Presentation is loading. Please wait.

Presentation is loading. Please wait.

Comparing Acid Strengths by Comparing Structures  Look at the stability of the conjugate base. The more stable the conjugate base, the stronger its acid.

Similar presentations


Presentation on theme: "Comparing Acid Strengths by Comparing Structures  Look at the stability of the conjugate base. The more stable the conjugate base, the stronger its acid."— Presentation transcript:

1 Comparing Acid Strengths by Comparing Structures  Look at the stability of the conjugate base. The more stable the conjugate base, the stronger its acid. Electronegativity Size/polarizability Resonance Stabilization Induction Hybrid orbital containing electrons

2 Which H is more acidic?

3 Where does the equilibrium lie?

4 Does an Acid-Base Reaction Occur? Write the products. (CH 3 ) 3 N + NH 2 - 

5 Bond Polarity - Part I  A bond is polar when the charge is not equally shared between the two atoms.  The more electronegative atom will have a partial negative charge ( δ - ). The arrow shows the dipole moment. Here we show partial charges.

6 Bond Polarity - Part II  A polar bond has a dipole moment μ : μ (in debyes) = 4.8 δ d δ is the charge at either end of the dipole d is the bond length in angstroms (charge separation) (1Å= m) bond length, d dipole moment, μ

7 Bond Polarity - Part II μ (in D) = 4.8 δ d  The dipole moment μ gives a quantitative measure of the polarity of a bond. C=O (2.4D) is more polar than C-O (0.86 D)

8 Bond Dipole Moments from Wade, 7 th ed (Table 2-1) C - O 0.86D C - N 0.22D C - F 1.51D C=O 2.4D C ≡ N 3.6D C - Cl 1.56D H - O 1.53D C - Br 1.48D H - N 1.31D H - C 0.3D C - I 1.29D

9 Bond Polarity - Part II μ (in D) = 4.8 δ d  Knowing μ and d allows the charge separation δ to be calculated. C=O has a dipole moment of 2.4D and a bond length of 1.21Å. δ = 2.4/(4.8x1.21)= 0.41 C-O has a dipole moment of 0.86D and a bond length of 1.43Å. δ = 0.86/(4.8x1.43)= 0.13

10 Molecular Polarity  The polarity (or dipole moment) of a molecule is the vector sum of the dipole moment for each bond in the molecule. A molecule with a significant dipole moment is polar. A molecule with little or no dipole moment is considered nonpolar.

11 Molecular Polarity  The dipole moment of a molecule can be measured. The dipole moments of the individual bonds can then be estimated.  Lone pairs contribute to the dipole moments.

12  arise from the charged nature of the subatomic particles (electrons and protons).  are responsible for the cohesiveness of materials.  are what determine physical properties of pure substances such as melting point, boiling point, vapor pressure, and solubility. Intermolecular Forces

13  Substances that are gases at room temperature have weak intermolecular forces.  Substances that are condensed (liquids or solids) at room temperature have much stronger intermolecular forces.  If intermolecular forces did not exist, all substances would be gases, even at extremely low temperatures. Intermolecular Forces

14  Dipole-dipole generally attractive  Hydrogen bonding a special category of very strong dipole-dipole force that involves the attraction between an electropositive H atom and nonbonding electrons on an electronegative atom (usually N, O, F, or Cl)  London dispersion force instantaneous dipole-induced dipole increases with increasing surface area of the molecule present in all molecules

15 Intermolecular Forces  Which will have the higher boiling point? or

16 Intermolecular Forces  Why does CCl 4 have the higher boiling point? chloroform, CHCl 3 ( μ = 1.0D) or carbon tetrachloride, CCl 4 ( μ = 0) bp CHCl 3 = 62°C bp CCl 4 = 77°C

17 Intermolecular Forces and Solubility  “Like dissolves like.” Polar substances dissolve in polar solvents. Nonpolar substances dissolve in nonpolar solvents. The other pairings (polar substance/nonpolar solvent and nonpolar substance / polar solvent) will not dissolve.

18 Intermolecular Forces and Solubility  For one substance to dissolve in another, there must be an attraction similar in magnitude to the forces holding the solvent together. In water, H bonding holds the molecules of water together pretty tightly. For a substance to dissolve in water, there must be an attraction between the substance and water that is close in magnitude to those H bonds.  Ions, alcohols, and ethers all dissolve in water…can you show why?

19 Intermolecular Forces and Solubility  Carbon tetrachloride does NOT dissolve in water. Water is held together by H bonds, a strong intermolecular interaction. Carbon tetrachloride is nonpolar. The only force of attraction between CCl 4 and H 2 O is dispersion, and that is not strong enough to push apart the H- bonded water molecules.

20 Intermolecular Forces  Which are soluble in water and why?

21 Phosphatidyl choline – a lipid found in cell membranes

22 Intermolecular Forces and the Cell Membrane R79TiUj80&feature=related


Download ppt "Comparing Acid Strengths by Comparing Structures  Look at the stability of the conjugate base. The more stable the conjugate base, the stronger its acid."

Similar presentations


Ads by Google