kof_6b_radikal_2013

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Radical

description

Radkal bebas

Transcript of kof_6b_radikal_2013

  • Radical

  • Reaction via radicalStructure of radicalRadical stability (BDE)Mekanise radicalOUTLINE

  • carried out at high temperature (400 C)CH4 + Cl2 CH3Cl + HClCH3Cl + Cl2 CH2Cl2 + HClCH2Cl2 + Cl2 CHCl3 + HClCHCl3 + Cl2 CCl4 + HClChlorination of Methane

  • contain unpaired electronsExamples: O2NO..Free Radicals

  • Most free radicals in which carbon bears the unpaired electron are too unstable to be isolated.Alkyl radicals are classified as primary, secondary, or tertiary in the same way that carbocations are.CAlkyl Radicals

  • Structure of methyl radicalMethyl radical is planar, which suggests that carbon is sp2 hybridized and that the unpaired electron is in a p orbital.

  • The order of stability of free radicals can be determined by measuring bond strengths. By "bond strength" we mean the energy required to break a covalent bond. A chemical bond can be broken in two different waysheterolytically or homolytically.Alkyl Radicals

  • In a homolytic bond cleavage, the two electrons in the bond are divided equally between the two atoms. One electron goes with one atom, the second with the other atom. In a heterolytic cleavage, one atom retains both electrons.+HomolyticHeterolytic

  • The species formed by a homolytic bond cleavage of a neutral molecule are free radicals. Therefore, measure energy cost of homolytic bond cleavage to gain information about stability of free radicals.The more stable the free-radical products, the weaker the bond, and the lower the bond-dissociation energy.Homolytic

  • Bond-dissociation energy measurements tell us that isopropyl radical is 13 kJ/mol more stable than propyl.CH3CH2CH3Measures of Free Radical Stability

  • Bond-dissociation energy measurements tell us that tert-butyl radical is 30 kJ/mol more stable than isobutyl.410(CH3)3CH(CH3)2CHCH2+H..380Measures of Free Radical Stability

  • *6.10 The Hammond PostulateIf carbocation intermediate is more stable than another, why is the reaction through the more stable one faster?

    The relative stability of the intermediate is related to an equilibrium constant (DG)

    The relative stability of the transition state (which describes the size of the rate constant) is the activation energy (DG)

    The transition state is transient and cannot be examined

  • *Transition State StructuresA transition state is the highest energy species in a reaction step

    By definition, its structure is not stable enough to exist for one vibration

    But the structure controls the rate of reaction

    So we need to be able to guess about its properties in an informed way

    We classify them in general ways and look for trends in reactivity the conclusions are in the Hammond Postulate

    The structure of the transition state resembles the structure of the nearest stable species. T.S. for Endergonic steps resemble products. T.S. for Exergonic steps resemble reactants.

  • *Examination of the Hammond PostulateA transition state should be similar to an intermediate that is close in energySequential states on a reaction path that are close in energy are likely to be close in structure - G. S. Hammond

  • *Competing Reactions and the Hammond PostulateNormal Expectation: Faster reaction gives stable intermediateIntermediate resembles transition state

  • *The rate of a reaction is dependent of the activation energy (Eact)There is no formal relationship between Gact and G What is the structure of a transition state? How can the structures of the reactants and products affect Gact

    The Hammond Postulate provides an intuitive relationshipBetween rate (Gact) and product stability (G). Typical reaction coordinate less common

  • *The Hammond Postulate: The structure of the transition state more closely resembles the nearest stable species (i.e., the reactant, intermediate or product)

    For an endothermic reaction (G > 0), the TS is nearer to the product. The structure of the TS more closely resembles that of the product. Therefore, factors that stabilize the product will also stabilize the TS leading to that product.

    For an exothermic reaction (G < 0), the TS is nearer to the reactant. The structure of the TS more closely resembles that of the reactants.

    G= 0G> 0G< 0TS is halfway betweenreactant and products onthe reaction coordinateTS lies closer to the products than thereactants on the reaction coordinateTS lies closer to thereactants than theproducts on the reaction coordinate

  • *Fig. 4.16

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