ÌâÄ¿ÄÚÈÝ

ÑõµªÔÓ׿ÊÇÐÂÒ©ÑÐÖƹý³ÌÖз¢ÏÖµÄÒ»ÀàÖØÒª»îÐÔÎïÖÊ£¬¾ßÓп¹¾ªØÊ¡¢¿¹Ö×Áö¡¢¸ÄÉÆÄÔȱѪµÈÐÔÖÊ¡£ÏÂÃæÊÇijÑо¿Ð¡×éÌá³öµÄÒ»ÖÖÑõµªÔÓ׿À໯ºÏÎïHµÄºÏ³É·Ïߣº

£¨1£©Ô­ÁÏAµÄͬ·ÖÒì¹¹ÌåÖУ¬º¬Óб½»·ÇҺ˴Ź²ÕñÇâÆ×ÖÐÓÐ5¸ö·åµÄ½á¹¹¼òʽΪ £¬
д³ö¸ÃÎïÖÊ´ß»¯Ñõ»¯·´Ó¦µÄ»¯Ñ§·½³Ìʽ     ¡£
£¨2£©¢ÚµÄ·´Ó¦ÀàÐÍÊÇ    ¡£Ô­ÁÏDÖк¬ÓеĹÙÄÜÍÅÃû³ÆÊÇ     ¡¢     ¡£
£¨3£©Ð´³ö·ûºÏÏÂÁÐÌõ¼þµÄÖмä²úÎïFµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ     ¡£
(i) ÄÜ·¢ÉúÒø¾µ·´Ó¦£»
(ii) ·Ö×ÓÖк¬ÓÐÈýÈ¡´úµÄ±½»·½á¹¹£¬ÆäÖÐÁ½¸öÈ¡´ú»ùÊÇ£º¡ªCOOCH3ºÍ£¬ÇÒ¶þÕß´¦ÓÚ¶Ôλ¡£
£¨4£©Ô­ÁÏBË×Ãû¡°ÂíÀ³ôû¡±£¬ËüÊÇÂíÀ³ËᣨHOOC-CH=CH-COOH£©µÄËáôû£¬ÇëÉè¼ÆÓÃÔ­ÁÏCH2OH-CH=CH-CH2OHºÏ³ÉÂíÀ³ËáµÄºÏ³É·Ïß¡£ºÏ³É·ÏßÁ÷³ÌͼʾÀýÈçÏ£º


½âÎöÊÔÌâ·ÖÎö£º£¨1£©Ô­ÁÏAµÄ·Ö×ÓʽΪC7H8O£¬ËüµÄº¬Óб½»·ÇҺ˴Ź²ÕñÇâÆ×ÖÐÓÐ5¸ö·åµÄͬ·ÖÒì¹¹Ìå½á¹¹¼òʽΪ¡£Ð´³ö¸ÃÎïÖÊ´ß»¯Ñõ»¯·´Ó¦µÄ»¯Ñ§·½³Ìʽ¡££¨2£©·´Ó¦¢ÚµÄ»¯Ñ§·´Ó¦ÀàÐÍΪõ¥»¯·´Ó¦£¨»òÈ¡´ú·´Ó¦£©¡£Ô­ÁÏDÁÚ°±»ù±½·ÓÖк¬ÓеĹÙÄÜÍÅΪôÇ»ù¡¢°±»ù¡££¨3£©FµÄ·ûºÏÒªÇóµÄͬ·ÖÒì¹¹Ìå¡¢¡¢¡¢
£¨4£©ÒÔÔ­ÁÏCH2OH-CH=CH-CH2OHºÏ³ÉÂíÀ³ËáµÄºÏ³É·ÏßΪ+HBr£»
 +O2 £»
+O2£»
+NaOH+NaBr+H2O¡£
¿¼µã£º¿¼²éÓлúÎï¹ÙÄÜÍÅ¡¢½á¹¹¼òʽ¡¢Í¬·ÖÒì¹¹ÌåµÄÊéд¼°ºÏ³ÉµÈ֪ʶ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø