本篇論文使用quantum chemistry/RRKM的計算方法研究O(1D)與氟
乙烷、聯胺分子在交叉分子束碰撞環境下的反應，研究各種產物生成的
反應機制及路徑，進而計算各反應路徑的速率常數及產物產率分配。在
氧+氟乙烷反應中，O(1D)會插入C－H鍵生成2-氟乙醇及1-氟乙醇，根據
氟乙烷分子結構組成，2-氟乙醇與1-氟乙醇的生成比例假設為2：3。計
算活化2-氟乙醇的分解路徑得到H、F、OH、HF、H2、H2O及CH2F產率
依序為2.4、0.3、9.6、58.5、0.4、1.2及27.7%，與先前氧+乙烷研究之計
算值H(6.7)、H2(0.8)、OH(7.9)、H2O(26.8)、CH3(60.1%)的產物趨勢相符，
其中H2O產率有明顯不同，主要是HF脫除路徑影響H2O的產率。至於1-
氟乙醇的斷鍵主產物為F自由基，脫除路徑則是HF為主。整體而言，氧
+氟乙烷的插入反應產物以HF最多(83.4%)，其次為F自由基(7.9%)，此
產物產率趨勢與氟甲烷的多重分解路徑的計算結果相符合。
在氧+聯胺反應方面，激發態氧原子可插入N－H與N－N，其中插
入N－H鍵形成活化H2NNHOH分子，由計算結果知悉此活化分子，是整
體碰撞反應產物的主要來源，其中H及OH產率為0.8、29.4%，兩自由基
生成趨勢與氧+氨的實驗值及計算值相符。整體分解路徑產物H、OH、
NH2、H2O、H2、NH3及NH產率分別是0.8、29.4、12.9、54.5、0.5、1.5及0.4%。在此反應中H2O是主產物，不似氧+氨的計算結果以OH較多，
此乃因OH基與α-氮或β-氮上的H進行脫除H2O的反應，當H2O的產率提
高時，相對地斷鍵產生OH基的比率則會降低。同時在此研究中，我們
亦計算活化N2H4O分子的H-shift反應路徑，發現此氫轉移路徑生成的主
要產物是NH3(產率1.5%)，以整體碰撞系統的產物而言，NH3不算是重要
的產物，但是NH3的生成可以確認是來自於氫轉移反應。

In the present thesis, we used Quantum Chemistry/RRKM
calculation to examine the O(1D)+C2H5F and O(1D)+N2H4
reactions in molecular beam collision-free environment. The
product branching ratio of various products formed through the
insertion mechanisms of these two reactions are presented. In the
O(1D)+C2H5F reaction, O(1D) can insert into a C−H bond to form
activated 2-fluoroethanol and 1-fluoroethanol. The product
branching ratio of activated 2-fluoroethanol decomposition are
58.5, 27.7, 9.6, 2.4, 1.2, 0.4, and 0.3% for the HF, CH2F, OH, H,
H2O, H2, and F atom formation channels, respectively. These
results are similar to ethanol decomposition results of 60.1, 26.8,
7.9, 6.7, and 0.8% for the CH3, H2O, OH, H, and H2 formation
channels, respectively, found in a previous study except the lower
percentage H2O formation due to large amount of HF formed. For
the 1-fluoroethanol, the main products are HF (87.1%) and F
radical (8.7%). These results are in good agreement with the
calculated result for the O(1D)+CH3F reaction found in a previous
study.
In the O(1D)+N2H4 reaction, both insertions of O(1D) into
N-H and N-N bonds were examined. The activated H2NNHOH
decomposition is assumed to be the major reaction channel in this
reaction. The product branching ratios are 54.5, 29.4, 12.9, 1.5,
0.8, 0.5, and 0.4 % for the H2O, OH, NH2, NH3, H, H2, and NH
formation channels, respectively. Compared with the calculated
results of O(1D)+NH3, large amount of H2O was produced, which
was not seen in O(1D)+NH3 reaction. This is because elimination
reaction of OH and β-H to form H2O is the major H2O formation
channel. In addition, the calculation gave a low percentage of 1.5 %
for the NH3 formation channel. The calculation showed the NH3
formation takes place by an H-shift reaction.

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