利用~1H-核磁共振谱计算环化丁苯橡胶的环化度和溶剂参与量

Calculation of cyclization degree and incorporated solvent content of cyclized SBR by^1H-NMR

对环化丁苯橡胶的1 H 核磁共振谱进行了归属分析 ,并推导出计算环化度和溶剂参与量的公式。设环化丁苯橡胶的1 H 核磁共振谱图中化学位移为 7 0 ,5 4,4 9和 3 0～ 0处的峰强度分别为A ,B ,C和D ,则环化丁苯橡胶中丁二烯总物质的量 (X) =(B +C +D - 2A) /7 1 ,未环化的丁二烯物质的量 (Y) =B/2 +C/4,参与反应的二甲苯物质的量 (Z) =( 1 7 5A -B -C -D) /46 5,环化度 (P) =1 -Y/X ,溶剂参与量 (W ) =91 6Z/X。

HNMR spectra of cyclized SBR was analysed and the equations to calculate the cyclization degree and the incorporated solvent content by 1HNMR method were derived.The total butadiene content in cyclized SBR (x)=(B+C+D-2A)/7.1,the uncyclized butadiene content (Y)=B/2+C/4 ,the incorporated dimethyl benzene content (Z)=(175A-B-C-D)/465,the cyclization degree (P)=1-Y/X and the incorporated solvent content(W)=916Z/X,if the peak densities at the 7.0,5.4,4.9 and 3.0～0 chemical shifts in 1HNMR spectra of cyclized SBR were A,B,C and D respectively.