高三有机化学题
\u9ad8\u4e09\u6709\u673a\u5316\u5b66\u9898CH4O\u53ea\u6709CH3OH\u4e00\u79cd\u7ed3\u6784
C3H8O\u6709
CH3CH2CH2OH CH3CH(OH)CH3
CH3OCH2CH3
\u8131\u6c34\u65f6\u53ef\u80fd\u53d1\u751f\u7684\u53cd\u5e94\u6709
CH3OH+CH3CH2CH2OH=CH3OCH2CH2CH3+H2O
CH3OH+CH3CH(OH)CH3=CH3CH(CH3)OCH3+H2O
CH3CH(OH)CH3= CH2=CHCH3+H2O
2CH3OH = CH3OCH3 +H2O
2CH3CH2CH2OH = CH3CH2CH2OCH2CH2CH3 +H2O
2CH3CH(OH)CH3 = CH3CH(CH3)OCH(CH3)CH3 +H2O
CH3CH2CH2OH+CH3CH(OH)CH3 = CH3CH(CH3)OCH2CH2CH3 +H2O
\u4e0a\u97627\u4e2a\u53cd\u5e94\uff0c\u4e00\u51717\u79cd\u4ea7\u7269
A\u4e0eNaHCO3\u53cd\u5e94\uff0c\u751f\u6210\u7684\u6c14\u4f53\u4f7f\u77f3\u7070\u6c34\u53d8\u6d51\u6d4a\u3002
\u6c22\u5143\u7d20\u7684\u8d28\u91cf\u5206\u6570 = 1- 68.9% -26.2% = 4.9 %
C\uff1aH\uff1aO = 68.9 /12 \uff1a 4.9 /1 \uff1a26.2 /16 = 5.74 \uff1a4.9\uff1a 1.64 \u2248 7\uff1a 6\uff1a 2
\u6240\u4ee5 \u5206\u5b50\u5f0fC7H6O2
\u7ecf\u68c0\u9a8c\uff0c\u5206\u5b50\u91cf = 122\uff0c\u5206\u5b50\u91cf\u5c0f\u4e8e150
\u7b54\u6848\uff1a\u5206\u5b50\u5f0fC7H6O2
CH3CH2Br + NaOH（水）=== CH3CH2OH + NaBr
CH3CH2OH + O2 === CH3COOH + H2O
CH3COOH + CH3CH2OH === CH3COOC2H5 + H2O
2、
先用上法中的第一个方程制取乙醇，配置氢氧化钠的乙醇溶液
CH3CH2Br + NaOH === NaBr + CH2=CH2 + H2O
CH2=CH2 + Br2 === CH2BrCH2Br
CH2BrCH2Br + 2 NaOH（水）=== CH2OH-CH2OH + 2 NaBr
CH2OH-CH2OH + 2 O2 === HOOC-COOH + 2 H2O
HOOC-COOH + CH2OH-CH2OH === 乙二酸乙二酯 + 2 H2O
1、
CH3CH2Br + NaOH（水）=== CH3CH2OH + NaBr
CH3CH2OH + O2 === CH3COOH + H2O
CH3COOH + CH3CH2OH === CH3COOC2H5 + H2O
2、
先用上法中的第一个方程制取乙醇，配置氢氧化钠的乙醇溶液
CH3CH2Br + NaOH === NaBr + CH2=CH2 + H2O
CH2=CH2 + Br2 === CH2BrCH2Br
CH2BrCH2Br + 2 NaOH（水）=== CH2OH-CH2OH + 2 NaBr
CH2OH-CH2OH + 2 O2 === HOOC-COOH + 2 H2O
HOOC-COOH + CH2OH-CH2OH === 乙二酸乙二酯 + 2 H2O
有点复杂慢慢看
绛旓細1銆丆H3CH2Br + NaOH锛堟按锛=== CH3CH2OH + NaBr CH3CH2OH + O2 === CH3COOH + H2O CH3COOH + CH3CH2OH === CH3COOC2H5 + H2O 2銆佸厛鐢ㄤ笂娉曚腑鐨勭涓涓柟绋嬪埗鍙栦箼閱囷紝閰嶇疆姘㈡哀鍖栭挔鐨勪箼閱囨憾娑 CH3CH2Br + NaOH === NaBr + CH2=CH2 + H2O CH2=CH2 + Br2 === CH2BrCH2Br CH2BrCH2Br +...
绛旓細瑙ｄ竴:宸х敤鏋侀檺 C3H7OH涓阿鐨勫惈閲 13.3% ,C6H12涓阿鐨勫惈閲 14.3% 鏁呴塁 瑙ｄ簩:宸х敤鍙樺紡 C3H6.H2O(C3H7OH)鍜孋6H12鐨勬贩鍚堢墿鍙互鐪嬫垚锛圕H2锛塶銆侶2O鍗充负C6H12鍜孒2O涓ら儴鍒嗙粍鎴愶紝鍥犱负鐨凮鍚噺涓8%锛屽垯H2O 鐨勫惈閲忎负18/16*8%=9 C6H12鐨勫惈閲忎负1-9%=91 H鐨勫惈閲忎负 9%-8%+91%*2/14=14...
绛旓細A涓嶯aHCO3鍙嶅簲锛岀敓鎴愮殑姘斾綋浣跨煶鐏版按鍙樻祽娴娿傛阿鍏冪礌鐨勮川閲忓垎鏁 = 1- 68.9% -26.2% = 4.9 C锛欻锛歄 = 68.9 /12 锛 4.9 /1 锛26.2 /16 = 5.74 锛4.9锛 1.64 鈮 7锛 6锛 2 鎵浠 鍒嗗瓙寮廋7H6O2 缁忔楠岋紝鍒嗗瓙閲 = 122锛屽垎瀛愰噺灏忎簬150 绛旀锛氬垎瀛愬紡C7H6O2 ...
绛旓細鐪嬪浘鍚с
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绛旓細2銆丆D A涓姞鍏ヤ箼閰稿拰涔欓唶閮芥槸浜掓憾 B涓姞鍏ヤ箼閱囦簰婧讹紝鍔犲叆涔欓吀涓拰浜掓憾锛屽姞鍏ヤ箼閰镐箼閰按瑙ｄ簰婧讹紝鐪嬩笉鍑烘潵鐨 C涓姞鍏ヤ箼閰告皵娉★紝鍔犲叆涔欓唶浜掓憾锛屽姞鍏ヤ箼閰镐箼閰垎灞 D涓姞鍏ヤ箼閰稿憟绾㈣壊锛屼箼閱囦簰婧讹紝涔欓吀涔欓叝鍒嗗眰 3銆丄 A閮芥槸浜掓憾鐨 B閮芥槸鍒嗗眰鐨勶紝浣嗚嫰鍦ㄤ笂灞傦紝鍚庤呭湪涓嬪眰 C鐨勮瘽鍔犲叆婧存按鏃犲彉鍖栵紝鍔犲叆鑻拰...
绛旓細鏈鍏抽敭鐐:50ml,骞朵笉鏄敓鎴愭按鐨勪綋绉 瀹屽叏鐕冪儳鍚庣殑浜х墿閫氳繃娴撶～閰革紝姘斾綋浣撶Н鍑忓皯浜50ml,骞朵笉鏄敓鎴愭按鐨勪綋绉 ,鑰屾槸鍓嶅悗浣撶Н鐨勫彉鍖 閫氳繃姘㈡哀鍖栭挔婧舵恫锛屼綋绉張鍑忓皯浜40ml锛屾墍浠ョ噧鐑х敓鎴愪簩姘у寲纰40ml 鎵浠ワ細CnHm + 锛坣+m/4锛塐2 鈫抧CO2 + m/2 H2O 螖v=鍙嶅簲鍓-鍙嶅簲鍚庝綋绉 1---n---1+锛坣+m/...
绛旓細m(CO2)=44*0.2=8.8g 婊ゆ恫璐ㄩ噺姣斿師鐭崇伆姘磋川閲忓噺灏戜簡5.8g锛屽垯鐢熸垚鐨刴(CO2)+m(H2O)=20-5.8=14.2g 鎵浠ョ敓鎴愮殑m(H2O)=14.2-8.8=5.4g n(H2O)=5.4/18=0.3mol 鎵浠鏈夋満鐗╀腑涓暟姣擟:H=0.2:(0.3*2)=1:3 鍙堝洜涓烘湁鏈虹墿鑳戒笌Na鍙嶅簲鐢熸垚姘斾綋锛屽洓涓夐」涓彧鏈塁婊¤冻瑕佹眰锛孋2H6O ...
绛旓細瑙ｆ瀽锛8. Ph-NH-NH2 + H3C-C-CH3 鈫 Ph-NH-N=C-CH3 + H2O || | O CH3 9. H3C-CH2-CH=CH2 --锛圔2H6銆丯aOH/H2O2锛夆啋 H3C-CH2-CH2-CH2-OH O - CH2 \ \
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