(xx·北京理综,17分)常用作风信子等香精的定香剂D以及可用作安全玻璃夹层的高分子化合物PVB的合成路线如下:D-|||-(C11H12O 2)-|||-浓H2SO4,△-|||--CFbot O-|||-稀 ,Delta B C square -|||-A (1)NaOH溶液,△-|||-(C2H4O) (2) + Ac-|||-PVAc-|||-① E ② square ③ square square -|||-+-|||-CH2-|||-CH -C(H)_(2) (PvB)-|||-O-|||-CH dfrac (pi )(2)-|||-2CH2CH3已知:Ⅰ.RCHO+R′CH2CHOD-|||-(C11H12O 2)-|||-浓H2SO4,△-|||--CFbot O-|||-稀 ,Delta B C square -|||-A (1)NaOH溶液,△-|||-(C2H4O) (2) + Ac-|||-PVAc-|||-① E ② square ③ square square -|||-+-|||-CH2-|||-CH -C(H)_(2) (PvB)-|||-O-|||-CH dfrac (pi )(2)-|||-2CH2CH3D-|||-(C11H12O 2)-|||-浓H2SO4,△-|||--CFbot O-|||-稀 ,Delta B C square -|||-A (1)NaOH溶液,△-|||-(C2H4O) (2) + Ac-|||-PVAc-|||-① E ② square ③ square square -|||-+-|||-CH2-|||-CH -C(H)_(2) (PvB)-|||-O-|||-CH dfrac (pi )(2)-|||-2CH2CH3 (R、R′表示烃基或氢)Ⅱ.醛与二元醇(如:乙二醇)可生成环状缩醛:D-|||-(C11H12O 2)-|||-浓H2SO4,△-|||--CFbot O-|||-稀 ,Delta B C square -|||-A (1)NaOH溶液,△-|||-(C2H4O) (2) + Ac-|||-PVAc-|||-① E ② square ③ square square -|||-+-|||-CH2-|||-CH -C(H)_(2) (PvB)-|||-O-|||-CH dfrac (pi )(2)-|||-2CH2CH3(1)A的核磁共振氢谱有两种峰.A的名称是________.(2)A与D-|||-(C11H12O 2)-|||-浓H2SO4,△-|||--CFbot O-|||-稀 ,Delta B C square -|||-A (1)NaOH溶液,△-|||-(C2H4O) (2) + Ac-|||-PVAc-|||-① E ② square ③ square square -|||-+-|||-CH2-|||-CH -C(H)_(2) (PvB)-|||-O-|||-CH dfrac (pi )(2)-|||-2CH2CH3合成B的化学方程式是_________________________________________________________________________________________________________________.(3)C为反式结构,由B 还原得到.C的结构式是____________________________.(4)E能使Br2的CCl4溶液褪色.N由A经反应①~③合成.a.①的反应试剂和条件是________________________________________________.b.②的反应类型是________.c.③的化学方程式是__________________________________________________.(5)PVAc由一种单体经加聚反应得到,该单体的结构简式是_____________________.(6)碱性条件下,PVAc完全水解的化学方程式是_____________________________________________________________________________________________________.