
討論關於聯吡啶基銅（bpyCu(II)，bipyridine Copper II）幫助催化二氟甲基化（Difluoromethylation）反應之路徑選擇性；過程中使用了二種類似卻有著不同配位基之化合物，結果發現二種催化劑皆得到相同產物，即使配位基中有Cl的銅催化試劑，也沒有氯化（Chlorination）的產物。在討論了自由基的反應路徑、Cu催化還原消去（Reductive Elimination）後，發現反應傾向先與催化劑進行反彈(Rebound)鍵結過程，此步驟活化能皆小於5 kcal/mol、且過渡態為介於開閉殼層之間(open-close shell)的特殊型態，此步驟也是反應選擇性的主因；而下ㄧ步的還原消去(Reductive Elimination)則是反應速率決定步驟（Rate Determining Step），進行二氟甲基化反應所需活化能較氯化反應之活化能低。探討其選擇性之原因，是因為Cu-Cl的鍵結電子能為61.6 kcal/mol、Cu-C的鍵結電子能為21.0 kcal/mol，二者鍵能相差近3倍，這也是反應不會進行氯化(Chlorination)反應的理由；本篇以理論計算，將實驗結果完整解釋及驗證機構可能性。 

第二個主題-金錯合物之位向選擇性反應：
討論關於金催化劑形成一錯合物的位向選擇，初始反應試劑R1結構中具有二反應位置，對於金試劑（R2）來說，二個位置都是有可能發生反應的，但產物僅得到X3。
過程中，先將反應路徑進行計算，結果發現路徑中未有單一反應的決定因素，反應路徑大致上分成三步:R1與R2反應、R3的鍵結、甲基喹啉機團(methyl quino group)的離去；第一步無活化能，僅只有熱力學之產物位能高低，代表二條路徑皆可能有產物發生、第二步為R3的鍵結，此步為速率決定步驟（RDS），此步活化能皆大於15 kcal/mol，二者活化能差僅2.2 kcal/mol、而第三步的離去基，活化能差距僅1 kcal/mol，也不是決定整個反應的選擇性主因。因此假設，在X1與Y1之間有一過渡態TS-X1Y1幫助二條路徑的轉換，因為此過渡態之能量僅14.6 kcal/mol，比X路徑或Y路徑的任一過渡態都都還容易達到，進而使得選擇反應傾向沿著X路徑得到X3產物。利用假設並搭配計算，成功解釋了整個Au催化劑中，反應產物僅有X路徑之產物，也構築了整個完整的能量曲面。

In organic chemistry, constructing mechanisms of the reactions is powerful to predict the selectivity. Density functional theory is a common and accurate method to form energy surface to understand the mechanisms of the organic reactions. This study separates into two parts.
In the first part, the benzylic radical compound undergoes the rebound process with bis(difluoromethyl) bpyCu or Chloro difluoromethyl bpyCu to form copper complexes. After reductive elimination, the final products are collected from difluoromethylation, no products of chlorination. The reductive elimination is rate determining step (RDS) in this reactions. The probable reason of selectivity is that Cu-Cl bond strength (61.6 kcal/mol, electronic energy) is stronger than that of Cu-CF2H (21.0 kcal/mol).
In the second part, the reagent including two triple-bond has two reactive sites, X and Y, and the gold reagent can select one to bind and get the two kinds of the final products after methyl quinoline N-oxide (R3) binding and methyl quinoline group leaving. We find that the transition state between X1 and Y1 is the important key to decide the only final product of the X-pathway. The energy barrier of the changing binding site (14.6 kcal/mol) is lower than undergoing next step, R3-binding, in pathway X (20.4 kcal/mol) or pathway Y (18.2 kcal/mol). Therefore, the binding site stays at the most stable compound (X1), and only the product of the pathway X is collected.

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