Chemistry 125: Lecture 38 January 10, 2011 Reaction Rates: Radical-Chain Halogenation, Bond Dissociation Energies, Reaction Rate Laws This For copyright notice see final page of this file

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Grad TAs Jon Miller Phillip Lichtor Senior Peer Tutors Eva Uribe Jack Qian Julia Rogers OPTIONAL!

Semester 1 : Bonds & Molecular Structure (with some thermodynamics) Semester 2 : Reaction Mechanisms & Synthesis (with some spectroscopy)

How Mechanisms are Discovered and Understood in Terms of Structure and Energy Simplest Reactions - Bond Cleavage & Make-as-You-Break Solvent Effects on Ionic Reactions Nucleophilic Substitution and Elimination: Proving Mechanisms Exam 5 – February 2 Free-Radical Substitution: Reactivity and Selectivity Electrophilic Addition to Alkenes and Alkynes (and the Role of Nucleophiles) Conjugation, Aromaticity, & Pericyclic Reactions Exam 6 – February 28 Polymers and their Properties

Spectroscopy for Structure and Dynamics: UV/VIS, IR, MRI & NMR Aromatic Substitution Carbonyl Chemistry Oxidation & Reduction Exam 7 – April 6 Acid Derivatives – Substitution at C=O  -Reactivity and Classical Condensations Carbohydrates and Fischer’s Glucose Proof Final Exam – May 6 Complex Synthesis of Unnatural and Natural Products Spectroscopy & Synthesis

Free energy determines what can happen (equilibrium) K = e -  G/RT = 10 -(3/4)  G room Temp But how quickly will it happen? (kinetics) Energy & Entropy

Studying Lots of Random Trajectories Provides Too Much Detail Summarize Statistically with Collective Enthalpy (H) & Entropy (S)

“Reaction Coordinate” Diagram (for a one-step atom transfer) Not a realistic trajectory, but rather a sequence of three species Starting Materials Products Transition “State” G each with H and S, i.e. Free Energy (G)

Free Energy determines what can happen (equilibrium) K = e -  G/RT = 10 -(3/4)  G room Temp and how rapidly (kinetics) k (/sec) = e -  G /RT ‡ ‡ = (3/4)  G room Temp Amount of ts (universal) Velocity of ts theory

Using Energies to Predict Equilibria and Rates for One-Step Reactions No reaction is conceptually simpler than breaking a bond in the gas phase to give atoms or free radicals.

BondDissn Energies Ellison’s values as of 2003 from Barney Ellison & his friends Coming in April Streitwieser, Heathcock, and Kosower (1992)

Ellison I Larger halogen  Poorer overlap with H (at normal bond distance) & less e-transfer to halogen H I H F less e-stabilization  weaker bond Diagram qualitative; not to scale.

Ellison II No special stabilization  SOMO orthogonal to  *) C - H bond unusually strong (good overlap from sp 2 C ) Vinyl C - H bond normal (sp 3 C, as in alkane) Allyl Special stabilization  SOMO overlaps  *) hard 111 Phenyl Ditto hard 113 easy 89 Ditto Benzyl easy 90 All H-Alkyl 100 ± 5 Same trend as H-Halogen Special Cases SOMO C   Are unusual BDE values due to unusual bonds or unusual radicals? (Compared to what?) or actually

H 3 C H + X X  H 3 C X + H X F Cl Br I ” Possibility of Halogenation (Equilibrium) CostReturnProfit

H 3 C H + X X  H 3 C X + H X Possibility of Halogenation (Equilibrium) F Cl Br I ” CostReturnProfit Is break-two-bonds-then-make-two a plausible Mechanism? at RT (~300K)? at ~3000K?  = /sec  = 250/sec How about rate (which depends on Mechanism)? No Way! Yes (unless there is a faster one)

H H 2 H 2 H HHHHHH H H H Henry Eyring (1935) Dissociation followed by association requires high activation energy. SLOW Make-as-you-break “displacement” is much easier. FAST

H CH 3 Cl H Cl CH 3 Cl CH 3 Cl Cl "free-radical chain" Make-as-you-break “displacement” is much easier. FAST

Free-Radical Chain Substitution X-HR-H X-X R-X X R cyclic machinery preserves “radicalness”

H 3 C-H + X 2  HX + H 3 CX F Cl Br I ” Possibility of Halogenation (Equilibrium) CostReturnProfit H 3 C-H  HX X X2  H3CXX2  H3CX H 3 C Step Step (Mechanism for Reasonable Rate) How can we predict activation energy?

