0  B11R0.4  O=C(OCC)CCCCCCBr BCP15664 [1,0,1,0,0,0|2,0,0]=4
1  B11R0.148  CCCCCCC\C=CC(O)=O BCP19828 [1,0,1,0,0,0|0,0,2]=4
2  B13R0.2  O=C(CC#N)OCCCCCCCC BCP12781 [0,1,1,0,0,0|1,0,1]=4