2014年4月青奥赛试卷及答案5B

2014年名校联盟《创新》冲刺四月卷-语文（一）参考答案：1．【答案D】解析：A项“粘”应为“zhān”；B项“合”应为“he”； C项“着”应为“zhāo”。

2.【答案B】解析：A项“重点”应为“重典”；C项“尊家”应为“尊驾”；D项“麋集”应为“麇集”。

3．【答案C】解析：A项“动辄”即“动不动就”，与后文的“就”部分重复，可改为“动不动”； B 项“比翼”即“翅膀挨着翅膀（飞）”，引申为“齐头并进”，或喻为“夫妻相伴不离”，人与天“比翼”讲不通，可改为“比高”；C项“夫复何言”即“还能再说什么话呢”，表示感慨很深，此处使用正确；D项“前车之鉴”即“前面翻车的教训”，比喻当做鉴戒的前人的失败教训，句中原意应为前人成功的经验。

4.【答案C】解析：A项表意不明，“已故知名人士和党外全国政协委员夫人”有歧义；B项不合逻辑，“凌晨11点左右”概念混乱，“凌晨”即“接近早晨”，一般指从午夜零时到天亮前的一段时间；D项语序不当，“坚强地在轮椅上”应为“在轮椅上坚强地”。

5. 【答案B】6.【答案示例】生活中切勿因冲动而妄下结论、盲目行动，要学会等待。

7.【答案示例】倘若你会在夜晚出现，我愿用双手把白天扯裂，如同孩子将稿纸撕碎。

8.【答案A】解析：题干语句涵义指向是强调智识的重要性，A项侧重知识。

9.【答案B】解析：B项有价值导向，属智识范畴。

10. ①互联网上的知识丰富易取，但真假莫辨、是非不分，缺乏意义和价值的导向。

②智识可以在纷繁复杂的网络信息中为学生提供道义、信仰的指引，避免知识害人。

11.①显示野鸭与周边环境之间的关系是相互信赖、和谐无间的。

②为下文狗与人悍然打破生态平衡作铺垫，并形成鲜明对比。

12. ①以“鸿沟”为喻，说明野鸭与“我们”之间的不可兼容性。

②以“青石板”与“巢穴”、“泳池”与“池塘”两组事物的对举，表明野生世界与家居世界的冲突。

13. ①看着失去鸭蛋后野鸭的眼睛，“我觉得自己像一个罪犯”，感到“残忍”、“残酷”，对野鸭充满了同情与愧怍。

开始时间：2014-06-13 00:21:56 结束时间：2014-06-13 01:19:58 得分：95.0[返回个人主页]单选题30题、每题1.5分；多选题20题、每题2分；判断题10题、每题1.5分各题对错情况已列出，其中蓝色加粗选项为正确答案一、单选题（共30题、每题1.5分）1. [单选题] 南京青奥会的开闭幕式将在____举行。

南京奥体中心体育场南京奥体中心体育馆南京青奥体育公园南京滨江公园2. [单选题] 青奥组委会应确保青奥会场馆内使用____标识引导所有青奥会客户进入某区域或指定位置。

英语西班牙英语法语法语日语英语葡萄牙语3. [单选题] 南京青奥会参赛队员人数为____。

58403808222028004. [单选题] 南京青奥会最早开始比赛的项目是____，将于 8 月 14 日开始，8 月 27 日结束。

足球排球击剑5. [单选题] 南京青奥会包括____大项____小项。

28 22222 22829 22228 2286. [单选题] 南京青奥会于 2014 年____开幕，于____闭幕。

8 月 16 日 8 月 28 日8 月 12 日 8 月 30 日8 月 10 日 8 月 26 日8 月 16 日 8 月 30 日7. [单选题] 在场馆内，行使职责的人员被授予场馆内具体分区的通行权限，使用颜色、数字或字母在注册卡上标明分区通行权限，注册卡上的____表明有比赛场地区通行权限。

蓝区白区红区紫区8. [单选题] 第二届冬季青年奥林匹克运动会将于 2016 年在____举行。

利勒哈默尔北京赫尔辛基9. [单选题] 为使整个南京城都能参与青奥会的庆祝活动，并为当地社区和游客提供参与青奥盛会的机会，青奥会期间，将在城市举行艺术展览、文化表演、模范运动员见面会及其他文化教育活动。

