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【答案】(1)羟基[文]、醛基(2分)加成[章](1分)(2)澳化[来]氢(HBr)(1分[自])CH2CHBr([知]1分)OHECH一[嘛]CH2(3)n HOCH—CH=C[答]H2一定条件CH-[案]OH(2分)OH([网]4)CHCOOH或[文]CHOOCH(任写[章]一种)(2分)CH[来];CH,OMgBr[自]OHCH2BrCH[知]:CHOH2O(5[嘛])CH:MgBrC[答]H2 CHCH,CH2C[案]HCH(3分)CH[网]O【解析】A与苯酚[文]发生加成反应得到B[章],B经氧化生成C([来]),乙炔与澳化氢加[自]成生成H2C-CH[知]Br,OHH2C一[嘛]CHBr在Mg和乙[答]醚的作用下生成F,[案]F与C发生已知信息[网]的第二步反应得到G[文],G发生水解得到H[章],H和乙酸酐在冰水[来]浴条件下反应得到W[自],据此分析解题。C[知]HO(1)C为,含[嘛]氧官能团的名称是羟[答]基、醛基；A→B发[案]生酚醛加成反应。O[网]H(2)由分析可知[文]D→E还需要的无机[章]试剂是溴化氢；E的[来]结构简式为CH2一[自]CHBr。OH(3[知])H含有碳碳双键，[嘛]H在一定条件下发生[答]聚合反应的化学方程[案]式为nH0《○》C[网]H-CH-CH2E[文]CH-CH2一定条[章]件CH-OH。OH[来](4)同时满足题给[自]条件的H的同分异构[知]体的结构简式为CH[嘛]COOH或《CHO[答]OCH.CH:CH[案]3(4)同时满足题[网]给条件的H的同分异[文]构体的结构简式为或[章]CHCOOHCHO[来]OCH。CHCH3[自](5)以乙醛、乙醚[知]和甲苯为原料制备C[嘛]H2CCH3,根据[答]已知信息可知CH2[案] BrMe乙醚OMg[网]BrCH:CHOC[文]H2MgBr,结合[章]C十F→G→H的反[来]应可得CH2 MgBrCH2CH[自]CHOHH20CH[知]2CHCH3。-，[嘛]1
Robert was fast sinking.I[答]t seemed as if Robert would be drowned.J[案]ust at the moment,He[网]nry happened tobe passing by.Hearin[文]g the screams,H[章]enry ran to the riverside[来],took off his clothes and jumped into the waterwith[自]out hesitatio[知]n.He reached Robert just as he was sinking the last time.With[嘛] great effort,an[答]d at the risk ofdrownin[案]g himself,h[网]e brought Robert to the shore.Thu[文]s,Robert's life was saved.Rob[章]ert and his new friends were ashamed of having called Henry a coward.Th[来]ey realized that Henry had true courage.T[自]hey said sorry to Henry for their rude behavior and asked for hisforgiv[知]eness.Hen[嘛]ry forgave them happily.H[答]e and Robert were friends again and they went home together asusual.F[案]rom this,Robe[网]rt learned what true courage was:never[文] be afraid to do good,but always be fearful ofdoing evil.
