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Showing 1 - 10 of 115 results
  • Updated 17 months ago
    If E is a strong electrophile, as in the first equation, it will attack the nucleophilic oxygen of the carboxylic acid directly, giving a positively charged intermediate which then loses a proton. If E is a weak electrophile, such as an alkyl halide, it is necessary to convert the carboxylic acid to the more nucleophilic carboxylate anion to facilitate the substitution. The reaction is easily followed by the evolution of nitrogen gas and the disappearance of the reagent's color.
    http://chemwiki.ucdavis.edu/Organic_Chemistry/Carboxylic_Acids/Reactions_of_Carboxylic_Acids/Reactions_of_Carboxylic_Acids/Substitution_of_the_Hydroxyl_Hydrogen
  • Updated 17 months ago
    Epoxidation Hydroxylation Oxidative Cleavage of Double Bonds Contributors William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
    http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Reactions_of_Alkenes/Stereoselectivity_in_Addition_Reactions_to_Double_Bonds/Oxidations
  • Updated 17 months ago
    The stereoselectivity of Brønsted acid addition is sensitive to experimental conditions such as temperature and reagent concentration. The selectivity is often anti, but reports of syn selectivity and non-selectivity are not uncommon. Of all the reagents discussed here, these strong acid additions (E = H in the following equation) come closest to proceeding by the proposed two-step mechanism in which a discrete carbocation intermediate is generated in the first step.
    http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Reactions_of_Alkenes/Stereoselectivity_in_Addition_Reactions_to_Double_Bonds/Br%C3%B8nsted_Acid_Additions
  • Updated 17 months ago
    Since boron is electron deficient (it does not have a valence shell electron octet) the reagent itself is a Lewis acid and can bond to the pi-electrons of a double bond by displacement of the ether moiety from the solvated monomer. Indeed, this hydride shift is believed to occur concurrently with the initial bonding to boron, as shown by the transition state drawn below the equation, so the discrete intermediate shown in the equation is not actually formed.
    http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Reactions_of_Alkenes/Addition_Reactions_of_Alkenes/Addition_of_Lewis_Acids_(Electrophilic_Reagents)
  • Updated 17 months ago
    The radical addition process is unfavorable for HCl and HI because one of the chain steps becomes endothermic (the second for HCl & the first for HI). RCH 2 (CH 3 )CH· + CH 3 CH=CH 2 —> RCH 2 (CH 3 )CH-CH 2 (CH 3 )CH· + CH 3 CH=CH 2 —> RCH 2 (CH 3 )CHCH 2 (CH 3 )CH-CH 2 (CH 3 )CH· —> etc.
    http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Reactions_of_Alkenes/Free_Radical_Reactions_of_Alkanes/Addition_of_Radicals_to_Alkenes
  • Updated 17 months ago
    A non-catalytic procedure for the syn-addition of hydrogen makes use of the unstable compound diimide, N 2 H 2 . This reagent must be freshly generated in the reaction system, usually by oxidation of hydrazine, and the strongly exothermic reaction is favored by the elimination of nitrogen gas (a very stable compound).
    http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Reactions_of_Alkenes/Stereoselectivity_in_Addition_Reactions_to_Double_Bonds/Hydrogenation_of_Alkenes
  • Updated 17 months ago
    If, for example, we wish to carry out an S N 2 reaction of an alcohol with an alkyl halide to produce an ether (the Williamson synthesis), it is necessary to convert the weakly nucleophilic alcohol to its more nucleophilic conjugate base for the reaction to occur. If a 1:1 ratio of amine to alkyl halide is used, only 50% of the amine will react because the remaining amine will be tied up as an ammonium halide salt (remember that one equivalent of the strong acid HX is produced).
    http://chemwiki.ucdavis.edu/Organic_Chemistry/Amines/Reactions_of_Amines/Amine_Reactions
  • Updated 17 months ago
    In basic solution the purple permanganate anion is reduced to the green manganate ion, providing a nice color test for the double bond functional group. A possible explanation is that an empty d-orbital of the electrophilic metal atom extends well beyond the surrounding oxygen atoms and initiates electron transfer from the double bond to the metal, in much the same fashion noted above for platinum.
    http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Reactions_of_Alkenes/Stereoselectivity_in_Addition_Reactions_to_Double_Bonds/Oxidations/Hydroxylation
  • Updated 17 months ago
    Thus, if one molar equivalent of 1,5-hexadiene is treated with one equivalent of bromine a mixture of 5,6-dibromo-1-hexene, 1,2,5,6-tetrabromohexane and unreacted diene is obtained, with the dibromo compound being the major product (about 50%). Note that resonance stabilization of the allyl cation is greater than comparable stabilization of 1,3-butadiene, because charge is delocalized in the former, but created and separated in the latter.
    http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Properties_of_Alkenes/Dienes/Addition_Reactions_of_Dienes
  • Updated 17 months ago
    Since the hydroboration procedure is most commonly used to hydrate alkenes in an anti-Markovnikov fashion, we also need to know the stereoselectivity of the second oxidation reaction, which substitutes a hydroxyl group for the boron atom. In the less complex alkenes used in earlier examples the plane of the double bond was often a plane of symmetry, and addition reagents could approach with equal ease from either side.
    http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Reactions_of_Alkenes/Stereoselectivity_in_Addition_Reactions_to_Double_Bonds/Addition_Reactions_involving_other_Cyclic_Onium_Intermediates
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