城市庆祝场地为：新街口南京河西万达广场鼓楼广场绿博园10. [单选题] 电话礼仪中正确的是。

四年级B 卷一、填空题（共10题，每题5分）1. 计算：99999×22222=________.【答案】2222177778【解析】原式=（100000－1）×22222 = 2222200000－22222 = 22221777782. 如下图，长方形AFEB 和长方形FDCE 拼成了长方形ABCD ，长方形ABCD 的长是20，宽是12，则它内部阴影部分的面积是 ．【答案】120FE C BA 【解析】根据面积比例模型可知阴影部分面积等于长方形面积的一半，为120121202⨯⨯=．3. 新上任的宿舍管理员拿着 20 把钥匙去开 20 个房间的门，他知道每把钥匙只能打开其中的一扇门，但不知道哪一把钥匙开哪一扇门．现在要打开所有关闭的 20 扇门，他最多要开 次.【答案】210【解析】从最不利的极端情况考虑：打开第一个房间要 20 次，打开第二个房间需要 19次……最多一共要开 20＋19＋18＋…＋1=210（次）．4. 在一个口袋里装有 15 张小卡片，卡片上分别写着 1 到 15．从袋中任意摸出若干张小卡片，然后算出这些卡片上的各数的和，再将这个和的后两位数写在一张新卡片上放入袋中，如果几个数的和是一位数就将和写在一张新卡片上放入袋中．经过若干次这样的操作后，袋中还剩下一张卡片，这张卡片上的数是 .【答案】20【解析】无论如何选取，选取前后，卡片上数字总和的后两位是不变的．所以，最终所剩卡片上的后两位，也就是原来所有卡片上数字之和的后两位．（1+2+…+15）=（1+15）×15÷2=120，所以最后所剩卡片上的数就是 20．5. A、B 两村相距2800 米，小健从A 村出发步行5 分钟后，小峰骑车从B 村出发，又经过10 分钟两人相遇，已知小峰骑车比小健步行每分钟多行130 米，小健每分钟行米.【答案】60【解析】（2800－130×10）÷（10+10+5）=60（米/分）．6. 判断下图中的3 个图形，哪个图形不能一笔画？图________（请填入相应序号）【答案】图3不可以【解析】图1 能一笔画，因为该图中所有的点都是偶点；图2 能一笔画，因为该图中有两个奇点；图3 不能一笔画，因为图中奇点数超过两个．7. 一筐苹果分成小盒包装，如果每盒装3 个，剩2 个；如果每盒装5 个，剩3 个．如果每盒装6 个，可能剩个？（写出所有可能性）【答案】2 或 5【解析】除以5 余 3 的数从小到大为3、8、13、18……，其中8÷3=2……2，所以除以3 余2，除以5 余 3 的数从小到大排列为8、23、38、53、……，其中8÷6=1……2,23÷6=3……5．因此剩 2 个或 5 个．8. 某台晚会有8 位明星参加演出：两位男歌唱家P 和Q，两位女歌唱家R 和V，两位男喜剧演员T 和W，两位女喜剧演员S 和U．每位明星都是单独表演，并且只表演一次，他们上台表演的次序受以下要求限制：（1）歌唱家和喜剧家必须交替上场；（2）第一个上台表演的必须是女的，并且第二个上台表演的必须是男的；（3）最后一个上台表演的必须是一位男歌唱家；如果已知R 是第四位上台演出的，那么第六位上台演出的一定是.【答案】V【解析】综合（1）（3）说明偶数位出场的都是歌唱家．所以第六位一定是P、Q、R、V 中的一人；综合（2）（3）说明男歌唱家P、Q 一定是在第2、8 位出场；因为R 是第四个出场，所以第六位上台演出的一定是剩下的V．9. 一个自然数，可以分拆成3 个连续自然数之和，也可以分拆成4 个连续自然数之和，还可以分拆成7 个连续自然数之和．这个自然数最小是.【答案】42【解析】由分析可知，所求的自然数为[2、3、7]=42．构造为：42=13+14+15=9+10+11+12=3+4+5+6+7+8+910. 四名棋手每两名选手都要比赛一局，规则规定胜一局得2 分，平一局得1 分，负一局得0 分．比赛结果，没有人全胜，并且每个人的总分都不相同，那么至少有局平局.【答案】1【解析】四人共需进行 6 场比赛，总分为12 分．没有人全胜，所以最高得分不能达到6 分，且四人得分各不相同，故有12=5+4+3+0 或者12=5+4+2+1 两种情况，得分有奇数，必然得有平局，所以最少平一局．两种情况均可构造出来．二、解答题（共5题，每题10分）11. 正方形ABCD 与长方形BEFG 如下图放置，AG=CE=2 厘米，那么正方形ABCD 的面积比长方形BEFG 的面积大多少平方厘米？【答案】412. 在一星期的七天中，狼在星期一、二、三讲假话，其余各天都讲真话；狐狸在星期四、五、六讲假话，其余各天都讲真话．（1）狼说：“昨天是我说谎的日子．”狐狸说：“昨天也是我说谎的日子．”那么今天星期几？（2）一天，狼和狐狸都化了装，使人不容易辨认它们．一个说：“我是狼．”另一个说：“我是狐狸．”那么先说的是狼还是狐狸？这一天是星期几？【答案】（1）星期四（2）狼，星期天【解析】（1）狼只在星期一和星期四才能说：“昨天是我说谎的日子．”因为狼在星期一说谎话，而星期天说真话；而在星期四说真话，在星期三说谎话．狐狸只有在星期四和星期日才能说：“昨天是我说谎的日子．”综合起来，今天是星期四．（2）如果先说的是狼，它讲的是真话，那么后说的就是狐狸，讲的也是真话．同样道理，先说的是狐狸，它讲了假话，那么后说的就是狼，讲的也是假话．因此，它们都讲真话，或者都讲假话．没有一天，狼和狐狸都讲假话，只有星期天，狼和狐狸都讲真话．这一天是星期天，先讲的是狼．13. 20 头驴与 16 匹马分成两队，共重 11000 千克．如果从两队中分别牵出 4 匹马和 4 头驴相交换，两队的体重就相等了，那么每匹马比每头驴重多少千克？【答案】125【解析】16 驴+4 马=12 马+4 驴，12 驴=8 马，即 3 驴=2 马．20 驴+16 马=20 驴+16÷2×3 驴=44 驴=11000，得 1 驴=250，1 马=250÷2×3=375，所以每匹马比每头驴重 375-250=125（千克）．14. 如图，有两个小正方形和一个大正方形，大正方形的边长是小正方形边长的 2倍，阴影部分三角形面积为 240，请问三个正方形的面积和是多少？【答案】360【解析】法 1：此题考查割补法求不规则图形面积．将此图补成正方形，假设边长是 3a ，图中两个非阴影三角形面积之和为 3a×a ，补出的等腰直角三角形面积为2a×a ，阴影面积为22229324240a a a a --==，260a = =60．三个正方形面积 2224360a a a ++=； 法 2：补成大正方形，分成 9 小块，数块数即可15. 将若干个边长为 1 的正六边形（即单位六边形）拼接起来，得到一个拼接图形．例如：那么，要拼接成周长等于 18 的拼接图形，需要多少个单位六边形？【答案】4 或 5 或 6 或 7 个【解析】两个周长 10；周长 10 的嵌进周长 14 的中心，注意只重合 3 条边；周长 12的嵌进周长 14 的中心；七个周长 6 把一个围在中间．。

单选题30题、每题1.5分；多选题20题、每题2分；判断题10题、每题1.5分各题对错情况已列出，其中蓝色加粗选项为正确答案一、单选题（共30题、每题1.5分）1. [单选题] 南京青奥会的开闭幕式将在____举行。

南京奥体中心体育场南京奥体中心体育馆南京青奥体育公园南京滨江公园2. [单选题] 南京青奥会吉祥物的名字是____。

砳砳石石磊磊欢欢3. [单选题] 南京青奥会的口号是____。

分享青春共筑未来青春活力参与共享文化融合智慧创意绿色低碳平安勤廉4. [单选题] 第一届冬季青年奥林匹克运动会于____年，在____举行。

2011 新加坡2012 南京2012 因斯布鲁克2014 东京5. [单选题] 南京青年奥林匹克运动会由____个场馆组成，其中，比赛（兼训练）场馆____个，____个训练场馆。

35 27 834 28 66. [单选题] 南京青奥会吉祥物砳砳，以____为创意源泉。

彩虹青奥会会徽雨花石青奥主题色7. [单选题] 为使整个南京城都能参与青奥会的庆祝活动，并为当地社区和游客提供参与青奥盛会的机会，青奥会期间，将在城市举行艺术展览、文化表演、模范运动员见面会及其他文化教育活动。

城市庆祝场地为：新街口南京河西万达广场鼓楼广场绿博园8. [单选题] 在场馆内，行使职责的人员被授予场馆内具体分区的通行权限，使用颜色、数字或字母在注册卡上标明分区通行权限，注册卡上的____表明有比赛场地区通行权限。

蓝区白区红区紫区9. [单选题] 南京青奥会包括____大项____小项。

28 22222 22829 22228 22810. [单选题] 涉外礼仪十二原则中错误的是____。

求同存异以左为尊11. [单选题] 志愿者要注意自己的公众形象，其中____是正确的。

打饭可以争先恐后吃饭时可以狼吞虎咽吃饭时可以插队用餐时要依次排序打饭12. [单选题] 指引姿势应用____指向方向，拇指自然张开，其余四指并拢，手心向上，手臂自然伸展与肩平行高度。

2014年青奥会知识竞赛试题及答案第一篇：2014年青奥会知识竞赛试题及答案2014年青奥会知识竞赛试题及答案1.____年10月，中国奥林匹克学会成立。

A．1990年B．1991年C．1992年D．1993年正确答案：C2、_______，是我国自行设计、自己建造的第一座铁路、公路双层两用大桥。

A．南京长江大桥B．南京长江第二大桥C．南京长江第三大桥D．南京长江第四大桥正确答案：A3、新加坡青奥会上，______名参赛运动员都给新加坡留下了其个人遗产——在瓷砖上画上了自己喜欢的图形。

P75A．3000B．3528C．30000D．4568 正确答案：B4．1982年4月19日南京市第八届人大常委会第八次会议讨论决定，命名_______为南京市市花。

A．梅花B．兰花C．菊花D．桂花正确答案：A5、2014年青奥会前，南京市出租车全部达到国_____标准，公交车全部达到国Ⅲ以上标准。

P222A.IIB.IIVC.IVD.EI 正确答案：C6、_______横贯南京腹地，是长江下游的一条支流，被南京人称为“母亲河”。

A．秦淮河B．古青溪C．进香河D．胭脂河正确答案：A7．夫子庙位于“十里秦淮”文德桥北岸，原来仅指供奉和祭祀中国古代儒家始祖孔子的庙宇，又称孔庙、文庙或_____庙。

A．文宣王庙B．孔夫子庙C．圣人庙D．儒庙正确答案：A8．辛亥革命胜利后_______领导革命党人在_______建立了中华民国。

A．袁世凯北京B．孙中山广州C．黎元洪北京D．孙中山南京正确答案：D9．莫愁湖位于南京主城区西部的水西门外秦淮河西侧。

相传六朝时______居此，故名。

A．李莫愁B．莫愁女C．莫愁公子D．莫愁道长正确答案：B10．钟山位于南京市区______部，属宁镇山脉西段中支，古称______。

A．东金陵山B．西富贵山C．南中山D．北灵岩山正确答案：A11．首届青奥会“文化与教育活动”的具体内容（1）与冠军交流（2）__________（3）集体活动（4）世界文化村（5）艺术与文化（6）探索之旅（7）海岛冒险——Pulan Ubin探险 P75A．自我发掘B．小组交流C．规划设计D．才艺表演正确答案：A12．明孝陵坐落在南京东郊钟山南麓独龙阜玩珠峰下，是明朝开国皇帝明太祖_______和_______的合葬墓。

得分：100.0[返回个人主页]单选题30题、每题1.5分；多选题20题、每题2分；判断题10题、每题1.5分各题对错情况已列出，其中蓝色加粗选项为正确答案一、单选题（共30题、每题1.5分）1. [单选题] 南京青奥会吉祥物的名字是____。

砳砳石石磊磊欢欢2. [单选题] 南京青年奥林匹克运动会由____个场馆组成，其中，比赛（兼训练）场馆____个，____个训练场馆。

35 27 834 28 628 22 628 20 83. [单选题] 在场馆内，行使职责的人员被授予场馆内具体分区的通行权限，使用颜色、数字或字母在注册卡上标明分区通行权限，注册卡上的____表明有比赛场地区通行权限。

蓝区白区红区紫区4. [单选题] 青奥组委会应确保青奥会场馆内使用____标识引导所有青奥会客户进入某区域或指定位置。

英语西班牙英语法语法语日语英语葡萄牙语5. [单选题] 南京青奥会参赛队员人数为____。

58403808222028006. [单选题] 第一届冬季青年奥林匹克运动会于____年，在____举行。

2011 新加坡2012 南京2012 因斯布鲁克2014 东京7. [单选题] 现任国际奥委会主席是____。

潘基文罗格托马斯.巴赫顾拜旦8. [单选题] 南京青奥会的开闭幕式将在____举行。

南京奥体中心体育场南京奥体中心体育馆南京青奥体育公园南京滨江公园9. [单选题] 第二届冬季青年奥林匹克运动会将于 2016 年在____举行。

利勒哈默尔北京赫尔辛基盐湖城10. [单选题] 在仪容仪表方面眼神要专注，注视对方的____。

两眼与下颌的倒三角区域眼睛鼻子嘴巴11. [单选题] 引领礼仪，志愿者应行走在对方前面____左右，协调对方的行进速度。

半米一米一米半二米12. [单选题] 涉外礼仪十二原则中错误的是____。

求同存异入乡随俗女士优先以左为尊13. [单选题] 制服着装要求中正确的是____。

二、填空题：（本大题共5小题，每小题3分，共15分，请将答案填入答题表二内，否则不给分）
11、一组数据3、8、8、19、19、19、19的众数是__。

12、图（1）（2）是根据某地近两年6月上旬日平均气温情况绘制的折线统计图，通过观察图表，可
以判断这两年6月上旬气温比较稳定的年份是__。

13、如图，已知，在△ABC和△DCB中，AC=DB，若不增加任何字母与辅助线，要使
△ABC≌△DCB，则还需增加一个条件是__。

14、已知：，，，……，若（a、b都是正整数），则a+b的最小值是__。

15、如图，口ABCD中，点E在边AD上，以BE为折痕，将△ABE向上翻折，点A正好落在CD
上的点F，若△FDE的周长为8，△FCB的周长为22，则FC的长为__。

三、解答题：（共7题，共55分）
16、（6分）计算：()0+()-1--|-1|
17、（6分）先化简，再求值：（）÷，其中x=2005
18、（8分）大楼AD的高为10米，远处有一塔BC，某人在楼底A处测得踏顶B处的仰角为60º，爬到楼顶D点测得塔顶B点的仰角为30º，求塔BC的高度。

单选题30题、每题1.5分；多选题20题、每题2分；判断题10题、每题1.5分各题对错情况已列出，其中蓝色加粗选项为正确答案一、单选题（共30题、每题1.5分）1. [单选题] 青年奥林匹克运动会运动员的年龄限制在____之间。

14—1714—1615—1715—182. [单选题] 南京青奥会期间，空气质量将达到中国国家____级标准，符合国际体育赛事举办要求。

一二三四3. [单选题] 第一届冬季青年奥林匹克运动会于____年，在____举行。

2011 新加坡2012 南京2012 因斯布鲁克2014 东京4. [单选题] 南京青奥会运动员村开村时间是2014 年8 月____日，闭村时间是2014 年8月____日。

10 2810 3012 2812 305. [单选题] 南京青奥会于2014 年____开幕，于____闭幕。

8 月16 日8 月28 日8 月12 日8 月30 日8 月10 日8 月26 日8 月16 日8 月30 日6. [单选题] 南京青奥会最早开始比赛的项目是____，将于8 月14 日开始，8 月27 日结束。

足球手球排球击剑7. [单选题] 南京青奥会圣火在____采集。

瑞典希腊南京紫金山新加坡8. [单选题] 南京青奥会的开闭幕式将在____举行。

南京奥体中心体育场南京奥体中心体育馆南京青奥体育公园南京滨江公园9. [单选题] 在场馆内，行使职责的人员被授予场馆内具体分区的通行权限，使用颜色、数字或字母在注册卡上标明分区通行权限，注册卡上的____表明有比赛场地区通行权限。

