ÌâÄ¿ÄÚÈÝ

14£®ÈçͼÊÇijÓлúÎïµÄÇò¹÷Ä£ÐÍ£¬¸ÃÎïÖÊÖ»º¬C¡¢H¡¢O¡¢NËÄÖÖÔªËØ£¬Çë»Ø´ð£º
£¨1£©¸ÃÓлúÎïµÄ·Ö×ÓʽΪC3H7NO2
£¨2£©¸ÃÓлúÎï²»¿ÉÄÜ·¢ÉúµÄ·´Ó¦ÓТڢܣ¨Ìî±àºÅ£©
¢ÙÈ¡´ú·´Ó¦¢ÚÏûÈ¥·´Ó¦¢ÛÑõ»¯·´Ó¦
¢Ü¼Ó¾Û·´Ó¦¢Ýõ¥»¯·´Ó¦¢ÞËõ¾Û·´Ó¦
£¨3£©¸ÃÎïÖÊÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ£¨ÓлúÎïÓýṹ¼òʽ±íʾ£©CH3CH£¨NH2£©COOH+OH-=CH3CH£¨NH2£©COO-+H2O
£¨4£©ÔÚŨÁòËá¼ÓÈÈÌõ¼þÏ£¬¸ÃÓлúÎï¿ÉÒÔÓë¼×´¼·¢Éú»¯Ñ§·´Ó¦£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£¨ÓлúÎïÓýṹ¼òʽ±íʾ£©CH3CH£¨NH2£©COOH+CH3OH$?_{¡÷}^{ŨH_{2}SO_{4}}$CH3CH£¨NH2£©COOCH3+H2O£®

·ÖÎö ÓÉÓлúÎïµÄÇò¹÷Ä£ÐÍ£¬¸ÃÎïÖÊÖ»º¬C¡¢H¡¢O¡¢NËÄÖÖÔªËØ£¬¿ÉÖª¸ÃÓлúÎï½á¹¹¼òʽΪ£¬º¬°±»ù¡¢-COOH£¬½áºÏ°±»ùËáµÄÐÔÖÊÀ´½â´ð£®

½â´ð ½â£º£¨1£©¸ÃÓлúÎï½á¹¹¼òʽΪ£¬·Ö×ÓʽΪC3H7NO2£¬¹Ê´ð°¸Îª£ºC3H7NO2£»
£¨2£©º¬-COOH¿É·¢ÉúÈ¡´ú¡¢õ¥»¯·´Ó¦£»°±»ù¡¢ôÈËá¿É·¢ÉúËõ¾Û·´Ó¦£»ÄÜȼÉÕ¿É·¢ÉúÑõ»¯·´Ó¦£¬²»ÄÜ·¢ÉúÏûÈ¥·´Ó¦¡¢¼Ó¾Û·´Ó¦£¬¹Ê´ð°¸Îª£º¢Ú¢Ü£»
£¨3£©¸ÃÎïÖÊÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪCH3CH£¨NH2£©COOH+OH-=CH3CH£¨NH2£©COO-+H2O£¬¹Ê´ð°¸Îª£ºCH3CH£¨NH2£©COOH+OH-=CH3CH£¨NH2£©COO-+H2O£»
£¨4£©¸ÃÓлúÎï¿ÉÒÔÓë¼×´¼·¢Éú»¯Ñ§·´Ó¦£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪCH3CH£¨NH2£©COOH+CH3OH $?_{¡÷}^{ŨH_{2}SO_{4}}$ CH3CH£¨NH2£©COOCH3+H2O£¬
¹Ê´ð°¸Îª£ºCH3CH£¨NH2£©COOH+CH3OH $?_{¡÷}^{ŨH_{2}SO_{4}}$ CH3CH£¨NH2£©COOCH3+H2O£®

µãÆÀ ±¾Ì⿼²éÓлúÎïµÄ½á¹¹ÓëÐÔÖÊ£¬Îª¸ßƵ¿¼µã£¬°ÑÎÕ¹ÙÄÜÍÅÓëÐÔÖʵĹØÏµ¡¢ÓлúÎïµÄÇò¹÷Ä£ÐÍΪ½â´ðµÄ¹Ø¼ü£¬²àÖØ°±»ùËáÐÔÖʼ°·ÖÎö¡¢Ó¦ÓÃÄÜÁ¦µÄ¿¼²é£¬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø