University:
Boston University
Course:
CAS CH 214 | Organic Chemistry with Qualitative Analysis
Academic year:

2022
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305

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15
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pubicswprv

Typical Bond Energies (kcal · mol-1) } 95 20 @ ~3 Å 4.7 Van der Waals interactions 0.07 Hydrophobic interactions Molecular Forces Important for Life The 4 NON-covalent bonds: ✓ H-bonds ✓ Salt bridges ✓ van der Waal’s interactions Hydrophobic effect } Electrostatic Hydrophobic Interactions Hydrophobic interactions: atoms “attracted” together due to water structure. Nonpolar Solutes Aggregate in Water Hydrophobic Interactions Structure: Nonpolar Solutes in Water Like ICE Hydrophobic Interactions Figure 2.12 Hydrophilic and Hydrophobic Hydrophobic Interactions Figure 2.12 Hydrophilic and Hydrophobic x5 Hydrophobic Interactions 3 “degrees” of freedom (DoF) Statistical mechanics has determined that each DoF is equal to 5-6 kcal/mol at 25 °C 6 “degrees” of freedom Hydrophobic Interactions For EXAMPLE 3 [CH4·10H2O] [(CH4)3 · 12H2O] + 18H2O DG DG?is definitely NEGATIVE: How? Enthalpy Enthalpy 30 H2O x 4 H-bonds each x 5 kcal/mol = 600 kcal/mol heat release to make H-bonds 18 H2O x 3.5 H-bonds each x 5 kcal/mol = 315 kcal/mol, plus 12x4x5 = 240 kcal/mol; Total = 555 kcal/mol heat released to make bonds DH = –555 – (–600) = +45 DH = +45 kcal/mol added in excess to break Hbonds TDS=114-18 TDS = +96 3 degrees of 19 degrees of =96 kcal/mol in freedom (DoF) x 6 kcal/mol at freedom (DoF) x 6 kcal/mol at entropic 25 °C per DoF = 18 kcal/mol 25 °C per DoF = 114 kcal/mol energy DG = DH – TDS gained = +45 – 96 = –51 kcal/mol Entropy Entropy ENTROPY DRIVEN Hydrophobic Interactions Transfer of Hydrocarbons to Nonpolar Solvents is Entropically Driven Note the bottom two: Reaction in reverse is the completely entropic positive values of +17-20 kJ/mol for putting benzene or toluene into water Bond Energies of Important Noncovalent Interactions (kcal · mol-1) } 95 20 @ ~3 Å 4.7 Van der Waals interactions 0.07 Hydrophobic interactions (per –CH2–) 26 6 Molecular Forces Important for Life The 4 NON-covalent bonds: ✓ H-bonds ✓ Salt bridges ✓ van der Waal’s interactions ✓ Hydrophobic effect } Electrostatic Amphipathic Interactions (both hydrophobic and hydrophilic) =O Example: amphiphiles like Fatty Acid Anions, detergents –O–S–O– · +Na =O Sodium dodecyl sulfate (C12H25O4S) (SDS) Amphiphiles Form Micelles & Bilayers Structure of a Micelle Protein interiors Disrupters of the Hydrophobic Effect Urea Guanidinium-HCl Physical Properties of Water Learning Goals: Concepts about water • Water molecules, which are polar, can form hydrogen bonds with other molecules. • In ice, water molecules are maximally hydrogen bonded in a crystalline array, but in liquid water, still with many hydrogen bonds, they rapidly break and re-form in irregular networks. • Polar and ionic substances can dissolve in water. • The attractive forces acting on biological molecules include ionic interactions, hydrogen bonds, and van der Waals interactions. • The hydrophobic effect explains the exclusion of nonpolar groups as a way to maximize the entropy of water molecules. • Amphiphilic substances form micelles or bilayers that hide their hydrophobic groups while exposing their hydrophilic groups to water. WATER The 4 S’s for Water: ✓ Shape ✓ Size Solubility ✓ Stability The discussion of water’s stability is a discussion of its ionization. H2 O 2H2O D D H+ + OH– H3O+ + OH– Ionization of Water Proton “Transfers” Occurs Rapidly H The Positive Charge jumps H H H H H + H Proton “transfers” are very rapid! 2H2O D H3O+ + OH– What is the concentration of OH– and H3O+ (protons)? Ionization of Water What is the concentration of OH–? 2H2O D H3O+ + OH– Equilibrium constant for this acid dissociation (aka deprotonation) reaction Keq is 1.8 x 10-16 M + – Keq = 1.8 x 10-16 M = [H ] [OH ] [H2O] [H2O]1.8 x 10-16 M = 55.5 M x 1.8 x 10-16 M = 55.5 M x 1.8 x 10-16 M = 1.0 x 10-14 M2 = 1.0 x 10-7 M = Take the –log of both sides: *1 L water = 1000 g ÷ 18 g/mol = 55.5 M [H2O] = 55.5 M* [H+] [OH–] + – [H ] [OH ] since [H+] = [OH–] [H+]2 [H+]2 [H+] call the –log[H+] = pH 7 = pH Ionization of Water Relationship of pH, [H+], and [OH-] 2H2O D H3O+ + OH– Ionization of Water pH Values of Common Substances Ionization of Water Definitions of Acids and Bases: HA HB D D H+ + A– H+ + B– If dissociation is >water: ACID If dissociation is < water: BASE When B– associates with protons, it leaves a net difference in [OH–] 2H2O D H3O+ + OH– In other words: A proton donor is an Acid (HA) A proton acceptor is a Base (B–) Likewise: The resulting anion from acid dissociation is called the conjugate base (A–) The resulting protonated base is called a conjugate acid (HB) These are Bronsted/Lowry definitions Ionization of Water Definitions of Acids and Bases: HA Strong Acid Strong Base 2H2O Weak Acid Weak Base HA HB ® H+ + A– And the [HA] added is the [H+] in solution. H+ + B– ® HB Protons come from pulling the water dissociation equilibrium And the [B–] added is the – + [OH–] in solution. D D D H3O + OH H+ + A– H+ + B– For weak acids/bases, all 3 species are in measurable concentrations. How do you calculate these concentrations and how are they related to the pH? Ionization of Water How do you calculate these concentrations and how are they related to the pH? Use the Equilibrium Constant (Keq) for the dissociation reaction HA D H+ + A– [H+] [A–] [HA] since [H+] is best expressed as pH, can take –log of each side: – pKa = pH – log [A ] [HA] [A–] pH = pKa + log [HA] Keq = Kd = Ka = This is called the HendersonHasselbalch equation Ionization of Water Ionization of Water Dissociation Constants and pKa Values of Some Acids Ionization of Water Buffers: Titration Curves of Weak Acids } } } NOTICE: Adding protons, but pH not changing very much = BUFFER Ionization of Water Buffers: Titration of a Polyprotic Acid H3PO4 D H+ + H2PO4– D H+ + HPO42– D H+ + PO43– pK1 pK2 pK3 12.4 6.8 2.2 Ionization of Water Does this make sense? …. Go back to the reaction: HA D H+ + A– At low pH, [H+] is high, forcing the equilibrium to the left = more HA, less A– Biological Buffer Systems • Maintenance of intracellular pH is vital to all cells. – Enzyme-catalyzed reactions have optimal pH. – Solubility of polar molecules depends on H-bond donors and acceptors. – Equilibrium between CO2 gas and dissolved H2CO3 and HCO3– depends on pH. • Buffer systems in vivo are mainly based on: – phosphate, concentration in millimolar (mM) range – bicarbonate, important for blood plasma; blood pH = 7.40 EXAMPLE: Clinical Importance pKa = 6.1 HCO3− + H + ↔ H 2CO3 Bicarbonate =24 mM Carbonic Acid =1.2 mM Patient comes to ER with hyperventilation and acidosis (pH = 7.03; [CO2] = 1.1 mM) [A–] What is the [HCO3–] and what % is used? pH = pKa + log [HA] 9.36/24 = 39% left or 61% used What do you do? 7.03 = 6.1 + log [HCO3–]/[CO2] 0.93 = log [HCO3–]/[CO2] (What is 100.93?) 8.5 = [HCO3–]/[CO2] 8.5 = [HCO3–]/1.1 9.36 mM = [HCO3–]

Properties of Water Lecture Note Typical Bond Energies

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