蓝区白区红区紫区10. [单选题] 送别客人时需要____。

挥手致意与客人一起走比客人先走大声呼喊道别11. [单选题] 在面对媒体时要注意____。

随意发表言论对赛事工作表态发言事先准备，及时报告上岗时接受采访12. [单选题] 涉外礼仪十二原则中错误的是____。

求同存异入乡随俗女士优先以左为尊13. [单选题] 日本韩国等东方国家忌讳的数字是____。

46th International Chemistry OlympiadJuly 25, 2014Hanoi, VietnamTHEORETICAL EXAMINATION WITH ANSWER SHEETS GRADINGCountry:Name as in passport:Student Code:Language:GENERAL INTRODUCTIONYou have additional 15 minutes to read the whole set.This booklet is composed of 9 problems. You have 5 hours to fulfill the problems. Failure to stop after the STOP command may result in zero points for the current task.Write down answers and calculations within the designated boxes. Give your work where required.Use only the pen and calculator provided.The draft papers are provided. If you need more draft paper, use the back side of the paper. Answers on the back side and the draft papers will NOT be marked.There are 52 pages in the booklet including the answer boxes, Cover Sheet and Periodic Table.The official English version is available on demand for clarification only.Need to go to the restroom – raise your hand. You will be guided there.After the STOP signal put your booklet in the envelope (do not seal), leave at your table. Do not leave the room without permission.Physical Constants, Units, Formulas and EquationsAvogadro's constant N A = 6.0221 × 1023 mol–1Universal gas constant R = 8.3145 J·K–1·mol–1Speed of light c = 2.9979 × 108 m·s–1Planck's constant h= 6.6261 × 10–34 J·sStandard pressure p° = 1 bar = 105 PaAtmospheric pressure 1 atm = 1.01325 × 105 Pa = 760 mmHg Zero of the Celsius scale 273.15 KMass of electron m e = 9.1094 × 10–31kg1 nanometer (nm) = 10–9 m ; 1 angstrom (Å) = 10–10 m1 electron volt (eV) = 1.6022 × 10–19 J = 96485 J·mol–1Problem 1. Particles in a box: polyenesIn quantum mechanics, the movement of π electrons along a neutral chain ofconjugated carbon atoms may be modeled using the ‘particle in a box’ method. The energy of the π electrons is given by the following equation:2228mLh n E n = where n is the quantum number (n = 1, 2, 3, …), h is Planck’s constant, m is the mass of electron, and L is the length of the box which may be approximated by L = (k + 2)×1.40 Å (k being the number of conjugated double bonds along the carbon chain in the molecule). A photon with the appropriate wavelength λ may promote a π electron from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). An approximate semi-empirical formula based on this model which relates the wavelength λ, to the number of double bonds k and constant B is as follows:λ (nm) = B )12()2(2++×k k Equation 11. Using this semi-empirical formula with B = 65.01 nm calculate the value of the wavelength λ (nm) for octatetraene (CH 2 = CH – CH = CH – CH = CH – CH = CH 2).Code:Question 1 2 3 4 5 Total Examiner Mark 3 7 6 4 7 27 Theoretical Problem 1 5.0 % of thetotalGrade2.Derive Equation 1 (an expression for the wavelength λ (nm) corresponding to the transfer of an electron from the HOMO to the LUMO) in terms of k and the fundamental constants, and hence calculate theoretical value of the constant B calc..3. We wish to synthesize a linear polyene for which the excitation of a π electron from the HOMO to the LUMO requires an absorption wavelength of close to 600 nm. Using your expression from part 2, determine the number of conjugated double bonds (k) in this polyene and give its structure. [If you did not solve Part 2, use the semi-empirical Equation 1 with B = 65.01 nm to complete Part 3.]Thus, k = 15.So, the formula of polyene is:CH 2 = CH – (CH = CH)13 – CH = CH 22 points4. For the polyene molecule found in Part 3, calculate the difference in energy between the HOMO and the LUMO, ΔE , (kJ·mol –1).In case Part 3 was not solved, take k = 5 to solve this problem.5. The model for a particle in a one-dimensional box can be extended to a three dimensional rectangular box of dimensions L x , L y and L z , yielding the following expression for the allowed energy levels:⎟⎟⎠⎞⎜⎜⎝⎛++=2222222,,8z z y y x x n n n L n L n L n m h E zy xThe three quantum numbers n x , n y , and n z must be integer values and are independentof each other.5.1 Give the expressions for the three different lowest energies, assuming that the box is cubic with a length of L .Levels with the same energy are said to be degenerate. Draw a sketch showing all the energy levels, including any degenerate levels, that correspond to quantum numbers having values of 1 or 2 for a cubic box.Problem 2. Dissociating Gas CycleDininitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide:N 2O 4(g) ⇌ 2NO 2(g)1.00 mole of N 2O 4 was put into an empty vessel with a fixed volume of 24.44 dm 3.The equilibrium gas pressure at 298 K was found to be 1.190 bar. When heated to 348 K, the gas pressure increased to its equilibrium value of 1.886 bar. 1a. Calculate ∆G 0 of the reaction at 298K, assuming the gases are ideal.1b. Calculate ∆H 0 and ∆S 0 of the reaction, assuming that they do not change significantly with temperature.Code: Question 1a 1b 2 3 TotalExaminerMark12 8 3 10 33Theoretical Problem 2 5.0 % of thetotalGrade∆S 0∆G 0348 = - 4.07 kJ = ∆H – 348∆S (1) ∆G 0298 = 4.72 kJ = ∆H – 298∆S (2) (2) - (1) → ∆S = 0.176 kJ·mol –1·K –1 ∆H 0∆H 0 = 4.720 + 298 × 0.176 = 57.2 (kJ·mol –1)4pts 4ptsIf you cannot calculate ∆H 0, use ∆H 0 = 30.0 kJ·mol –1 for further calculations.The tendency of N 2O 4 to dissociate reversibly into NO 2 enables its potential use in advanced power generation systems. A simplified scheme for one such system is shown below in Figure (a). Initially, "cool" N 2O 4 is compressed (1→2) in a compressor (X ), and heated (2→3). Some N 2O 4 dissociates into NO 2. The hot mixture is expanded (3→4) through a turbine (Y ), resulting in a decrease in both temperatureand pressure. The mixture is then cooled further (4→1) in a heat sink (Z ), to promote the reformation of N 2O 4. This recombination reduces the pressure, thus facilitates the compression of N 2O 4 to start a new cycle. All these processes are assumed to take place reversibly.out(a)To understand the benefits of using reversible dissociating gases such as N 2O 4, we will focus on step 3 → 4 and consider an ideal gas turbine working with 1 mol of air (which we assume to be an inert, non-dissociating gas). During the reversible adiabatic expansion in the turbine, no heat is exchanged .2. Give the equation to calculate the work done by the system w(air) during the reversible adiabatic expansion for 1 mol of air during stage 3 → 4. Assume that C v,m(air) (the isochoric molar heat capacity of air) is constant, and the temperature changes from T3 to T4.∆U = q + w; work done by turbine w(air)=-w 1 ptq = 0, thus w(air) = ∆U = C v,m(air)[T3-T4] 2 pts3.Estimate the ratio w(N2O4)/w(air), in which w(N2O4) is the work done by the gas during the reversible adiabatic expansion process 3 → 4 with the cycle working with 1 mol of N2O4, T3 and T4 are the same as in Part 2. Take the conditions at stage 3 to be T3 = 440 K and P3 = 12.156 bar and assume that:(i) the gas is at its equilibrium composition at stage 3;(ii) C v,m for the gas is the same as for air;(iii) the adiabatic expansion in the turbine takes place in a way that the composition of the gas mixture (N2O4 + NO2) is unchanged until the expansion is completed.Oxidation number of Ag1 : ……….+1 Oxidation number of Ag2 : ……… +3 2 pointsCode: Question 1 2 3 4 Total ExaminerMarks 8 14 2 12 36 Theoretical Problem 3 9.0 % of the totalGrade1c.What is the coordination number of O atoms in the lattice of A?The coordination number of O atoms =……… 3 1 point1d.How many Ag I and Ag III bond to one O atom in the lattice of A?Number of Ag I = (1)Number of Ag III = ……. 