第二十一講——要點復(fù)習1、醚的制備——威廉姆合成的原料選擇。2、環(huán)氧乙烷開環(huán)反應(yīng)——用于制備伯醇。3、取代環(huán)氧乙烷在不同介質(zhì)中的開環(huán)位置。CH3CH2–X+NaO–C–(CH3)3OCH2—CH2CH2–CH2ROMgXH2OCH2–CH2ROHR–MgXCH3–CH—CH2OCH3OHCH3–CH—CH2OHOCH3OCH3–CH–CH2+CH3O–HOH-H+一、醛酮的分類和結(jié)構(gòu)分析二、醛酮的化學性質(zhì)(重點)三、醛酮的制備四、重要的醛酮第九章醛酮在鏈端O–C–H醛在鏈中RO–C–R酮O–C–羰基醛酮的官能團RO–C–?；鵕O–C–HRO–C–RRO–C–OHRO–C–OR′醛酮酸酯…………分析官能團OR—CH—C—H(R’)H一、醛酮的分類和結(jié)構(gòu)分析①羰基親核加成反應(yīng)②-H的反應(yīng)③氧化(以醛為主)與還原反應(yīng)飽和醛酮不飽和醛酮脂肪族醛酮脂環(huán)族醛酮芳香族醛酮根據(jù)烴基分類CH3CH=CH-CHO共軛結(jié)構(gòu)–C–CH3O–CHO羰基直接連芳環(huán)O根據(jù)官能團分類OCH3–C–CH2–C–CH3O2,4-戊二酮(-二酮)CH3–C–CH3OCH3CHO–CH2–C–CH3O加成試劑：HCN；NaHSO3；ROH；NH2–Y；RMgX；Ph3P=CRR′二、醛酮的化學性質(zhì)（一）羰基的親核加成反應(yīng)C＝O+-E+(親電)C—OE+Nu-(親核)C—O-Nu√影響因素①電子效應(yīng)——供電基團，不利于反應(yīng)。②空間效應(yīng)——基團越大，越不利于反應(yīng)。③試劑親核能力——不同試劑反應(yīng)范圍不同。反應(yīng)活性：醛＞酮；甲基酮＞非甲基酮；脂肪族＞芳香族C=OCH3CH3C=ORCH3＞C=OHHC=ORH＞C=OArH＞＞＞C=OArCH3＞C=ORRC=OArAr＞氨的衍生物魏悌希試劑重點一：反應(yīng)形式和產(chǎn)物結(jié)構(gòu)重點二：在合成上的應(yīng)用1、加H-CNC=O+H–CN⑴反應(yīng)范圍：醛、脂肪族甲基酮、8個C以下脂環(huán)酮⑵反應(yīng)條件：堿性，pH≈8.0，有利于CN-離解。⑶應(yīng)用：H2OOHCH3CH3–C–COOHCH3CH2=C–COOHH2SO4△CH3CH2=C–COOCH3CH3OH-甲基丙烯酸甲酯CH3–C–CH3O+H–CNCH3–C–CNOHCH3丙酮羥腈(丙酮氰醇)NaOH-羥基腈(氰醇)C–OHCN

