HRID1112621

反应详情

EQUATION

反应方程式

HRID 1112621 的结构方程式

PROCEDURE

实验过程

"C6H6 concenses with (Me2C(OH) 2 only if 2-2.5 moles AlCl3 are used at 80°, yielding 8% 1,1,2-trimethyl-indan and some 50% Me2CAc. CH2 (CMe2OH)2 with 1.5 moles AlCl3 at 50°-70° yields 52.5% 1,1,3,3-tetramethyl-indan and 18% 1,2,3,5,6,7-hexahydro-1,1,3,3,5,5,7,7-octamethyl-indacene, m. 214°, while 2,5-dimethyl-2,5-hexanediol similarly gives 57.3% 1,1,4,4-tetramethyl-1,2,3,4-tetrahydronaphthalene and 1,1,4,4,5,5,8,8-octamethyl-1,2,3,4,5,6,7,8-octahydroanthracene (20%). Reaction with MeCH-(OH)CH2C(OH)Me2 with 1.5 moles AlCl3 at 50°-60° gave 20% Me2CPhCH2CH(OH)Me and 25% 1,1,3-trimethyl-indan. Me2C(OH)CH2COMe with 2 moles AlCl3 at 20° gave 61% PhCH2CHMeCH2COMe and a little unknown product, m. 127°. The mechanism is probably analogous to that given for ROH (Byull. Sredneaziat. Gosudarst. Univ. No. 25, 45(1947), with additional possibility of intermediate formation of oxido structures."

WORKUP

后处理

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