ÌâÄ¿ÄÚÈÝ

ÒÑÖª·´Ó¦£º
¡¢
ÏÖÓÐÎïÖÊA~IµÄת»¯¹ØϵÈçÏÂͼ£º

ÈôBµÄ·Ö×ÓʽΪC8H8O£¬Æä±½»·ÉϵÄһԪȡ´úÎïÖ»ÓÐÁ½ÖÖ£»GΪ¸ß·Ö×Ó»¯ºÏÎï¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)д³öÏÂÁз´Ó¦µÄ·´Ó¦ÀàÐÍ£º·´Ó¦¢Ü               £¬·´Ó¦¢Û               ¡£
(2)д³öÏÂÁÐÎïÖʵĽṹ¼òʽ£ºF               £¬I               £¬A               ¡£
(3)д³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢ÙB¡úC£º                                                            £»
¢ÚC+D¡úH£º                                                         £»
¢ÛF¡úG£º                                                            £»
(4)CµÄͬËØ·ÖÒì¹¹ÌåÇÒÊôÓÚõ¥ÀàµÄ·¼Ïã×廯ºÏÎï¹²ÓР      ÖÖ£¬Çëд³öÆäÖÐÒ»ÖÖͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º                                ¡£

(1)Ëõ¾Û£»õ¥»¯£»(2) HOCH2¡ª¡ªCOOH£»HOOC¡ª¡ªCOOH£»¡ªCH3£»
(3)¢ÙH3C¡ª¡ªCHO + 2Cu(OH)2 H3C¡ª¡ªCOOH + Cu2O¡ý+ 2H2O£»
¢ÚH3C¡ª¡ªCOOH+HOCH2¡ª¡ªCH3            H3C¡ª¡ªCOOCH2¡ª¡ªCH3+H2O

¢ÛnHOCH2¡ª¡ªCOOH                                 + nH2O
(4)6£»

                              ÖÐÈÎдһÖÖ¡£

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÒÑÖª·´Ó¦£º

¡¢

ÏÖÓÐÎïÖÊA~IµÄת»¯¹ØϵÈçÏÂͼ£º

ÈôBµÄ·Ö×ÓʽΪC8H8O£¬Æä±½»·ÉϵÄһԪȡ´úÎïÖ»ÓÐÁ½ÖÖ£»GΪ¸ß·Ö×Ó»¯ºÏÎï¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)д³öÏÂÁз´Ó¦µÄ·´Ó¦ÀàÐÍ£º·´Ó¦¢Ü                £¬·´Ó¦¢Û                ¡£

(2)д³öÏÂÁÐÎïÖʵĽṹ¼òʽ£ºF                £¬I                £¬A                ¡£

(3)д³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

¢ÙB¡úC£º                                                             £»

¢ÚC+D¡úH£º                                                          £»

¢ÛF¡úG£º                                                             £»

(4)CµÄͬËØ·ÖÒì¹¹ÌåÇÒÊôÓÚõ¥ÀàµÄ·¼Ïã×廯ºÏÎï¹²ÓР       ÖÖ£¬Çëд³öÆäÖÐÒ»ÖÖͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º                                 ¡£

 

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

¾«Ó¢¼Ò½ÌÍø