解析:本题考查有机合成与推断,意在考查考生处理新信息的能力及有机合成中逆推和顺推相结合的推理能力.C2H4O的核磁共振氢谱有两种峰,又能与苯甲醛在碱性、加热条件下反应,结合已知反应 Ⅰ,确定A为乙醛,B的结构简式为D-|||-(C11H12O 2)-|||-浓H2SO4,△-|||--CFbot O-|||-稀 ,Delta B C square -|||-A (1)NaOH溶液,△-|||-(C2H4O) (2) + Ac-|||-PVAc-|||-① E ② square ③ square square -|||-+-|||-CH2-|||-CH -C(H)_(2) (PvB)-|||-O-|||-CH dfrac (pi )(2)-|||-2CH2CH3,因C为反式结构且由B还原得到,确定B到C的过程中只有醛基被还原为羟基,碳碳双键不被还原,则C的结构式为D-|||-(C11H12O 2)-|||-浓H2SO4,△-|||--CFbot O-|||-稀 ,Delta B C square -|||-A (1)NaOH溶液,△-|||-(C2H4O) (2) + Ac-|||-PVAc-|||-① E ② square ③ square square -|||-+-|||-CH2-|||-CH -C(H)_(2) (PvB)-|||-O-|||-CH dfrac (pi )(2)-|||-2CH2CH3.乙醛自身也能发生已知反应 Ⅰ 生成CH3CH===CHCHO,即为E,E能与溴水反应,根据已知反应 Ⅱ 和PVB的结构确定N的结构简式为CH3CH2CH2CHO,E发生还原反应生成F(正丁醇),F经催化氧化生成CH3CH2CH2CHO.高聚物PVB的结构简式中出现D-|||-(C11H12O 2)-|||-浓H2SO4,△-|||--CFbot O-|||-稀 ,Delta B C square -|||-A (1)NaOH溶液,△-|||-(C2H4O) (2) + Ac-|||-PVAc-|||-① E ② square ③ square square -|||-+-|||-CH2-|||-CH -C(H)_(2) (PvB)-|||-O-|||-CH dfrac (pi )(2)-|||-2CH2CH3,不难确定PVA的结构简式为.根据PVAc转化成PVA和M的反应条件,确定该过程为酯的水解,M为羧酸,C中含有羟基,二者发生酯化反应生成D,结合C的结构简式和D的分子式确定M为乙酸.进而推出PVAc的单体为CH3COOCH===CH2,书写PVAc在碱性条件下水解的化学方程式时,n不能漏掉,且有乙酸钠生成.4.(xx·新课标全国Ⅰ理综,6分)分子式为C5H10O2的有机物在酸性条件下可水解为酸和醇,若不考虑立体异构,这些醇和酸重新组合可形成的酯共有( )A.15种 B.28种C.32种 D.40种解析:选D 本题考查同分异构体、意在考查考生同分异构体的书写与判断能力。分子式为C5H10O2的酯可能是HCOOC4H9、CH3COOC3H7、C2H5COOC2H5、C3H7COOCH3,其水解得到的醇分别是C4H9OH(有4种)、C3H7OH(有2种)、C2H5OH、CH3OH,水解得到的酸分别是HCOOH、CH3COOH、CH3CH2COOH、C3H7COOH(有2种)。C4H9OH与上述酸形成的酯有4+4+4+4×2=20种;C3H7OH与上述酸形成的酯有2+2+2+2×2=10种;C2H5OH、CH3OH与上述酸形成的酯都是1+1+1+1×2=5种,以上共有40种。5.(xx·大纲理综,6分)某单官能团有机化合物,只含碳、氢、氧三种元素,相对分子质量为58,完全燃烧时产生等物质的量的CO2和H2O。它可能的结构共有(不考虑立体异构)( )A.4种 B.5种C.6种 D.7种解析:选B 本题考查有机化学基础知识,意在考查考生的推理能力。由于该有机物完全燃烧时产生等物质的量的CO2和H2O,所以分子中C和H的原子个数比为1∶2,再结合该分子的相对分子质量可得其分子式为C3H6O。又由于是单官能团有机化合物,所以它可能的结构有:CH3CH2CHO、CH3COCH3、D-|||-(C11H12O 2)-|||-浓H2SO4,△-|||--CFbot O-|||-稀 ,Delta B C square -|||-A (1)NaOH溶液,△-|||-(C2H4O) (2) + Ac-|||-PVAc-|||-① E ② square ③ square square -|||-+-|||-CH2-|||-CH -C(H)_(2) (PvB)-|||-O-|||-CH dfrac (pi )(2)-|||-2CH2CH3,共5种,B项正确。6.(xx·新课标全国理综,6分)分子式为C5H12O且可与金属钠反应放出氢气的有机化合物有(不考虑立体异构)A.5种 B.6种C.7种 D.8种解析:分子式为C5H12O且可与钠反应放出氢气的有机物是醇,可表示为C5H11—OH,由于C5H11—有3种结构,再结合—OH的取代位置可知,符合题意的同分异构体有8种。.