摘要
甲醇与一氧化碳(CO)反应制芳烃是煤制芳烃技术重大突破。以杂化轨道理论和对含Si、Al元素催化剂酸中心认识为理论依据和研究方法,以反应物分子与催化剂B酸中心相互作用为切入点,分析其相互作用,揭示甲醇与CO制芳烃反应机理。结果发现:甲醇与CO制芳烃反应过程分为5步:(1)B酸中心释放H+攻击甲醇分子中C-O键,C-O键断裂生成甲基碳正离子(CH3^+),B酸中心失活;(2)失活B酸中心从CH3^+获得H+,反应生成中间体C↑↓〇H2,C↑↓〇H2中C原子外层有1对孤对电子(e-)和1个空轨道,失活的B酸中心获得H+后恢复活性;(3)C↑↓〇H2与CO反应生成中间体C↑↓↑〇H和C↑,↑↑H;(4)6个C↑↓↑〇H或6个C↑,↑↑H可以形成苯环结构,但构成方式不同;(5)C↑↓〇H2与苯环反应生成混合芳烃。
Producing aromatics through the reaction of methanol with carbon monoxide is a major breakthrough in coal-based aromatics production technology.Based on the hybrid orbital theory,acid center theory of Si-Al catalysts and the mutual interaction between reactant and catalyst B-acid center,the molecular structure of methanol and B-acid center in the interaction process was studied.The reaction mechanism of methanol with carbon monoxide to aromatics was therefore further understood.It was found that the reaction of methanol with carbon monoxide to aromatics on the action of B-acid center can be five steps:(1)B-acid center provides hydrogen proton(H+)to attack C-O bond in the methanol molecular,and then C-O bond is broken to form methyl carbocation(CH3^+),but B-acid center loses its activity after releasing hydrogen proton(H+).(2)Inactivated B-acid center obtains hydrogen proton(H+)from CH3^+,and produces a new intermediate C↑↓〇H2,which has a pair of lone pairs of electrons(e-)and an empty orbit.In this step,inactive B-acid center recovers its activity after obtaining hydrogen proton(H+).(3)CO reacts with C↑↓〇H2 to form intermediates C↑↓↑〇H and C↑,↑↑H.(4)6 C↑↓↑〇H or 6 C↑,↑↑H forms benzene ring structure which has a special configuration.(5)Benzene reacts with the C↑↓〇H2 to form different aromatics.
作者
贺振富
HE Zhenfu(CHN ENERGY,Beijing 100011,China)
出处
《石油学报（石油加工）》
EI
CAS
CSCD
北大核心
2020年第4期857-865,共9页
Acta Petrolei Sinica(Petroleum Processing Section)
