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[求助]
請教幾個tripos原子類型的問題
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本人剛學(xué)SYBYL ，對tripos原子類型不是很熟悉，請問圖片1中左上方苯環(huán)上CH2NH2處的N是N.3還是N.pl3，因為我用Sketch模塊搭建分子時，sybyl最初自動判斷是N.pl3；吡啶環(huán)旁邊的NH2的N應(yīng)該是N.pl3吧；右邊哌啶環(huán)上的N與乙�；噙B時是N.3還是N.am，因為我搭建分子時，sybyl最初自動判斷是N.am； 圖片2中左上方苯環(huán)上的NHCOMe處的N應(yīng)該是N.pl3還是N.am，因為我用Sketch模塊搭建分子時，sybyl最初自動判斷是N.pl3。 本人覺得網(wǎng)上搜到的Tripos公司給的原子類型說明太粗略，沒有結(jié)合太多復(fù)雜的實際例子來講解；而有一個例子是我搜到的，卻很粗略，希望大家提供更好的例子資源來講解。 圖片1.jpg 圖片2.jpg |
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2013-07-25 00:21:18, 211.77 K
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我從網(wǎng)上摘了一段，不知道對不對 1.6 If element_symbol is C then 1.6.1 If num_bond .ge. 4 .AND. all bonds are single then atom_type is C.3 1.6.2 If num_bond .eq. 3 .AND. all bonds are acyclic .AND. all bonds are to nitrogen .AND. each nitrogen forms bonds to 2 other atoms both of which are not oxygen then atom_type is C.cat 1.6.3 If num_bond .ge. 2 .AND. 2 bonds are aromatic then atom_type is C.ar 1.6.4 If ( num_bond .eq. 1 .OR. num_bond .eq. 2 ) .AND. one bond is triple then atom_type is C.1 1.6.5 If element_symbol is C and none of the above then atom_type is C.2 1.7 If element_symbol is O then 1.7.1 If num_nonmet .eq. 1 then 1.7.1.1 If bond is to carbon .AND. carbon forms a total of 3 bonds, 2 of which are to an oxygen forming only 1 non-metal bond then atom_type is O.co2 1.7.1.2 If bond is to phosphorus .AND. phosphorus forms at least 2 bonds to an oxygen forming only 1 non-metal bond then atom_type is O.co2 1.7.2 If num_bond .ge. 2 .AND. all bonds are single then atom_type is O.3 1.7.3 If element_symbol is O and none of the above then atom_type is O.2 1.8 If element_symbol is N then 1.8.1 If num_nonmet .eq. 4 .AND. all bonds are single then atom_type is N.4 1.8.2 If num_bond .ge. 2 .AND. 2 bonds are aromatic then atom_type is N.ar 1.8.3 If num_nonmet .eq. 1 .AND. bond is triple then atom_type is N.1 1.8.4 If num_nonmet .eq. 2 .AND. ( bonds are double, double .OR. bonds are single, triple ) then atom_type is N.1 1.8.5 If num_nonmet .eq. 3 .AND. one bond is to C=O or C=S then atom_type is N.am 1.8.6 If num_nonmet .eq. 3 then 1.8.6.1 If one bond is not single then atom_type is N.pl3 1.8.6.2 If all bonds are single then 1.8.6.2.1 If one single bond is to an atom that forms a bond of type double, triple, aromatic or delocalised .AND. one other single bond is to H then atom_type is N.pl3 1.8.6.2.2 If one single bond is to an atom that forms a bond of type double, triple, aromatic or delocalised .AND. neither of the other single bonds are to H .AND. sum_of_angles around N .ge. 350 deg then atom_type is N.pl3 1.8.6.3 If num_nonmet .eq. 3 otherwise then atom_type is N.3 1.8.7 If element_symbol is N and none of the above then atom_type is N.2 1.9 If element_symbol is S then 1.9.1 If num_nonmet .eq. 3 .AND. 1 bond is to an oxygen with only one non-metal bond then atom_type is S.o 1.9.2 If num_nonmet .eq. 4 .AND. 2 bonds are to an oxygen with only one non-metal bond then atom_type is S.o2 1.9.3 If num_bond .ge. 2 .AND. all bonds are single then atom_type is S.3 1.9.4 If element_symbol is S and none of the above then atom_type is S.2 |
2樓2013-07-25 00:47:15
3樓2013-07-25 01:11:20
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