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Energetics RH + X2 → RX + HX 4.14 Halogenation of Alkanes RH + X2 → RX + HX explosive for F2 exothermic for Cl2 and Br2 endothermic for I2 Chlorination of Methane carried out at high temperature (400 °C) 4.15 Chlorination of Methane CH4 + Cl2 → CH3Cl + HCl CH3Cl + Cl2 → CH2Cl2 + HCl CH2Cl2 + Cl2 → CHCl3 + HCl CHCl 3 + Cl2 → CCl4 + HCl Free Radicals 4.16 Structure and Stability of Free Radicals contain unpaired electrons .. .. : O O: Examples: O2 . . .. . : N O: NO .. : Cl . Cl .. Figure 4.17 Structure of Methyl Radical Alkyl Radicals R . C R R Most free radicals in which carbon bears the unpaired electron are too unstable to be isolated. Methyl radical is planar, which suggests that carbon is sp2 hybridized and that the unpaired electron is in a p orbital. Alkyl radicals are classified as primary, secondary, or tertiary in the same way that carbocations are. Alkyl Radicals Alkyl Radicals H R . C . C R R H 3C H less stable than H Methyl radical The order of stability of free radicals is the same as for carbocations. H 3C less stable than . C CH3 . C H Ethyl radical (primary) H 3C A chemical bond can be broken in two different ways— ways—heterolytically or homolytically. tert-Butyl tert-Butyl radical (tertiary) 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. Heterolytic CH3 CH3 Homolytic Alkyl Radicals By "bond strength" we mean the energy required to break a covalent bond. . C less stable than H Isopropyl radical (secondary) The order of stability of free radicals can be determined by measuring bond strengths. H – + Homolytic The species formed by a homolytic bond cleavage of a neutral molecule are free radicals. Therefore, measure enthalpy 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. Measures of Free Radical Stability Bond-dissociation enthalpy measurements tell us that isopropyl radical is 13 kJ/mol more stable than propyl. . CH3CH2CH2 . + CH3CHCH 3 . H + 410 397 H. CH3CH2CH3 Measures of Free Radical Stability Bond-dissociation enthalpy measurements tell us that tert-butyl tert-butyl radical is 30 kJ/mol more stable than isobutyl. . (CH3)2CHCH 2 + (CH3)3C . H. + 410 380 H. 4.17 Mechanism of Chlorination of Methane (CH3)3CH Mechanism of Chlorination of Methane Free-radical chain mechanism. Initiation step: (Light or Heat is Necessary) .. .. .. .. : Cl: Cl : : Cl . + . Cl .. : .. .. .. The initiation step "gets the reaction going" by producing free radicals— radicals—chlorine atoms from chlorine molecules in this case. Initiation step is followed by propagation steps. Each propagation step consumes one free radical but generates another one. Mechanism of Chlorination of Methane First propagation step: H 3C : H + .. . Cl: Cl: .. Second propagation step: .. .. : H3C . + : Cl: Cl .. .. H 3C . .. Cl + H : Cl: .. : .. .. Cl: H3C: Cl: Cl: + . Cl: .. .. Mechanism of Chlorination of Methane First propagation step: H 3C : H + .. . Cl: Cl: .. Second propagation step: .. .. : H3C . + : Cl: Cl .. .. .. .. : Cl : H3C : H + : Cl .. .. Almost all of the product is formed by repetitive cycles of the two propagation steps. First propagation step: H 3C . .. Cl + H : Cl: .. : .. .. Cl: H3C: Cl: Cl: + . Cl: .. .. .. .. H3C: Cl : + H : Cl: Cl .. .. : Termination Steps stop chain reaction by consuming free radicals .. H3C . + . Cl: Cl: .. .. H3C: Cl: Cl: .. hardly any product is formed by termination step because concentration of free radicals at any instant is extremely low H 3C : H + .. . Cl: Cl: .. H 3C . Second propagation step: .. .. : H3C . + : Cl: Cl .. .. .. Cl + H : Cl: .. : .. .. Cl: H3C: Cl: Cl: + . Cl: .. .. .. .. H3C: Cl : + H : Cl: Cl .. .. : .. .. : Cl : H3C : H + : Cl .. .. Question 11 The step shown below is a _____________ step of the free-radical chlorination of chloromethane. A) initiation B) propagation C) chain-terminating D) bond cleavage Question 16 For the free-radical reaction below, light is involved in which of the following reaction steps? 4.18 Halogenation of Higher Alkanes A) Initiation only B) Propagation only C) Termination only D) Initiation and propagation Chlorination of Alkanes can be used to prepare alkyl chlorides from alkanes in which all of the hydrogens are equivalent to one another 420° 420°C CH3CH3 + Cl2 CH3CH2Cl + HCl (78%) hν + Cl2 Cl + HCl Chlorination of Alkanes Major limitation: Chlorination gives every possible monochloride derived from original carbon skeleton. Not much difference in reactivity of different hydrogens in molecule. (73%) Example Chlorination of butane gives a mixture of 1-chlorobutane and 2-chlorobutane. Percentage of Product that Results from Substitution of Indicated Hydrogen if Every Collision with Chlorine Atoms is Productive (28%) 10% CH3CH2CH2CH2Cl CH3CH2CH2CH3 Cl2 10% 10% 10% hν CH3CHCH 2CH3 (72%) 10% Cl 18% 4.6% 4.6% 18% 4.6% 4.6% 4.6% 18% 18% 4.6 =1 4.6 18 4.6 4.6% 10% Relative Rates of Hydrogen Atom Abstraction divide by 4.6 18% 10% 10% 10% Percentage of Product that Actually Results from Replacement of Indicated Hydrogen 4.6% 10% = 3.9 A secondary hydrogen is abstracted 3.9 times faster than a primary hydrogen by a chlorine atom. Similarly, chlorination of 2-methylbutane gives a mixture of isobutyl chloride and tert-butyl tert-butyl chloride (63%) Question 10 CH3 CH3CCH2Cl CH3 CH3CCH3 H Cl2 A) two B) three hν H How many monochlorination products do you expect to obtain from the chlorination of 2-methylbutane? CH3 (37%) C) four D) five CH3CCH3 Cl Percentage of Product that Results from Replacement of Indicated Hydrogen Answer 10 How many monochlorination products do you expect to obtain from the chlorination of 2-methylbutane? A) two 7.0% B) three C) four 37% D) five Relative Rates of Hydrogen Atom Abstraction Selectivity of Free-radical Halogenation R3CH divide by 7 7.0 7 =1 37 7 chlorination: = 5.3 bromination: 5 1640 > R2CH2 > RCH3 4 1 82 1 Chlorination of an alkane gives a mixture of every possible isomer having the same skeleton as the starting alkane. Useful for synthesis only when all hydrogens in a molecule are equivalent. A tertiary hydrogen is abstracted 5.3 times faster than a primary hydrogen by a chlorine atom. Bromination is highly regioselective for substitution of tertiary hydrogens. Major synthetic application is in synthesis of tertiary alkyl bromides. Synthetic Application of Chlorination of an Alkane Cl Cl2 hν (64%) Chlorination is useful for synthesis only when all of the hydrogens in a molecule are equivalent. Question 20 An alkane with a molecular formula of C8H18 reacts with Cl2 in the presence of light and heat to give a single monochloride C8H17Cl. What is the most reasonable structure for the starting alkane? alkane? A) CH3CH2CH2CH2CH2CH2CH2CH3 B) (CH3CH2)2CHCH 2CH2CH3 C) (CH3)2CHCH 2CH2CH(CH 3)2 D) (CH3)3CC(CH3)3 Synthetic Application of Bromination of an Alkane H CH3CCH2CH2CH3 CH3 Br2 hν Br CH3CCH2CH2CH3 CH3 (76%) (A) Barbamide, a cyanobacterial peptide containing a trichloromethyl group and (B) dysidenin, a barbamide-related compound isolated from a sponge-cyanobacterial association Question 19 Which of the following best describes a mechanistic feature of the free-radical bromination (Br2, light) of 2methylpropane? A) The initiation step involves cleavage of a C-H bond. B) The free-radical (CH3)3C· is produced in one propagation step and reacts with Br 2 in another. C) The reaction is characterized by the homolytic cleavage of the C-Br bond. D) The reaction is concerted; i.e., it occurs in a single step. Bromination is highly selective for substitution of tertiary hydrogens. Major synthetic application is in synthesis of tertiary alkyl bromides.