Ester - Lipid - vận dụng

Home Lớp 12 Hoá học Lớp 12 - Sách KNTT Bài 1: Ester - LipidBài tập nâng cao

{"common":{"save":0,"post_id":"10114","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"9097","post_id":"10114","mon_id":"1160346","chapter_id":"1158865","question":"Ch\u1ea5t X tác d\u1ee5ng v\u1edbi dung d\u1ecbch NaOH v\u1eeba \u0111\u1ee7, thu \u0111\u01b0\u1ee3c hai ch\u1ea5t Y và Z. Cho Z tác d\u1ee5ng v\u1edbi dung d\u1ecbch AgNO\u2083 trong NH\u2083 thu \u0111\u01b0\u1ee3c ch\u1ea5t h\u1eefu c\u01a1 T. Cho T tác d\u1ee5ng v\u1edbi dung d\u1ecbch NaOH l\u1ea1i thu \u0111\u01b0\u1ee3c ch\u1ea5t Y. Ch\u1ea5t X là?<\/p>","options":["A. CH\u2083COOCH=CH\u2082","B. HCOOCH\u2083","C. CH\u2083COOCH=CH–CH\u2083","D. HCOOCH=CH\u2082"],"correct":"2","level":"3","hint":"G\u1ee3i ý:<\/strong> Hãy \u0111\u1ec3 ý chu\u1ed7i ph\u1ea3n \u1ee9ng: X + NaOH → Y + Z. Z có th\u1ec3 tráng b\u1ea1c t\u1ea1o T. T tác d\u1ee5ng NaOH → l\u1ea1i thu \u0111\u01b0\u1ee3c Y. V\u1eady Z là HCHO (có ph\u1ea3n \u1ee9ng tráng b\u1ea1c), T là HCOONa. Y chính là CH\u2083OH.<\/p>","answer":"\u0110áp án \u0111úng: B. HCOOCH\u2083 (metyl fomat).<\/span><\/strong><\/p>Gi\u1ea3i thích:<\/strong><\/p><ul data-end=\"844\" data-start=\"251\"> <li data-end=\"654\" data-start=\"251\">N\u1ebfu X = HCOOCH\u2083 (metyl fomat):<\/p> <ul data-end=\"654\" data-start=\"288\"> <li data-end=\"348\" data-start=\"288\">Th\u1ee7y phân ki\u1ec1m: HCOOCH\u2083 + NaOH → HCOONa (Y) + CH\u2083OH (Z).<\/p><\/li> <li data-end=\"424\" data-start=\"351\">CH\u2083OH không cho tráng b\u1ea1c → sai? Xem l\u1ea1i: \u0110\u1ec1 nói Z tráng b\u1ea1c → t\u1ea1o T.<\/p><\/li> <li data-end=\"654\" data-start=\"427\">Th\u1ef1c t\u1ebf: v\u1edbi HCOOCH\u2083, s\u1ea3n ph\u1ea9m là HCOONa (mu\u1ed1i, g\u1ecdi Y) và CH\u2083OH (r\u01b0\u1ee3u, g\u1ecdi Z). \u1ede \u0111ây \u0111\u1ec1 tráo ký hi\u1ec7u Y, Z. N\u1ebfu coi Z = HCOONa thì ph\u1ea3n \u1ee9ng v\u1edbi AgNO\u2083\/NH\u2083 t\u1ea1o Ag và HCOONH\u2084 (T). T tác d\u1ee5ng v\u1edbi NaOH l\u1ea1i cho HCOONa (Y). H\u1ee3p lý.<\/p><\/li> <\/ul> <\/li> <li data-end=\"844\" data-start=\"656\">Các ch\u1ea5t khác:<\/p> <ul data-end=\"844\" data-start=\"677\"> <li data-end=\"765\" data-start=\"677\">CH\u2083COOCH=CH\u2082 hay CH\u2083COOCH=CH–CH\u2083 th\u1ee7y phân cho s\u1ea3n ph\u1ea9m không có ph\u1ea3n \u1ee9ng tráng b\u1ea1c.<\/p><\/li> <li data-end=\"844\" data-start=\"768\">HCOOCH=CH\u2082 th\u1ee7y phân cho HCOOH và r\u01b0\u1ee3u vinyl, không th\u1ecfa chu\u1ed7i bi\u1ebfn \u0111\u1ed5i.<\/p><\/li> <\/ul> <\/li><\/ul>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-10-04 09:18:51","option_type":"txt","len":2},{"id":"9098","post_id":"10114","mon_id":"1160346","chapter_id":"1158865","question":"\u0110\u1ed1t cháy hoàn toàn 0,05 mol este X thu \u0111\u01b0\u1ee3c 3,7185 lít khí CO\u2082 (\u0111kc) và 2,7 gam H\u2082O. Công th\u1ee9c phân t\u1eed c\u1ee7a X là:<\/p>","options":["A. C\u2083H\u2084O\u2082 \n ","B. C\u2082H\u2084O\u2082 \n ","C. C\u2083H\u2086O\u2082 \n ","D. C\u2084H\u2086O\u2082"],"correct":"3","level":"3","hint":"G\u1ee3i ý:<\/strong> T\u1eeb s\u1ed1 mol CO\u2082 và H\u2082O, xác \u0111\u1ecbnh s\u1ed1 nguyên t\u1eed C, H trong phân t\u1eed X. So sánh v\u1edbi s\u1ed1 mol ch\u1ea5t ban \u0111\u1ea7u \u0111\u1ec3 tìm công th\u1ee9c phân t\u1eed phù h\u1ee3p.<\/p>","answer":"\u0110áp án \u0111úng: C. C\u2083H\u2086O\u2082<\/span><\/strong><\/p>Gi\u1ea3i thích:<\/strong><\/p><ul data-end=\"347\" data-is-last-node=\"\" data-is-only-node=\"\" data-start=\"166\"> <li data-end=\"202\" data-start=\"166\">nCO\u2082 = 3,7185 : 22,4 = 0,166 mol<\/p><\/li> <li data-end=\"233\" data-start=\"203\">nH\u2082O = 2,7 : 18 = 0,15 mol<\/p><\/li> <li data-end=\"261\" data-start=\"234\">S\u1ed1 C = 0,166 : 0,05 ≈ 3<\/p><\/li> <li data-end=\"347\" data-is-last-node=\"\" data-start=\"262\">S\u1ed1 H = 0,15 × 2 : 0,05 = 6 → Công th\u1ee9c g\u1ea7n \u0111úng C\u2083H\u2086O\u2082, \u0111úng d\u1ea1ng este CnH2nO2.<\/p><\/li><\/ul>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-10-04 09:20:33","option_type":"txt","len":1},{"id":"9099","post_id":"10114","mon_id":"1160346","chapter_id":"1158865","question":"Th\u1ee7y phân hoàn toàn h\u1ed7n h\u1ee3p ethyl propionate và ethyl formate trong dung d\u1ecbch NaOH, thu \u0111\u01b0\u1ee3c s\u1ea3n ph\u1ea9m g\u1ed3m<\/p>","options":["A. 1 mu\u1ed1i và 1 alcohol.","B. 2 mu\u1ed1i và 2 alcohol. ","C. 1 mu\u1ed1i và 2 alcohol.","D. 2 mu\u1ed1i và 1 alcohol."],"correct":"2","level":"0","hint":"G\u1ee3i ý:<\/strong> M\u1ed7i este khi th\u1ee7y phân trong dung d\u1ecbch ki\u1ec1m t\u1ea1o ra m\u1ed9t mu\u1ed1i c\u1ee7a axit t\u01b0\u01a1ng \u1ee9ng và m\u1ed9t r\u01b0\u1ee3u.<\/p>","answer":"\u0110áp án \u0111úng: B. 2 mu\u1ed1i và 2 alcohol.<\/span><\/strong><\/p>Gi\u1ea3i thích:<\/strong><\/p><ul data-end=\"364\" data-is-last-node=\"\" data-is-only-node=\"\" data-start=\"168\"> <li data-end=\"233\" data-start=\"168\">Ethyl propionat (CH\u2083CH\u2082COOC\u2082H\u2085) + NaOH → CH\u2083CH\u2082COONa + C\u2082H\u2085OH<\/p><\/li> <li data-end=\"364\" data-is-last-node=\"\" data-start=\"234\">Ethyl fomat (HCOOC\u2082H\u2085) + NaOH → HCOONa + C\u2082H\u2085OH → T\u1ed5ng c\u1ed9ng thu \u0111\u01b0\u1ee3c 2 mu\u1ed1i (CH\u2083CH\u2082COONa, HCOONa) và 2 r\u01b0\u1ee3u (\u0111\u1ec1u là C\u2082H\u2085OH).<\/p><\/li><\/ul>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-10-04 13:46:19","option_type":"txt","len":2}]}