2 2 points1e.Predict the magnetic behaviour of A. Check the appropriate box below.S2O82-(aq) + 2Ag+(aq) + 2H2O (l) 2SO42-(aq) + Ag I Ag III O2 (s) + 4H+(aq)1 point2. Among the silver oxides which have been crystallographically characterized, the most surprising is probably that compound A is not a Ag II O. Thermochemical cycles are useful to understand this fact. Some standard enthalpy changes (at 298 K) are listed:Atom Standard enthalpyof formation(kJ·mol–1)1st ionization(kJ·mol–1)2nd ionization(kJ·mol–1)3rd ionization(kJ·mol–1)1st electronaffinity(kJ·mol–1)2nd electronaffinity(kJ·mol–1)Cu(g) 337.4 751.7 1964.1 3560.2Ag(g) 284.9 737.2 2080.2 3367.2O(g)249.0 -141.0844.0Compounds ΔH o f (kJ ·mol –1) Ag I Ag III O 2 (s) –24.3 Cu II O (s) –157.3The relationship between the lattice dissociation energy (U lat ) and the lattice dissociation enthalpy (ΔH lat ) for monoatomic ion lattices is: nRT U H lat lat +=Δ, where n is the number of ions in the formula unit.2a. Calculate U lat at 298 K of Ag I Ag III O 2 and Cu II O. Assume that they are ionic compounds. U lat of Ag I Ag III O 2 Calculations:ΔH lat (Ag I Ag III O 2) = 2 ΔH o f (O 2-) + ΔH o f (Ag +) + ΔH o f (Ag 3+) –ΔH o f (Ag I Ag III O 2) = (2×249 – 2 × 141 + 2 × 844) + (284.9 + 737.2) + (284.9 + 737.2 + 2080.2 + 3367.2 ) – (–24.3)= +9419.9 (kJ·mol –1)U lat (Ag I Ag III O 2) = ΔH lat (Ag I Ag III O 2) – 4RT= + 9419.9 – 10.0 = + 9409.9 (kJ·mol –1)3 points(no penalty if negative sign)U lat of Cu II OCalculations for: U lat of Cu II OΔH lat (Cu II O) = ΔH o f (O 2–) + ΔH o f (Cu 2+) – ΔH o f (Cu II O)= (249 – 141 + 844) + (337.4 + 751.7 + 1964.1) – (–157.3)= 4162.5 (kJ ·mol –1)U lat (Cu II O) = ΔH lat (Cu II O) – 2RT = 4162.5 – 5.0 = 4157.5 (kJ ·mol –1)3 points(no penalty if negative sign)If you can not calculate the U lat of Ag I Ag III O 2 and Cu II O, use following values forfurther calculations: U lat of Ag I Ag III O 2 = 8310.0 kJ·mol –1; U lat of Cu II O = 3600.0 kJ·mol –1.The lattice dissociation energies for a range of compounds may be estimated using this simple formula:311C ⎟⎟⎠⎞⎜⎜⎝⎛×=m lat V UWhere: V m (nm 3) is the volume of the formula unit and C (kJ·nm·mol –1) is an empirical constant which has a particular value for each type of lattice with ions of specified charges.The formula unit volumes of some oxides are calculated from crystallographic data as the ratio between the unit cell volume and the number of formula units in the unit cell and listed as below:Oxides V m (nm 3)Cu II O 0.02030 Ag III 2O 3 0.06182 Ag II Ag III 2O 4 0.089852b. Calculate U lat for the hypothetical compound Ag II O. Assume that Ag II O and Cu II O have the same type of lattice, and that V m (Ag II O) = V m (Ag II Ag III 2O 4) – V m (Ag III 2O 3).2c. By constructing an appropriate thermodynamic cycle or otherwise, estimate the enthalpy change for the solid-state transformation from Ag II O to 1 mole of Ag I Ag III O 2. (Use U lat Ag II O = 3180.0 kJ·mol -1 and U lat Ag I Ag III O 2 = 8310.0 kJ·mol -1 if you cannot calculate U lat Ag II O in Part 2b).2Ag IIO (s)Ag I Ag III O 2(s)2Ag 2+(g)+2O 2-(g)Ag +(g)+Ag 3+(g)+2O 2-(g)H rxn2U lat (AgO)+4RT-U lat (Ag I Ag III O)-4RTIE 3(Ag)-IE 2(Ag)Calculations:ΔH rxn = 2U lat (Ag II O) + 4RT + IE 3 – IE 2 – U lat (Ag I Ag III O 2) – 4RT= 2 × 3733.6 + 3367.2 – 2080.2 – 9409.9= – 655.7 (kJ/mol) or - 663.0 kJ/mol using given U lat values 4 pts2d. Indicate which compound is thermodynamically more stable by checking the appropriate box below.3. When Ag I Ag III O 2 is dissolved in aqueous HClO 4 solution, a paramagnetic compound (B ) is first formed then slowly decomposes to form a diamagnetic compound (C ). Given that B and C are the only compounds containing silver formed in these reactions, write down the equations for the formation of B and C .For B :Ag I Ag III O 2 (s) + 4 HClO 4 (aq) 2Ag(ClO 4)2 (aq) + 2 H 2O (l) 1 pointFor C : 4Ag(ClO 4)2 (aq) + 2 H 2O (l)4 AgClO 4 (aq) + 4 HClO 4 (aq) + O 2 (g) 1 point4. Oxidation of Ag+ with powerful oxidizing agents in the presence of appropriate ligands can result in the formation of high-valent silver complexes. A complex Z is synthesized and analyzed by the following procedures:An aqueous solution containing 0.500 g of AgNO3 and 2 mL of pyridine (d = 0.982 g/mL) is added to a stirred, ice-cold aqueous solution of 5.000 g of K2S2O8. The reaction mixture becomes yellow, then an orange solid (Z) is formed which has a mass of 1.719 g when dried.Elemental analysis of Z shows the mass percentages of C, H, N elements are38.96%, 3.28%, 9.09%, respectively.A 0.6164 g Z is added to aqueous NH3. The suspension is boiled to form a clear solution during which stage the complex is destroyed completely. The solution is acidified with excess aqueous HCl and the resulting suspension is filtered, washed and dried (in darkness) to obtain 0.1433 g of white solid (D). The filtrate is collected and treated with excess BaCl2 solution to obtain 0.4668 g (when dry) of white precipitate (E).4a.Determine the empirical formula of Z and calculate the percentage yield in the preparation.4b. Ag (IV) and Ag (V) compounds are extremely unstable and found only in few fluorides. Thus, the formation of their complexes with organic ligands in water can be discounted. To confirm the oxidation number of silver in Z, the effective magnetic moment (µeff ) of Z was determined and found to be 1.78 BM. Use the spin only formula to determine the number of unpaired electrons in Z and the molecular formula of Z. (Z contains a mononuclear complex with only one species of Ag and only one type of ligand in the ligand sphere.)4c. Write down all chemical equations for the preparation of Z, and its analysis.Formation of Z:2Ag+(aq) + 8Py (l) + 3S2O82–(aq) 2[Ag II(Py)4](S2O8) (s) + 2SO42–(aq) 2 ptsDestruction of Z with NH3:[Ag II(Py)4](S2O8) (s) + 6NH3(l) [Ag(NH3)2]+(aq) + ½ N2(g) + 2SO42-(aq)+3NH4+ + 4Py (l) 2 pts(aq)(All reasonable N –containing products and O2 are acceptable)Formation of D:[Ag(NH3)2]+(aq) + 2H+(aq) + Cl– (aq) AgCl (s) + 2NH4+(aq) 1 ptFormation of E:Ba2+(aq) + SO42– (aq) BaSO4(s)1ptProblem 4. Zeise’s Salt1. Zeise's salt, K[PtCl 3C 2H 4], was one of the first organometallic compounds to bereported. W. C. Zeise, a professor at the University of Copenhagen, prepared this compound in 1827 by reacting PtCl 4 with boiling ethanol and then adding potassium chloride (Method 1). This compound may also be prepared by refluxing a mixture of K 2[PtCl 6] and ethanol (Method 2). The commercially available Zeise's salt is commonly prepared from K 2[PtCl 4] and ethylene (Method 3).1a. Write balanced equations for each of the above mentioned preparations of Zeise's salt, given that in methods 1 and 2 the formation of 1 mole of Zeise’s salt consumes 2 moles of ethanol.PtCl 4 + 2 C 2H 5OH → H[PtCl 3C 2H 4] + CH 3CH=O + HCl + H 2O H[PtCl 3C 2H 4] + KCl → K[PtCl 3C 2H 4] + HClK 2[PtCl 6] + 2 C 2H 5OH → K[PtCl 3C 2H 4] + CH 3CH=O + KCl + 2 HCl + H 2O K 2[PtCl 4] + C 2H 4 → K[PtCl 3C 2H 4] + KCl1pt for each (2 pts if the first two reactions combined), total of 4 pts1b. Mass spectrometry of the anion [PtCl 3C 2H 4]– shows one set of peaks with mass numbers 325-337 au and various intensities.Calculate the mass number of the anion which consists of the largest natural abundance isotopes (using given below data).Code: Question 1a 1b 2a 3a 3b 3c Total ExaminerMark 4 1 10 2 6 4 27Theoretical Problem 4 4.0 % of the totalGradeIsotopePt 19278Pt 19478Pt 19578Pt 19678Pt 19878C 126C136Natural abundance,% 0.8 32.9 33.8 25.3 7.2 75.8 24.2 98.9 1.1 99.99Calculations:195 + 3×35 + 2×12 + 4×1 = 328 1 pt2. Some early structures proposed for Zeise’s salt anion were:In structure Z1, Z2, and Z5 both carbons are in the same plane as dashed square. [You should assume that these structures do not undergo any fluxional process byinterchanging two or more sites.]2a. NMR spectroscopy allowed the structure for Zeise’s salt to be determined as structure Z4. For each structure Z1-Z5, indicate in the table below how many hydrogen atoms are in different environments, and how many different environments of hydrogen atoms there are, and how many different environments of carbon atoms there are?StructureNumber of differentenvironments of hydrogen Number of differentenvironments of carbonZ121pt 21 ptZ22 1pt 21 ptZ321pt 21 ptZ41 1pt 11 ptZ52 1pt 11 pt3. For substitution reactions of square platinum(II) complexes, ligands may be arranged in order of their tendency to facilitate substitution in the position trans to themselves (the trans effect). The ordering of ligands is:CO , CN- , C2H4 > PR3 , H- > CH3- , C6H5- , I- , SCN- > Br- > Cl- > Py > NH3 > OH- , H2OIn above series a left ligand has stronger trans effect than a right ligand.Some reactions of Zeise’s salt and the complex [Pt2Cl4(C2H4)2] are given below.3a.Draw the structure of A, given that the molecule of this complex has a centre of symmetry, no Pt-Pt bond, and no bridging alkene.Structure of A2 pt3b.Draw the structures of B, C, D, E, F and G.B1 ptCPtCl NH2C6H5Cl1 ptD1 ptE1 ptF1 ptG1 pt3c.Suggest the driving force(s) for the formation of D and F by choosing one or more of the following statements (for example, i and ii):i) Formation of gasii) Formation of liquidiii) Trans effectiv) Chelate effectStructure D FDriving force(s) i iii and iv2 pts 2 ptsProblem 5. Acid-base Equilibria in WaterA solution (X) contains two weak monoprotic acids (those having one acidicproton); HA with the acid dissociation constant of K HA = 1.74 × 10–7, and HB with the acid dissociation constant of K HB = 1.34 × 10–7. The solution X has a pH of 3.75.1. Titration of 100 mL solution X requires 100 mL of 0.220 M NaOH solution for completion.Calculate the initial (total) concentration (mol·L –1) of each acid in the solution X . Use reasonable approximations where appropriate. [K W = 1.00 × 10–14 at 298 K.]HAH HBH OH Code:Question 1 2 3 4 TotalExaminer Mark 6 4 4 6 20 Theoretical Problem 5 6.5 % of thetotalGrade2. Calculate the pH of the solution Y which