第三單元例題解析

1、BrA A. . C6H5CH2Br按按S SN N1 1和和S SN2 2機理都很容易，按機理都很容易，按S SN N1 1形形成中間體成中間體 ，穩(wěn)定，按，穩(wěn)定，按S SN2 2機理形成過渡機理形成過渡態(tài)是共軛體系，能量低。態(tài)是共軛體系，能量低。 CH2+ B(C6H5)2CHBr 按按SN1機理很容易，形成中間體機理很容易，形成中間體(C6H5)2CH很穩(wěn)定；按很穩(wěn)定；按SN2機理進行反應(yīng)難，形成機理進行反應(yīng)難，形成過渡態(tài)時空間阻力大。過渡態(tài)時空間阻力大。 BrNO2BrNO2BrCH3CH3N(CH3)3BrCl+- N(CH3)3BrNO2BrNO2BrCH3CH3Cl+-Et CH

2、BrCHDEt主要產(chǎn)物：主要產(chǎn)物： 少量少量 穩(wěn)定： 不穩(wěn)定：EtCHCDEtEtCHCDEtHDEtHBrEtHDEtHBrEt解：解：因為反因為反-2-氯環(huán)己醇堿性水解有鄰基參與，而氯環(huán)己醇堿性水解有鄰基參與，而順式?jīng)]有。順式?jīng)]有。 ClOHClOOOH2OHOH-OH-Cl-OHCOOHOHOHO2NNO2OHOHOMe解：解： A C D B E 因為因為A為酸，為酸，C的的NO2有有C效應(yīng)、效應(yīng)、I效應(yīng)，效應(yīng)， 穩(wěn)定，穩(wěn)定， D的的NO2有有I效應(yīng)，效應(yīng)，E的的OCH3有有C效應(yīng)。效應(yīng)。OO2N-(CH3)3COCH3Et2OCH3I (CH3)3COHCH3OH (CH3)3CI無

3、水 HI濃HI+ORIH+( )+( )OHRIOHH2SO4Br2+(濃)( )+( )HBrBrHHBrBrHBrHCH2OH+( )BrCH3CH2CCH3BrCH3CH2CH CH2ONa+( )+( )H3CCHC(CH3)2CH2CCH3C2H5OMe(CH3)3COOH2( )-(CH3)3COHCH2OCH3( )+( )CH2OHCH2OHBrH+CH2BrCH2OHCH2OHBrCOHCOHPhPhPhCH3H+( )CCOPhPhCH3PhOHHOCH2CH2CH2CH2Cl( )OCH3CH3O CH2CH CHCH3( )*CH3CH3OHCH2CH CHCH3*(

4、)+( )CH3HICH3CH3OH25 C。CH3HCH3OMeCH3HCH3OMeCH3OHCH3HIHCH3O50 C。( )+-CH3CH2CHCH2CH3OHCH2CHCH2CH3OHMgXCHCH2CH3CH2O+BrMg+Br2+CH2CHCH2CH3H3CCOOOH+Br2BrBrHFeCl3+BrMgMgBr+醚CH2CHCH2CH3H3CCOOOH+CHCH2CH3CH2OCH3COOH+MgXCH2CHCH2CH3OMgBrOH3OCH2CHCH2CH3OH+CCCH3CH2HCH2CH2OHHCH3CH2CC CH2CH2OHCH3CH2CCNaCH2CH2O+NaCC

5、NaCH3CH2Br+CHCHNaCCHCHCHNaNH2NH3CH3CH2BrCH3CH2CCHNH3NaNH2CH3CH2CCNaCH2CH2OCH3CH2CC CH2CH2OHHCCCH3CH2HCH2CH2OHH+液液2Lindlar Pt.低溫低溫高溫高溫OCCOO F：OCCOO E： D:CH2CHCHCH3OH C:CH2CHCHCH3Cl B:ClCH2ClCHCH2CH3 A:CH2CHCH2CH3低溫低溫高溫高溫OCCOO F：OCCOO E： D:CH3CHCHCH2OH C:CH3CHCHCH2Cl B:ClCHCH3CH3ClCH A:CH3CHCHCH3 A可能為可能為:OH(CH