Nishimura H, Saito S, Kishida F, Matsuo M
Environmental Health Science Laboratory, Sumitomo Chemical Co. Ltd.
Sangyo Igaku. 1994 Sep;36(5):314-23. doi: 10.1539/joh1959.36.5_314.
Acute oral toxicity (LD50-value) of organic chemicals to rats was analyzed by using solubility parameter (delta c), a thermodynamic parameter, of the chemicals. Certain parabolic correlations were established between logarithm of LD50-value (mmol/kg body weight, rats) and delta c of all the collected chemicals (n = 144, R = 0.578), alcohols (n = 29, R = 0.587), ketones (n = 7, R = 0.962), aldehydes (n = 9, R = 0.621), ethers (n = 5, R = 0.890), acetates (n = 7, R = 0.670) and aromatics (n = 84, R = 0.736). Introducing molar volume (Vc) to the above equations could not improve the correlation. In the study, we assumed that as for acute toxicity, chemicals taken into the mammals through biological membrane first disturb the homeostasis, which causes certain biological reactions (i.e. death) and that amounts of the chemicals intaken are regulated by their solubility in the membrane. Based on the assumption, we drew a theoretical equation, which describes LD50 by a parabolic function of delta c. A regression analysis using the equation gave significant correlations as stated above, which incarnates the assumption. A solubility parameter of 2.30 x 10(4) (J/m3)1/2 was also determined for the biological membrane (absorption site) of rats. For comparison, log P was used to describe LD50 of all the chemicals, but no correlation was established (R = 0.164-0.443).
利用有机化学品的热力学参数溶解度参数(δc)分析了有机化学品对大鼠的急性经口毒性(LD50值)。在所收集的所有化学品(n = 144,R = 0.578)、醇类(n = 29,R = 0.587)、酮类(n = 7,R = 0.962)、醛类(n = 9,R = 0.621)、醚类(n = 5,R = 0.890)、醋酸酯类(n = 7,R = 0.670)和芳烃类(n = 84,R = 0.736)中,LD50值(mmol/kg体重,大鼠)的对数与δc之间建立了一定的抛物线相关性。将摩尔体积(Vc)引入上述方程并不能改善相关性。在该研究中,我们假设对于急性毒性而言,通过生物膜进入哺乳动物体内的化学品首先会扰乱体内平衡,从而引发某些生物反应(即死亡),并且摄入的化学品量由其在膜中的溶解度调节。基于该假设,我们推导了一个理论方程,该方程用δc的抛物线函数描述LD50。使用该方程进行回归分析得到了如上所述的显著相关性,这体现了该假设。还确定了大鼠生物膜(吸收部位)的溶解度参数为2.30×10⁴(J/m³)¹/²。为作比较,用log P描述所有化学品的LD50,但未建立相关性(R = 0.164 - 0.443)。