initially contains 6.00×10-2 M of NaA and 4.00×10-2 M of NaB.Solution:Solution Y contains NaA 0.06 M and NaB 0.04 M. The solution is basic, OH– was produced from the reactions:NaA + H 2O HA + OH–K b,A = K w/K HA = 5.75 ×10-8NaB + H 2O HB + OH– K b,B = K w/K HB = 7.46 ×10-8H 2O H+ + OH–K w = 1.00 10-14and we have:3. Adding large amounts of distilled water to solution X gives a very (infinitely) dilute solution where the total concentrations of the acids are close to zero. Calculate the percentage of dissociation of each acid in this dilute solution.Solving the equation gives: α = 0.573- The percentage of dissociation of HA = 65.5 %- The percentage of dissociation of HB = 57.3 % 2 points4. A buffer solution is added to solution Y to maintain a pH of10.0. Assume no change in volume of the resulting solution Z.Calculate the solubility (in mol·L–1) of a subtancce M(OH)2 in Z, given that the anions A– and B– can form complexes with M2+:M(OH)2 M2+ + 2OH–K sp = 3.10 ×10-12M2+ + A– [MA]+K 1= 2.1 × 103[MA]+ + A– [MA 2] K2 = 5.0 × 102M2+ + B– [MB]+K’1 = 6.2 × 103[MB]+ + B– [MB 2] K’2 = 3.3 × 102MO H[MB][MBSolve this equation: [A -] = 8.42× 10 –3 M Substitute this value into Eq. 3 and Eq. 4:[MA +] = 0.651 × [A –] = 5.48 × 10 –3 M [MA 2] = 325.5 × [A –]2 = 2.31 × 10 –2 MSimilarly, [B –]total = 0.04 M][92.1][1010.3102.6]][[][432'1−−−−++×=××××==B B B M K MB Eq. 6222'2'12][3.634]][[][−−+×==B B M K K MB Eq.7[B –]total = [B -] + [MB +] + 2 × [MB 2] = 0.04 M Eq. 8 2ptsSubstitute Eq. 6 and Eq. 7 into Eq. 8: [B –] + 1.92 × [B –] + 2 × 634.3 × [B –]2 = 0.04 Solve this equation: [B –] = 4.58 × 10–3 M Substitute this value into Eq. 6 and Eq. 7: [MB +] = 1.92 ×[B –] = 8.79 × 10 –3 M [MB 2] = 634.3 ×[B –]2 = 1.33 × 10–2 MThus, solubility of M(OH)2 in Z is s’s’ = 3.10×10 – 4 + 5.48×10 – 3 + 2.31×10 – 2 + 8.79 × 10 – 3+ 1.33 ×10 – 2 = 5.10×10 – 2 M Answer: Solubility of M(OH)2 in Z = 5.10×10 – 2 M. 2 pointsProblem 6. Chemical KineticsThe transition-metal-catalyzed amination of aryl halides has become one of the mostpowerful methods to synthesize arylamines. The overall reaction for the nickel-catalyzed amination of aryl chloride in basic conditions is:in which NiLL’ is the nickel complex catalyst. The reaction goes through several steps in which the catalyst, reactants, and solvent may be involved in elementary steps.6a. To determine the reaction order with respect to each reactant, the dependence of the initial rate of the reaction on the concentrations of each reagent was carried out with all other reagents present in large excess. Some kinetic data at 298 K are shown in the tables below. (Use the grids if you like)Code: Question 6a6b 6c 6d 6e Total ExaminerMarks 6 8 4 12 2 32 Theoretical Problem 6 7.0 % of thetotalGradeDetermine the order with respect to the reagents assuming they are integers. -Order with respect to [ArCl] = = 1-Order with respect to [NiLL’] = = 1-Order with respect to [L’] = = -1 6pts6b. To study the mechanism for this reaction, 1H, 31P, 19F, and 13C NMR spectroscopy have been used to identify the major transition metal complexes in solution, and the initial rates were measured using reaction calorimetry. An intermediate, NiL(Ar)Cl, may be isolated at room temperature. The first two steps of the overall reaction involve the dissociation of a ligand from NiLL’ (step 1) at 50 o C, followed by the oxidation addition (step 2) of aryl chloride to the NiL at room temperature (rt):Using the steady state approximation, derive an expression for the rate equation for the formation of [NiL(Ar)Cl].(4 pts for rate calculation)The next steps in the overall reaction involve the amine (RNH2) and t BuONa. To determine the order with respect to RNH2 and t BuONa, the dependence of the initial rates of the reaction on the concentrations of these two reagents was carried with the other reagents present in large excess. Some results are shown in the tables below.6c . Determine the order with each of these reagents, assuming each is an integer. (Use the grids if you like)- Order with respect to [NaO t Bu] = 0 2 pts- Order with respect to [RNH 2] = 02 ptsDuring a catalytic cycle, a number of different structures may be involved which include the catalyst. One step in the cycle will be rate-determining.A proposed cycle for the nickel-catalyzed coupling of aryl halides with amines is as follows:6d. Use the steady-state approximation and material balance equation to derive the rate law for d[ArNHR]/dt for the above mechanism in terms of the initial concentration of the catalyst [NiLL’]0 and concentrations of [ArCl], [NH 2R], [NaO t Bu], and [L’].NiLLNiLLApply the steady-state approximation to the concentrations for the intermediates:[NiL][L’] + k [NiL(Ar)HNR] (Equation 1) 1pt(Equation 2) 1pt6e.Give the simplified form of the rate equation in 6d assuming that k1 is very small. d[ArNHR]/dt = - d[ArCl]/dt =k2[ArCl] [NiL] = k1k2 [ArCl][NiLL’]0 / k-1[L’] (i.e. consistent with all the orders of reaction as found in the beginning) 2 ptsProblem 7. Synthesis of Artemisinin(+)-Artemisinin, isolated from Artemisia annua L.(Qinghao, Compositae ) is a potent antimalarial effective against resistant strains of Plasmodium . A simple route for the synthesis of Artemisinin is outlined below.First, pyrolysis of (+)-2-Carene broke the cyclopropane ring forming, among other products, (1R )-(+)-trans -isolimonene A (C 10H 16), which then was subjected to regioselective hydroboration using dicyclohexylborane to give the required alcohol B in 82% yield as a mixture of diastereoisomers. In the next step, B was converted to the corresponding γ,δ-unsaturated acid C in 80% yield by Jones’ oxidation.7a. Draw the structures (with stereochemistry) of the compounds A-C .A B CMeMeHHO4 pts (2 pts if wrong stereochemistry) 4 pts 4 ptsCode: Question 7a 7b 7c 7d 7e 7f Total ExaminerMark128 8 12 12 12 64Theoretical Problem 7 8.0 % of thetotalGradeThe acid C was subjected to iodolactonization using KI, I2 in aqueous. NaHCO3solution to afford diastereomeric iodolactones D and E (which differ in stereochemistry only at C3 ) in 70% yield.7b. Draw the structures (with stereochemistry) of the compounds D and E.The acid C was converted to diastereomeric iodolactones D and E (epimeric at the chiralcenter C3). Look at the number-indicated in the structure F in the next step.D E4 pts 4ptsThe iodolactone D was subjected to an intermolecular radical reaction with ketoneX using tris(trimethylsilyl)silane (TTMSS) and AIBN (azobisisobutyronitrile) in acatalytic amount, refluxing in toluene to yield the corresponding alkylated lactone F in72% yield as a mixture of diastereoisomers which differ only in stereochemistry at C7along with compound G (~10%) and the reduced product H, C10H16O2(<5%).7c. Draw the structures (with stereochemistry) of compound H and the reagent X.Because alkylated lactone F is known, we can deduce the reagent X as methyl vinylketone. H is the reduced product of D.X H2 pts 6 ptsThe keto group of F reacted with ethanedithiol and BF3•Et2O in dichloromethane(DCM) at 0o C to afford two diastereomers: thioketal lactones I and J in nearlyquantitative yield (98%). The thioketalization facilitated the separation of the majorisomer J in which the thioketal group is on the opposite face of the ring to the adjacentmethyl group.7d.Draw the structures (with stereochemistry) of the compounds I and J.The keto group of lactone F reacted with ethanedithiol and BF3·Et2O in dichloromethaneto afford thioketal lactones, I and the major isomer J.I J6 pts (3 pts if I and J are swapped) 6 pts (3 pts if I and J are swapped)The isomer J was further subjected to alkaline hydrolysis followed byesterification with diazomethane providing hydroxy methyl ester K in 50% yield. Thehydroxy methyl ester K was transformed into the keto ester L using PCC (P yridiumC hloro C hromate) as the oxidizing agent in dichloromethane (DCM).A two-dimensional NMR study of the compound L revealed that the twoprotons adjacent to the newly-formed carbonyl group are cis to each other andconfirmed the structure of L.7e. Draw the structures (with stereochemistry) of the compounds K and L .Hydrolysis followed by esterification of J provided hydroxy ester K .Oxidation of the hydroxy group in K by PCC resulted in the keto ester L in which two protons adjacent to the carbonyl group are cis-oriented.K L6 pts 6 ptsThe ketone L was subjected to a Wittig reaction with methoxymethyl triphenylphosphonium chloride and KHMDS (P otassium H exa M ethyl D i S ilazid - a strong, non-nucleophilic base) to furnish the required methyl vinyl ether M in 45% yield. Deprotection of thioketal using HgCl 2, CaCO 3 resulted in the key intermediate N (80%). Finally, the compound N was transformed into the target molecule Artemisinin by photo-oxidation followed by acid hydrolysis with 70% HClO 4.LMN323231. O 2, h υ4。