聚合有機玻璃2、加亞硫酸氫鈉(飽和NaHSO3)C=O+Na–SO3H⑴反應(yīng)范圍：醛、脂肪族甲基酮、8個C以下脂環(huán)酮⑵應(yīng)用：①鑒別；②分離SO3HC–ONa-羥基磺酸鈉(晶體)SO3NaC–OH鹽析效應(yīng)練習1：鑒別(分離)環(huán)己酮和3-己酮OCH3CH2CCH2CH2CH3ONaHSO3(飽和)×↓HOSO3NaNa2CO3HClNaCl+SO2+H2ONa2SO3+CO2+H2O強酸②H+O①過濾3、加醇ROHC=O+H–ORORC–OR+H2O縮醛干HClH–OR——縮醛化反應(yīng)半縮醛OHC–OR干HClC=O+H+C=OH+OHC–OR-H+HOROHC–ORH+很活潑，不穩(wěn)定⑵反應(yīng)條件：無水H+條件催化C=O+CH3HH–OC2H5H–OC2H5結(jié)構(gòu)似醚，穩(wěn)定縮醛水解CCH3HOC2H5OC2H5乙醛縮二乙醇H+-H2OH–OCH2H–OCH2CCH3HO–CH2O–CH2乙醛縮乙二醇H+C=O+CH3H-H2O⑴反應(yīng)范圍：醛。酮較難環(huán)己酮縮乙二醇、-羥基醛發(fā)生分子內(nèi)縮醛化反應(yīng)，可得到5、6圓環(huán)狀穩(wěn)定的半縮醛。CH2CH2CH2CHOOH-羥基丁醛干HClCH2–CH2OCH2CHOHCH2–CH2O-HCH2CHOOOO+HOCH2CH2OH+H2O水油分離器加苯69℃苯+水91%+9%苯水H+2HOC2H5CH2=CHCHOC2H5OC2H5丙烯醛縮二乙醇CH2=CHCHOCH2–CHCHOOHOHCH2–CHCHOC2H5OC2H5OHOHKMnO4H+H2OKMnO4?×⑶應(yīng)用：保護醛基練習2：CH3–CH2CHOC2H5OC2H5Ni/H2CH3–CH2CHONi/H2?×H+/H2O4、與氨的衍生物加成縮合C=O+H–NH-YOHC—N–YHC＝N–Y-H2OC=O+H2N–YC＝N–Y-H2O羰基化試劑用于分離、提純、鑒定結(jié)構(gòu)H2N–OHH2N–NH2H2N–NH–PhH2N–NH–NO2NO2(羥胺)(肼)(苯肼)H2N–NH–C–NH2O(2,4-二硝基苯肼)(氨基脲)C＝N–NH–C–NH2O(縮氨脲)C＝N–NH–NO2NO2(2,4-二硝基苯腙)C＝N–NH–Ph(苯腙)C＝N–NH2(腙)C＝N–OH(肟)具有固定熔點的晶體練習3：分析下列物質(zhì)，是由哪些醛酮轉(zhuǎn)變得到的①CH3CH2CSO3NaOHCH3–CH＝N–OH②NNHCONH2③OOHH④苯甲醛肟環(huán)己酮縮氨脲CH2CH2CH2CH2CHOOH-羥基戊醛CH3CH2CCH3+NaHSO3O–CHOH2N＝N–OHOH2N–NH–C–NH2O5、與RMgX(格利雅試劑)加成C=O+R–MgXOMgXC—R干醚H2OOHC—R醇OHC—RHH伯醇HOHC—RR仲醇OHC—RRR叔醇-+C=OHH甲醛C=ORH醛C=ORR酮–MgCl+C=OHH–CH2OH①干醚②H2O–MgCl+C=OCH3CH3①干醚②H2O–C–OHCH3CH3環(huán)己基甲醇2-環(huán)己基-2-丙醇練習4：P382—14(3)用不多于四個C的醇和鹵代烴合成CH3CHCH2–C–CH2CH3CH3OHCH3伯醇仲醇叔醇R–CH2–CH2–OHR2–CH–OHR1R2–C–OHR3R1原料分析：RMgX+C—CORCH2MgX+HCHOR2CHOR1MgX+R2MgX+R1CHOOR2–C–R3R1MgX+CH3CHCH2–MgXCH3OCH3–C–CH2CH3OR1–C–R3R2MgX+OR1–C–R2R3MgX+？？練習5：分析下列各醇用什么原料合成CH3CH2CH2CH2OH③①OH–C–CH3CH3②–CH–CH2CH3OHCH3–C–CH3O+–MgX–C–CH3O+CH3MgX–CHO+CH3CH2MgX–MgXCH3CH2CHO+HCHO+CH3CH2CH2MgX+CH3CH2MgXCH2–CH2OC–C–C–C–CCCOH-H2OC–C=C–C–CCC練習6：2-甲基-1-丙醇、異丙醇2,4-二甲基-2-戊烯HO–C–C–CCOHC–C–CC–C–C–MgXCOC–C–C練習7：苯、乙烯、丙烯3-甲基-1-苯基-2-丁烯–C–C=C–CC–C–C–MgXOC–C–C–C–C–C–CCOH-H2O–C–C–OH–MgXCH2–CH2OC–C–C–C–CCCOH-H2O–C–C–C–CCOH-H2O？以羰基化合物和鹵代烴為原料，通過魏悌希試劑，制備結(jié)構(gòu)較復(fù)雜的烯烴。6、魏悌希(Witting)反應(yīng)(C6H5)3P三苯基膦季鏻鹽+CH3X(C6H5)3P–CH3X+

-強堿性條件消去-H和鹵素(C6H5)3P＝CH2(C6H5)3P–CH2+

-磷葉立德(ylide)yl—基團；ide—鹽鄰位兩性離子CH2O+(C6H5)3P–CH2+

-三苯基亞甲基膦C=O+Ph3P=C(C6H5)3P＝CH2R4N–Cl(季銨鹽)+-NH4Cl＝＝＝＝＝＝C=O+Ph3P=CRR′C＝CRR′Ph3P–CHXRR′+

-季鏻鹽Ph3P=CRR′HO--HX魏悌希試劑Ph3P+CHXRR′(含H)鹵代烴三苯基膦練習8：P379—5(2)用魏悌希試劑合成(CH3)2C＝CHC6H5①(CH3)2CO+Ph3P＝CHC6H5②C6H5CHO+Ph3P＝C(CH3)2Ph3P+CH3CHCH3BrPh3P+C6H5CH2–Br練習7：苯、乙烯、丙烯3-甲基-1-苯基-2-丁烯C–C–C=C–CCC–C–C=C–CC練習6：2-甲基-1-丙醇、異丙醇2,4-二甲基-2-戊烯HO–C–C–CCOHC–C–CC–C–CHOC+PPh3＝C(CH3)2–C–CHO+PPh3＝C(CH3)2–MgXCH2–CH2O+BrC–C–C+Ph3PBrC–C–C+Ph3P–C–C–OHC–C–C＝PPh3C+O＝C(CH3)2–C–C＝PPh3+O＝C(CH3)2第二十二講——總結(jié)1、加NaHSO3——鑒別醛、脂肪族甲基酮、小環(huán)酮2、加ROH——用于保護醛基保護的含義：①轉(zhuǎn)化；②復(fù)原3