alken izomerlari va bromlash

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powerpoint presentation mavzu: akenlarning olinishi va xossalariga doir masalalar yechish 1- masala. tarkibi c4h8 bo’lgan alken izomerlari bromlaganda qanday moddalar hosil bo’ladi. c4h8 ning izomerlari: 1) buten-1 ch2=ch-ch2-ch3 2) 2-metilpropen ch2=c(ch3)-ch3 3) sis-buten-2 ch3-ch=ch-ch3 4) trans-buten-2 ch3-ch=ch-ch3 1) ch2=ch-ch2-ch3 + br2 → ch2(br)-chbr-ch2-ch3 buten-1 1,2-dibrombutan 2) ch2=c(ch3)-ch3 + br2 → ch2(br)-c(br)(ch3)-ch3 2-metilpropen 1,2-dibrom-2-metilpropan 3) ch3-ch=ch-ch3 + br2 → ch3-ch(br)-ch(br)-ch3 sis va trans buten-2 2,3-dibrombutan 2- masala. 2,3,3-trimetil-2-xlorbutanga natriy gidroksidning spirtdagi eritmasi bilan ta’sirlashidan olingan maxsulotga dastlab vodorod bromid, so’ngra natriy metali ta’sir ettirildi. oxirgi moddani aniqlang. ch3-(br)c(ch3)-c(ch3)2-ch3+naoh(spirt) → ch2=c(ch3)-c(ch3)2-ch3+ hbr 2,3,3-trimetil-2-xlorbutan 2,3,3-trimetilbuten-1 → ch3-(br)c(ch3)-c(ch3)2-ch3 + 2na → ch3-c(ch3)2-c(ch3)2-c(ch3)2-c(ch3)2-ch3 2,3,3-trimetil-2-brombutan 2,2,3,3,4,4,5,5-oktametilgeksan 3- masala. metan, etan va etendan iborat 22,4 litr (n.sh.da) aralashma 32g bromni biriktiradi. gazlar aralashmasining vodorodga nisbatan zichligi 11,3 ga teng bo’lsa, boshlang’ich aralashmadagi etanning massa ulushini aniqlang? 1) ch4 + br2 = birikish reaksiyasi bormaydi; c2h6 + br2 = birikish reaksiyasi bormaydi x = 0,2mol 32g ch2 = …
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6) + (c2h4) = = 8 + 9 + 5,6 = 22,6 g aralashma ω% (c2h6) = 9/22,6 = 0,3982 yoki 39,82% 4- masala. nomalum alkenning 3,36 g massasi kaliy permanganatning suvli eritmasi bilan oksidlanganda 6,96 g massali cho`kma hosil bo`ldi. alkenni aniqlang. 3,36g 6,96g 3cnh2n + 2kmno4 + h2o = 3cnh2n(oh)2 + 2koh + 2mno2↓ 3x= 84 174 x=28. mr(cnh2n)=28: 12n + 2n = 14n: 14n=28: n=2: c2h4-etilen 5- masala. miqdor 0,1 mol bo`lgan nomalum organik modda kaliy permanganatning kislotali eritmasi bilan tasirlashishi natijasida 4,48 l (nsh) karbonat angirid 36,24 g mnso4, 20,88g k2so4 va suv hosil bo`ldi. unglevodorodni aniqlang. 0,1 mol 0,2 mol 0,12mol 0,24mol cnh2n + kmno4 + h2so4 = nco2 + k2so4 + mnso4 + h2o 1 n=2 n=2 c2h4-etilen -2 -2 +7 +4 +2 ch2=ch2 + kmno4 + h2so4 = co2 + k2so4 + mnso4 + h2o 2c-2 -12e- → 2c+4 5 mn+7 + 5e- …
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iyaga kirishmagan propilenning massasi. 79g --------------100% x------------------6% x=4,74g kmno4 x= 1,89g 4,74g 3ch2=ch-ch3+2kmno4+4h2o = 3ch2oh-choh-ch3 +2koh+2mno2 126 316 2) 10,5g – 1,89g = 8,61g propilen polimerlangan 8,61 x=8,61 nch2=ch-ch3 → (- ch2-ch(ch3)- )n 42n 42n 3) 6,02·1023 --------------- 42g 0,008827·1023 ----------x = 0,0616g polipropilen 4) n= 8,61 / 0,0616 = 140 ta molekula polimerlangan 5) mr(polietilen)= 140·42 = 5880 g/mol (- ch2-ch(ch3)- )n mustaqil ishlash uchun masalalar 1. nomalum alkil bromid kaliy gidroksidning spirtli eritmasi bilan ishlov berilib hosil bo`lgan unglevodorodga vodorod bromid qo`shildi. bunda hosil bo`lgan galogenli birkma natriy metali qo`shilib qizdirilganda 2,3-dimetil butan hosil bo`ldi. boshlang`ich alkil bromid(lar)ni aniqlang. 2. nomalum alkil bromid kaliy gidroksidning spirtli eritmasi bilan ishlov berilib hosil bo`lgan unglevodorodga vodorod bromid qo`shildi. bunda hosil bo`lgan galogenli birkma natriy metali qo`shilib qizdirilganda 3,3,4,4-tetrametil butan hosil bo`ldi. boshlang`ich alkil bromid(lar) ni aniqlang. 3. propan, etan va propendan iborat 22,4 litr (n.sh.da) aralashma 48g bromni biriktiradi. gazlar …
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g? 8. noma’lum alken kaliy permanganatning kislotali eritmasida oksidlaganda karbonat angidrid va metilpropilketon hosil qilsa, alkenni toping. 9. buten-2 va metilsiklopropandan iborat 5 litr aralashma va mo’l miqdorda olingan kislorodda portlatilgandan so’ng hosil bo’lgan suv bug’lari kondensatlandi. qolgan gazlar aralashmasi 30 litrni tashkil etsa, reaksiya uchun olingan kislorodni hajmini toping? 10. 25,2g propilenning polimerlanishidan olingan namunada 2,2073·1021 ta makromolekula bor. reaksiyaga kirishmagan propilen 160g 5% li bromli suvni rangsizlantiradi. polipropilenni o’rtacha molekulyar masassini toping.
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powerpoint presentation mavzu: akenlarning olinishi va xossalariga doir masalalar yechish 1- masala. tarkibi c4h8 bo’lgan alken izomerlari bromlaganda qanday moddalar hosil bo’ladi. c4h8 ning izomerlari: 1) buten-1 ch2=ch-ch2-ch3 2) 2-metilpropen ch2=c(ch3)-ch3 3) sis-buten-2 ch3-ch=ch-ch3 4) trans-buten-2 ch3-ch=ch-ch3 1) ch2=ch-ch2-ch3 + br2 → ch2(br)-chbr-ch2-ch3 buten-1 1,2-dibrombutan 2) ch2=c(ch3)-ch3 + br2 → ch2(br)-c(br)(ch3)-ch3 2-metilpropen 1,2-dibrom-2-metilpropan 3) ch3-ch=ch-ch3 + br2 → ch3-ch(br)-ch(br)-ch3 sis va trans buten-2 2,3-dibrombutan 2- masala. 2,3,3-trimetil-2-xlorbutanga natriy gidroksidning spirtdagi eritmasi bilan ta’sirlashidan olingan maxsulotga dastlab vodorod bromid, so’ngra natriy metali ta’sir ettirildi. oxirgi moddani aniqlang. ch3-(br)c(ch3)-c(ch3)2-ch3...

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