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CHEM 494 Lecture 7 CHEM 494 University of Illinois Special Topics in Chemistry at Chicago UIC CHEM 494 - Lecture 7 Prof. Duncan Wardrop October 22, 2012 CHEM 494 University of Illinois Special Topics in Chemistry at Chicago UIC Halogenation of Alkanes Methods and Mechanism Chapter 39 Structure of Alkyl Radical Intermediates • radicals are high energy intermediates; 7 valence electrons; cannot be isolated • sp2-hybridized; contain one empty p- orbital; unpaired electron in the p- C H orbital; H C 3 H • approximately planar: three bonds to carbon are at ~120º angles from each other and ~90º to half-!lled p-orbital alkyl radicals can be stabilized by inductive effects and hyperconjugation; • stabilized by inductive effects and similar to carbocations hyperconjugation • Stability: 3º > 2º >> 1º > CH3 University of Slide CHEM 494, Spring 2010 3 Illinois at Chicago UIC Lecture 7: October 22 Stabilizing Effects on Alkyl Radicals • smallest inductive effect • largest inductive effect • no hyperconjugation • most hyperconjugation University of Slide CHEM 494, Spring 2010 4 Illinois at Chicago UIC Lecture 7: October 22 Bromination is More Selective Than Chlorination H H † Cl † ∆Ea (chlorination) Hammond Postulate H3C CH3 ‡ • chlorine radicals are higher in energy than bromine radicals = ∆Ea (bromination) H H ‡ • transition states in chlorination are earlier= Br H3C CH3 • look more like reactants = H H • less difference in TS energy = H3C CH2 • less selective = H • greater mixture † = early transition state structures ‡ = late transition state structures H3C CH3 • bromine radicals are lower in energy than ∆Ea (bromination) > ∆Ea (chlorination) chlorine radicals = Bromination is more selective. • transition states in bromination are later= Relative Rates (krel) of Halogenation • look more like products (radical interm.) = R3CH R2CH2 RCH3 • greater difference in TS energy = (tertiary, 3º) (secondary, 2º) (primary, 1º) more selective = chlorination 5.2 3.9 1.0 • bromination 1640 82 1.0 • less of a mixture University of Slide CHEM 494, Spring 2010 5 Illinois at Chicago UIC Lecture 7: October 22 Quantifying Selectivity Relative Rates (krel) of Halogenation R3CH R2CH2 RCH3 (krel) x (statistical factor) (tertiary, 3º) (secondary, 2º) (primary, 1º) % = chlorination 5.2 3.9 1.0 total bromination 1640 82 1.0 Predicted Product Ratios Product Relative Yield Absolute Yield H A (2 2º H’s) 2 x 3.9 = 7.8 7.8/13.8 = 57% H H Cl2 H Cl H + H3C CH3 H3C CH3 H3C CH2 B (6 1º H’s) 6 x 1 = 6.0 6.0/13.8 = 43% Cl A: 57% B: 43% Sum 13.8 100% chlorination A (2 2º H’s) 2 x 82 = 164 164/170 = 96% H H Br2 H Br H H + B (6 1º H’s) 6 x 1 = 6.0 6.0/170 = 4% H3C CH3 H3C CH3 H3C CH2 Br Sum 170 100% A: 96% B: 4% bromination University of Slide CHEM 494, Spring 2010 6 Illinois at Chicago UIC Lecture 7: October 22 Self Test Question Determine the predicted product distribution for A in the following chlorination. Br2 A. 99% + Br Br B. 97% C. 95% A B D. 93% Relative Rates (krel) of Halogenation R3CH R2CH2 RCH3 E. 91% (tertiary, 3º) (secondary, 2º)(primary, 1º) chlorination 5.2 3.9 1.0 bromination 1640 82 1.0 University of Slide CHEM 494, Spring 2010 7 Illinois at Chicago UIC Lecture 7: October 22 De"ning Regioselectivity Regioselectivity (regioselective) A regioselective reaction is one in which one direction of bond making or breaking occurs preferentially over all other possible directions. Reactions are termed completely (100%) regioselective if the discrimination is complete, or partially (<100%), if the product of reaction at one site predominates over the product of reaction at other sites. The discrimination may also semi-quantitatively be referred to as high or low regioselectivity. IUPAC Compendium of Chemical Terminology 2nd Edition (1997) Br O N O CCl , hν 4 Cl Cl Cl Regioselective Chlorination........ .......not stereoselective! University of Slide CHEM 494, Spring 2010 8 Illinois at Chicago UIC Lecture 7: October 22 Mechanism of Alkane Chlorination homolysis n o i Cl Cl Cl + Cl t a i light (hν) t i 7 valence n electrons I hydrogen H abstraction H Cl H C CH3 C CH3 + HCl n H H o i t 7 valence a ~1,000,000 cycles electrons g per initiation step a p o r halogen P H abstraction H Cl Cl C CH 3 Cl C CH3 + Cl H H University of Slide CHEM 494, Spring 2010 9 Illinois at Chicago UIC Lecture 7: October 22 Factors Governing Regioselectivity Cl Cl2 + Cl hν 45% 55% • Strength of C-H Bond Broken Cl2 Cl + Cl • Number & Type of Substrate C-H Bonds hν 63% 37% • Strength of C-X Bond Formed and X-X Bond Broken Br Br2 + Br hν >99% <1% University of Slide CHEM 494, Spring 2010 10 Illinois at Chicago UIC Lecture 7: October 22 Bromination is More Selective Than Chlorination H H † Cl † ∆Ea (chlorination) H3C CH3 Hammond Postulate ‡ • chlorine radicals are higher in energy than bromine radicals = ∆Ea (bromination) H H ‡ • transition states in chlorination are earlier = Br look more like reactants = H3C CH3 • H H • less difference in TS energy = less selective = H C CH • 3 2 • greater mixture H † = early transition state structures ‡ = late transition state structures H3C CH3 • bromine radicals are lower in energy than ∆Ea (bromination) > ∆Ea (chlorination) chlorine radicals = Bromination is more selective. • transition states in bromination are later = Relative Rates (krel) of Halogenation • look more like products (radical interm.) = R3CH R2CH2 RCH3 • greater difference in TS energy = (tertiary, 3º) (secondary, 2º) (primary, 1º) • more selective = chlorination 5.2 3.9 1.0 bromination 1640 82 1.0 • less of a mixture University of Slide CHEM 494, Spring 2010 11 Illinois at Chicago UIC Lecture 7: October 22 Quantifying Selectivity of Halogenation Relative Rates (krel) of Halogenation R3CH R2CH2 RCH3 (krel) x (statistical factor) (tertiary, 3º) (secondary, 2º) (primary, 1º) % = chlorination 5.2 3.9 1.0 total bromination 1640 82 1.0 Predicted Product Ratios Product Relative Yield Absolute Yield H A (2 2º H’s) 2 x 3.9 = 7.8 7.8/13.8 = 57% H H Cl2 H Cl H + H3C CH3 H3C CH3 H3C CH2 B (6 1º H’s) 6 x 1 = 6.0 6.0/13.8 = 43% Cl A: 57% B: 43% Sum 13.8 100% chlorination A (2 2º H’s) 2 x 82 = 164 164/170 = 96% H H Br2 H Br H H + B (6 1º H’s) 6 x 1 = 6.0 6.0/170 = 4% H3C CH3 H3C CH3 H3C CH2 Br Sum 170 100% A: 96% B: 4% bromination University of Slide CHEM 494, Spring 2010 12 Illinois at Chicago UIC Lecture 7: October 22 Self Test Question Determine the predicted product distribution for A in the following chlorination. Br2 A. >99% + Br Br B. 97% C. 95% A B D. 93% Relative Rates (krel) of Halogenation R3CH R2CH2 RCH3 E. 91% (tertiary, 3º) (secondary, 2º)(primary, 1º) chlorination 5.2 3.9 1.0 bromination 1640 82 1.0 University of Slide CHEM 494, Spring 2010 13 Illinois at Chicago UIC Lecture 7: October 22 CHEM 494 University of Illinois Special Topics in Chemistry at Chicago UIC Nomenclature & Stereoisomerism in Alkenes The Terms Alkene and Ole"n are Synonymous H H • Alkenes are hydrocarbons that are C C characterized by a C-C double bond • Also called “ole!ns” H H • General molecular formula = CnH2n • Described as unsaturated since they have two fewer H atoms than ethylene equivalent alkanes C2H4 University of Slide CHEM 494, Spring 2010 15 Illinois at Chicago UIC Lecture 7: October 22 Index of Hydrogen De"ciency (IHD) IHD is synonymous with degrees of unsaturation and indicates the number of double/triple bonds and/or rings in a molecule IHD = 1 (1 double bond) IHD = 4 (3 double bonds & 1 ring) IHD = 2 (2 rings) IHD = 1 (one ring) University of Slide CHEM 494, Spring 2010 16 Illinois at Chicago UIC Lecture 7: October 22 Calculating IHD 1. Hydrocarbons (CnHx) and Oxygenates (CnHxOy): Ignore O (2n+2) - X IHD = 2 2. Compounds with N (CnHxNy): Subtract # N from # H (2n+2) - (X-Y) IHD = 2 3. Halogens (CnHxXy): Add # Halogens to # H (2n+2) - (X+Y) IHD = 2 University of Slide CHEM 494, Spring 2010 17 Illinois at Chicago UIC Lecture 7: October 22 IHD & The Chemistry of Margarine Employing equation 1 (slide 12)........ O linolenic acid OH m.p. -11 °C C18H30O2 IHD = 4 H2 / Ni O linoleic acid OH m.p. -5 °C C18H32O2 IHD = 3 H2 / Ni O oleic acid OH m.p. 16 °C C18H34O2 IHD = 2 H2 / Ni O stearic acid OH m.p. 71 °C C18H36O2 IHD = 1 University of Slide CHEM 494, Spring 2010 18 Illinois at Chicago UIC Lecture 7: October 22 IUPAC Naming of Alkenes Cl H H 2 C C H CH CHCH 1 4 2 3 3 5 Cl 4-chloro-1-pentene or 4-chloropent-1-ene Steps: 1. Number longest chain that includes both alkene carbons so that they have the lowest locants. 2. Replace -ane ending of parent alkane with -ene (alkene). 3. In the name, list the locant of only the !rst alkene carbon. Conventions/Rules: • alkenes have higher priority over alkane and halide substituents when numbering the longest chain • alcohols have higher priority over alkenes when numbering the longest chain • locant may be placed in front of parent name (e.g. 2-pentene) or in front of suffix (e.g. pent-2-ene). substituent 1st alkene locant parent base “ene” alphabetically locant University of Slide CHEM 494, Fall 2012 19 Illinois at Chicago UIC Lecture 7: October 22 IUPAC Naming of Alkenols OH 1 2 H3C CH2CHCH3 5 3 C C 6 4 OH H3C H 4-hexen-2-ol or hex-4-en-2-ol Steps: 1.
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