ÌâÄ¿ÄÚÈÝ

4£®À±½·ËØÊÇÀ±½·µÄ»îÐԳɷ֣¬¿ÉÒÔÔ¤·ÀÐÄÔಡ£¬Ò²ÄÜ»º½â¼¡Èâ¹Ø½ÚÌÛÍ´£®À±½·ËØÖÐõ¥À໯ºÏÎïµÄ½á¹¹¿ÉÒÔ±íʾΪ£º£¨RΪÌþ»ù£©£®ÆäÖÐÒ»ÖÖÀ±½·ËØõ¥À໯ºÏÎïJµÄºÏ³É·ÏßÈçÏ£º

ÒÑÖª£º
¢ÙA¡¢BºÍEΪͬϵÎÆäÖÐBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª44£¬AºÍBºË´Å¹²ÕñÇâÆ×ÏÔʾ¶¼ÓÐÁ½×é·å£»
¢Ú»¯ºÏÎïJµÄ·Ö×ÓʽΪC15H22O4£»
¢Û
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©GËùº¬¹ÙÄÜÍŵÄÃû³ÆÎªôÇ»ù¡¢ÃѼü£®
£¨2£©ÓÉAºÍBÉú³ÉCµÄ»¯Ñ§·½³ÌʽΪ£¨CH3£©3CCHO+CH3CHO$¡ú_{¡÷}^{Ï¡NaOH}$£¨CH3£©3CCH=CHCHO+H2O£®
£¨3£©ÓÉCÉú³ÉDµÄ·´Ó¦ÀàÐÍΪ¼Ó³É·´Ó¦£¬DµÄ»¯Ñ§Ãû³ÆÎª4£¬4-¶þ¼×»ù-1-Îì´¼£®
£¨4£©ÓÉHÉú³ÉIµÄ»¯Ñ§·½³ÌʽΪ£®
£¨5£©JµÄ½á¹¹¼òʽΪ£®
£¨6£©GµÄͬ·ÖÒì¹¹ÌåÖУ¬±½»·ÉϵÄÒ»ÂÈ´úÎïÖ»ÓÐÒ»ÖֵĹ²ÓÐ8ÖÖ£¨²»º¬Á¢ÌåÒì¹¹£©£¬ºË´Å¹²ÕñÇâÆ×ÏÔʾ2×é·åµÄÊÇ£¨Ð´½á¹¹¼òʽ£©£®

·ÖÎö DÑõ»¯µÃµ½E¡¢EÑõ»¯µÃµ½F£¬ÔòDÖдæÔÚ-CH2OH½á¹¹£¬Eº¬ÓÐ-CHO£¬Fº¬ÓÐ-COOH£¬FÓëI·´Ó¦µÃµ½J£¬JµÄ·Ö×ÓʽΪC15H22O4£¬½áºÏJµÄ½á¹¹Ìص㣬¿ÉÖªFΪC6H13COOH£®A¡¢BºÍEΪͬϵÎÆäÖÐBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª44£¬ÔòBΪCH3CHO£¬A·Ö×ÓÖÐ̼ԭ×ÓÊýĿΪ5£¬AµÄºË´Å¹²ÕñÇâÆ×ÏÔʾÓÐÁ½×é·å£¬ÔòAΪ£¨CH3£©3CCHO£¬¸ù¾ÝÐÅÏ¢¢Û¿ÉÖª£¬AÓëB·´Ó¦µÃµ½C£¬CÓëÇâÆø·´Ó¦µÃµ½D£¬ÔòCΪ£¨CH3£©3CCH=CHCHO£¬DΪ£¨CH3£©3CCH2CH2CH2OH£¬EΪ£¨CH3£©3CCH2CH2CHO£¬FΪ£¨CH3£©3CCH2CH2COOH£®GÓëÂÈÆø¹âÕÕ·¢Éú²àÁ´Ìþ»ùÉÏÈ¡´ú·´Ó¦Éú³ÉH£¬H·¢Éú±´úÌþË®½â·´Ó¦µÃµ½I£¬Ôò
¿ÉÖªIΪ£¬GΪ£¬HΪ£¬FºÍI·¢Éúõ¥»¯·´Ó¦Éú³ÉJΪ£¬¾Ý´Ë½â´ð£®