青奥考试题（B套）及参考答案单选题1 南京青奥会运动员村开村时间是2014年8月____日，闭村时间是2014年8月____日。

（2分）A. 10 28B. 10 30C. 12 28D. 12 30标准答案：D2 南京青奥组委必须实施兴奋剂检测计划，至少进行相当于参赛运动员数量____的检测。

（2分）A. 30%B. 50%C. 70%D. 80%标准答案：A3 第一届夏季青年奥林匹克运动会于____年，在____举行。

（2分）A. 2007 危地马拉B. 2010 新加坡C. 2012 因斯布鲁克D. 2011 柏林标准答案：B4 在场馆内，行使职责的人员被授予场馆内具体分区的通行权限，使用颜色、数字或字母在注册卡上标明分区通行权限，注册卡上的____表明有比赛场地区通行权限。

（2分）A. 蓝区B. 白区C. 红区D. 紫区标准答案：A5 非注册人员临时进入南京青奥会场馆，须办理____。

（2分）A. 一日卡B. 升级卡C. 访客卡D. 注册卡标准答案：C6 为使整个南京城都能参与青奥会的庆祝活动，并为当地社区和游客提供参与青奥盛会的机会，青奥会期间，将在____城市举行艺术展览、文化表演、模范运动员见面会及其他文化教育活动。

城市庆祝场地为：（2分）A. 新街口B. 南京河西万达广场C. 鼓楼广场D. 绿博园标准答案：B7 现任国际奥委会主席是____。

（2分）A. 潘基文B. 罗格C. 托马斯.巴赫D. 顾拜旦标准答案：C8 OBS将对南京青奥会____个大项比赛进行现场直播。

（2分）A. 19B. 28C. 22D. 20标准答案：A9 南京青奥会有____个国家和地区参加。

（2分）A. 204B. 205C. 214D. 21510 南京青奥会的口号是____。

（2分）A. 分享青春共筑未来B. 青春活力参与共享C. 文化融合智慧创意D. 绿色低碳平安勤廉标准答案：A11 南京青奥会包括____大项____小项。