Even if we could predict the rate of Step 1 or Step 2, how would we reckon the overall rate with two reaction steps? We must learn to cope with such Complex Reactions

Digression on Reaction Order & Complex Reactions The kinetic analogue of the Law of Mass Action (i.e. dependance of rate on concentrations) can provide insight about reaction mechanism.

Could use a single tap “twice” as large Rate (amount per second) Doubled Rate Chemists can also change [Concentration]

Rate “Laws”: Kinetic Order Rate = d [Prod] / d t 0 th Order: Rate = k Simple One-Step Reactions = k  concentration(s) ? Dependent on Mechanism Discovered by Experiment

0 th Order Kinetics Would more sheep give a faster rate? NO! (saturation) Catalyst e.g. enzyme “Substrate” Rate  But if the catalysis was not initially recognized. [Catalyst] [Substrate] 0 1  Photo: Antonio Vidigal by permission But first-order in substrate at low concentration. [Substrate] 1 for high [Substrate]

Rate “Laws”: Kinetic Order 1 st Order: Rate = k [A] Rate = d [Prod] / d t 0 th Order: Rate = k Simple One-Step Reactions = k  concentration(s) ? Dependent on Mechanism Discovered by Experiment (Reasonable)

Product Time (sec) Concentration First-Order Kinetics k = 0.69/sec

Product Time (sec) Concentration First-Order Kinetics Starting Material 1/2 1/4 1/8 1/16 Exponential Decay  Constant “Half Life” = 0.69 / k k = 0.69/sec

Reversible First-Order Kinetics Starting MaterialProduct k -1 k1k1 at Equilibrium forward rate = reverse rate k 1  [Starting Material] = k -1  [Product] = k -1 k1k1 [Product] [Starting Material] K

Time (sec) Concentration Reversible First-Order Kinetics Starting Material Product k 1 = 0.69/sec k -1 = 0.23/sec Exponential Decay to Equilibrium Mixture Half Life = 0.69 / (k 1 + k -1 ) Starting MaterialProduct k -1 k1k1 ( K = 3 )

Rate Laws: Kinetic Order 2 nd Order: Rate = k [A] 2 “1 st Order in A” Rate = d [Prod] / d t 1 st Order: Rate = k [A] 0 th Order: Rate = k Simple One-Step Reactions = k  concentration(s) ? Dependent on Mechanism Discovered by Experiment or Rate = k [A] [B] “Pseudo” 1 st Order If [B] is (effectively) constant k or [B] >> [A] e.g. [B] a catalyst

Time (sec) Concentration Second- vs First-Order Kinetics First Order Second Order Slows Faster Not Exponential No Constant Half Life

Rate Laws: Kinetic Order Rate = d [Prod] / d t Complex Reactions = k  concentration(s) ? Dependent on Mechanism Discovered by Experiment The Rate-Limiting Step Who Cares? Rapid pre- equilibrium reactive intermediate(low concentration) “ ” with starting material

Starting MaterialIntermediate k -1 k1k1 Product k2k2 Actual as if 2 nd TS were sole barrier as if 1 st TS were sole barrier SM Int Flaky Excel Program Available Prod TS1 TS2 Once Int reaches steady-state “equilibrium” with SM, it yields Prod 1/10 as fast as it is formed. k 2 / k ≈ 1/9 k 1 / k ≈ 1/9 Once Int reaches steady-state “equilibrium” with SM, SM / Int ≈ 9

Rate Laws: Kinetic Order Rate = d [Prod] / d t Fractional Order Complex Reactions = k  concentration(s) ? Dependent on Mechanism Discovered by Experiment The Rate-Limiting Step

End of Lecture 38 Jan. 10, 2011 Copyright © J. M. McBride Some rights reserved. Except for cited third-party materials, and those used by visiting speakers, all content is licensed under a Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0).Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0) Use of this content constitutes your acceptance of the noted license and the terms and conditions of use. Materials from Wikimedia Commons are denoted by the symbol. Third party materials may be subject to additional intellectual property notices, information, or restrictions. The following attribution may be used when reusing material that is not identified as third-party content: J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0