½â´ð ½â£ºDÑõ»¯µÃµ½E¡¢EÑõ»¯µÃµ½F£¬ÔòDÖдæÔÚ-CH2OH½á¹¹£¬Eº¬ÓÐ-CHO£¬Fº¬ÓÐ-COOH£¬FÓëI·´Ó¦µÃµ½J£¬JµÄ·Ö×ÓʽΪC15H22O4£¬½áºÏJµÄ½á¹¹Ìص㣬¿ÉÖªFΪC6H13COOH£®A¡¢BºÍEΪͬϵÎÆäÖÐBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª44£¬ÔòBΪCH3CHO£¬A·Ö×ÓÖÐ̼ԭ×ÓÊýĿΪ5£¬AµÄºË´Å¹²ÕñÇâÆ×ÏÔʾÓÐÁ½×é·å£¬ÔòAΪ£¨CH3£©3CCHO£¬¸ù¾ÝÐÅÏ¢¢Û¿ÉÖª£¬AÓëB·´Ó¦µÃµ½C£¬CÓëÇâÆø·´Ó¦µÃµ½D£¬ÔòCΪ£¨CH3£©3CCH=CHCHO£¬DΪ£¨CH3£©3CCH2CH2CH2OH£¬EΪ£¨CH3£©3CCH2CH2CHO£¬FΪ£¨CH3£©3CCH2CH2COOH£®GÓëÂÈÆø¹âÕÕ·¢Éú²àÁ´Ìþ»ùÉÏÈ¡´ú·´Ó¦Éú³ÉH£¬H·¢Éú±´úÌþË®½â·´Ó¦µÃµ½I£¬Ôò
¿ÉÖªIΪ£¬GΪ£¬HΪ£¬FºÍI·¢Éúõ¥»¯·´Ó¦Éú³ÉJΪ£¬
£¨1£©GΪ£¬Ëùº¬¹ÙÄÜÍŵÄÃû³ÆÎªôÇ»ù¡¢ÃѼü£¬
¹Ê´ð°¸Îª£ºôÇ»ù¡¢ÃѼü£»
£¨2£©ÓÉAºÍBÉú³ÉCµÄ»¯Ñ§·½³ÌʽΪ£º£¨CH3£©3CCHO+CH3CHO$¡ú_{¡÷}^{Ï¡NaOH}$£¨CH3£©3CCH=CHCHO+H2O£¬
¹Ê´ð°¸Îª£º£¨CH3£©3CCHO+CH3CHO$¡ú_{¡÷}^{Ï¡NaOH}$£¨CH3£©3CCH=CHCHO+H2O£»
£¨3£©ÓÉCÉú³ÉDµÄ·´Ó¦ÀàÐÍΪ¼Ó³É·´Ó¦£¬DΪ£¨CH3£©3CCH2CH2CH2OH£¬Ãû³ÆÎª£º4£¬4-¶þ¼×»ù-1-Îì´¼£¬
¹Ê´ð°¸Îª£º¼Ó³É·´Ó¦£»4£¬4-¶þ¼×»ù-1-Îì´¼£»
 £¨4£©ÓÉHÉú³ÉIµÄ»¯Ñ§·½³ÌʽΪ£º£¬
¹Ê´ð°¸Îª£º£»
£¨5£©JµÄ½á¹¹¼òʽΪ£º£¬
¹Ê´ð°¸Îª£º£»
£¨6£©G£¨£©µÄͬ·ÖÒì¹¹ÌåÖУ¬±½»·ÉϵÄÒ»ÂÈ´úÎïÖ»ÓÐÒ»ÖÖ£¬¿ÉÒÔº¬ÓÐ2¸ö-CH2OHÇÒ´¦ÓÚ¶Ô룬»òº¬ÓÐ2¸ö-OCH3ÇÒ´¦ÓÚ¶Ô룬¿ÉÒÔº¬ÓÐ4¸öÈ¡´ú»ù£¬Îª2¸ö-OH¡¢2¸ö-CH3£¬µ±2¸ö¼×»ùÏàÁÚʱ£¬ÁíÍâ2¸ö-OH·Ö±ðÓë¼×»ùÏàÁÚ£¬»ò·Ö±ðÓë¼×»ù´¦ÓÚ¼ä룬µ±2¸ö¼×»ù´¦ÓÚ¼äλʱ£¬ÁíÍâ2¸ö-OH´¦ÓÚ¶ÔλÇÒ·Ö±ðÓë¼×»ùÏàÁÚ¡¢Ïà¼ä£¬µ±2¸ö¼×»ù¶Ôλʱ£¬2¸ö-OH·Ö±ðÓë2¸ö¼×»ùÏàÁÚÇÒ2¸öôÇ»ù´¦ÓÚ¶Ô룬»òÕß2¸öôÇ»ù´¦ÓÚ¶Ô룬»òÕß2¸ö-OH·Ö±ðÓë1¸ö¼×»ùÏàÁÚ£¬¹Ê·ûºÏÌõ¼þµÄͬ·ÖÒì¹¹¹²ÓÐ8ÖÖ£¬ºË´Å¹²ÕñÇâÆ×ÏÔʾ2×é·åµÄÊÇ£¬
¹Ê´ð°¸Îª£º8£»£®

µãÆÀ ±¾Ì⿼²éÓлúÎïÍÆ¶ÏÓëºÏ³É£¬³ä·ÖÀûÓ÷´Ó¦Ìõ¼þ¡¢JµÄ½á¹¹ÌصãÓë·Ö×Óʽ½øÐÐÍÆ¶Ï£¬ÐèҪѧÉúÊìÁ·ÕÆÎÕ¹ÙÄÜÍŵÄÐÔÖÊÓëת»¯£¬²àÖØ¿¼²éѧÉú·ÖÎöÍÆÀíÄÜÁ¦£¬ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø