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Januar 2003, Opgave I, svar. 2-Buten dames i starrst rmngde, da ...

Januar 2003, Opgave I, svar. 2-Buten dames i starrst rmngde, da ...

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<strong>Januar</strong> <strong>2003</strong>, <strong>Opgave</strong> I, <strong>svar</strong>.2-<strong>Buten</strong> <strong><strong>da</strong>mes</strong> i <strong>starrst</strong> <strong>rmngde</strong>, <strong>da</strong> denne isomer indeholder den hqst substituerededobbeltbinding og dermed er mere stabil end I-buten.g. Ketonernefo'srtsaettes naeste side


fol-tsat fra forrige sideIslutproduktH3C-CH2-11C-OH0


<strong>Januar</strong> <strong>2003</strong>, Bpgave 2, <strong>svar</strong>.ACyclohexanCEster1 CH3COCI, AlCk.2 1) NaBH4, 2) H~O+3 - PBr3.4 1) Mg i tPII. ether, 2) CH3COCH3, 3)H30+1 Elektrophil aromatisk substitution (Friedel-Cmfts' reaktion)2 Reduktion.


<strong>Januar</strong> <strong>2003</strong>, <strong>Opgave</strong> 3, <strong>svar</strong>.MarkoMikovs regel: Under addition af HX ti1 en C-C-dobbeltbinding vil Het lcnyttes ti1 det C-atom,der har fzrrest alkylsubstituenter og X'et ti1 det C-atom, der har flest.Reglen forklares ved, at den indledende addition af en proton til en C-C-dobbeltbinding leder til<strong>da</strong>nnelsen af den mest stabile carbokation. Stabilitetec 3f simple carbokationer varierer pi Mgendemide:tertiser > sekundzr > prim=g. Kationen kan ikke optmde i stereoisomere (her enantiomere) former, <strong>da</strong> tilstedevaerelsen afde to ens allcylgrupper pa N (methyl) betyder, at den er identisk med sit spejlbillede.

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