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Organic Chemistry

Organic Chemistry. Review Information for Unit 1 Intermolecular Forces Empirical and Molecular Formulas. Intermolecular Forces. Attractive forces between molecules are very important in “condensed” phases (solids and liquids) molecules are constantly in contact with each other

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Organic Chemistry

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  1. Organic Chemistry Review Information for Unit 1 Intermolecular Forces Empirical and Molecular Formulas

  2. Intermolecular Forces • Attractive forces between molecules are very important in “condensed” phases (solids and liquids) • molecules are constantly in contact with each other • important in determining the BP, MP, and solubility of compounds

  3. Intermolecular Forces • 3 major kinds of intermolecular (attractive) forces in solids and liquids: • dipole-dipole interactions • polar molecules • hydrogen bonding • molecules with -OH or -NH groups • London dispersion forces • all molecules

  4. + - + - Intermolecular Forces • Polar molecules have permanent dipoles • positive end • negative end • The most stable arrangement of polar molecules is the one in which the positive end of one molecule is oriented toward the negative end of the other molecule. Dipole-dipole interaction (force)

  5. Intermolecular Forces • Dipole-Dipole Interaction (Force) • an attractive intermolecular force resulting from the attraction of the positive and negative ends of the dipole moments of polar molecules • When a polar liquid vaporizes, the dipole-dipole interactions must be overcome. • More polar molecules have stronger dipole-dipole interactions. • large DHvap • high BP

  6. Intermolecular Forces • Organic compounds containing O-H and/or N-H bonds exhibit hydrogen bonding: • a strong dipole-dipole interaction between a hydrogen atom that is covalently bonded to either O, N, or F and a lone pair of electrons on a different O, N, or F atom

  7. d+ d- d- d+ O - H N - H Intermolecular Forces • Due to the large difference in electronegativity, O-H and N-H bonds are highly polar • H has a partial positive charge in such bonds • The H atom is an electrophile and has a strong affinity for nonbonding electrons on a N or O atom (other than the one it is covalently bonded to).

  8. Intermolecular Forces Hydrogen bonding between 2 methyl amine molecules: Hydrogen bonding between 2 ethyl alcohol molecules: d+ d+ Hydrogen bond d- d+ d- d+ d+ d- d- d+

  9. Intermolecular Forces • Although H-bonding is a strong attractive force, it is weaker than covalent bonds: • H-bond • ~20 kJ/mol to break • C-H, N-H, O-H covalent bonds: • ~400 kJ/mol to break • H-bonding has a large effect on physical properties.

  10. Intermolecular Forces • Impact of H-bonding on BP: • H-bonding leads to higher BP • Alcohols have stronger H-bonds than amines • alcohol BP > amine BP (when # of C is the same) • As the number of H’s on a N increases, H-bonding increases: • amine BP increases with increasing # of H’s on the N

  11. Intermolecular Forces • In nonpolar molecules, London dispersion forces are the principal attractive force. • Arise from temporary dipole moments induced in one molecule by other nearby molecules • electrons in molecules are displaced from symmetrical arrangement • temporary dipole established • last for a fraction of a second • change continuously

  12. Intermolecular Forces London dispersion forces in nonpolar molecules:

  13. Intermolecular Forces • London dispersion forces require close surface contact between two (or more) molecules. • The strength of the London dispersion forces is roughly proportional to surface area. • As surface area increases, LDF increase and BP increases

  14. Intermolecular Forces • For simple hydrocarbons with the same total number of carbons: • Unbranched (linear) has greater surface area • higher BP • branched has less surface area (more spherical) • lower BP 98.4oC 81oC

  15. Intermolecular Forces Example: Which of the following should have the highest BP? N-pentane 2-methyl butane neopentane

  16. Intermolecular Forces Example: Which of the following should have the highest BP? N-pentane 2-methyl butane neopentane Since all of the molecules are nonpolar, LDF are the most important attractive force. As surface area increases, BP increases. For simple hydrocarbons, surface area is greatest for unbranched hydrocarbons. Answer:

  17. Intermolecular Forces • Given a series of compounds, you should be able to rank them in order of their BP based on these intermolecular forces and explain why. • To predict relative boiling points, look for differences in: • Molecular weight • Hydrogen bonding • Polarity • Surface area

  18. Intermolecular Forces • If two compounds have similar molecular weights, then the IMFs with the greatest influence on BP are: • hydrogen bonding > dipole-dipole> LDF 60 g/mol 60 g/mol 58 g/mol 58 g/mol 89oC 56.5oC -11.7oC

  19. Intermolecular Forces • If two compounds have drastically different molecular weights, then molecular weight may be more important than hydrogen bonding or polarity in determining their relative BPs. 32 g/mol 86 g/mol 142 g/mol 64.7oC 69oC 174.1oC

  20. Intermolecular Forces Example:Arrange the following in order of increasing boiling point. Explain why. 2,2-dimethylbutane hexane 1-hexanol dipropylamine 1,6-hexanediol n-hexylamine

  21. Intermolecular Forces Neither hexane or 2,2-dimethylbutane are capable of hydrogen bonding so they will have the two lowest BP’s. • 2,2-dimethylbutane is more branched • lower BP • hexane is linear • higher BP hexane 2,2-dimethylbutane

  22. Intermolecular Forces • 1-hexanol, 1,6-hexanediol, dipropylamine, and n-hexylamine all have groups capable of forming hydrogen bonds. • amines form weaker hydrogen bonds than alcohols • dipropylamine and n-hexylamine will have lower BP’s than the alcohols • Dipropylamine has one N-H bond (less H-bonding) while n-hexylamine has 2 N-H bonds (more H-bonding)

  23. Intermolecular Forces • So far: 2,2-dimethylbutane < hexane < dipropylamine < n-hexylamine • 1,6-hexanediol has two OH groups while 1-hexanol has one OH group • 1,6-hexanediol has higher BP than 1-hexanol

  24. Intermolecular Forces So: 2,2-dimethylbutane < hexane < dipropylamine < n-hexylamine < 1-hexanol < 1,6-hexanediol 49.7oC 68.95oC 109.4oC 130oC 158oC 250oC

  25. Intermolecular Forces • Intermolecular forces also determine the solubility of organic compounds. • “Like dissolves like” • polar compounds dissolve in polar solvents • nonpolar compounds dissolve in nonpolar solvents • Ionic compounds generally dissolve readily in water because water hydrates or solvates the individual ions

  26. Intermolecular Forces • Polar organic compounds dissolve in polar solvents due to: • dipole-dipole interactions • H-bonding Hydrogen bonding between methyl amine and water.

  27. Intermolecular Forces • Nonpolar compounds do not dissolve in water because they cannot break the H-bonding network that exists in a sample of water

  28. Calculating Empirical & Molecular Formulas • The empirical formula for an unknown compound can be determined using quantitative elemental analysis. • Often, the % composition for all elements except oxygen is reported • Example: 40.0% C and 6.67% H • Remainder (unless it adds up to very close to 100%) is assumed to be oxygen • In previous example: 53.3% O

  29. Calculating Empirical & Molecular Formulas • Calculate empirical formula by: • % to mass • mass to mole • divide by smallest • multiply ‘til whole

  30. Calculating Empirical & Molecular Formulas Example: Calculate the empirical formula for a compound containing 81.8% C and 18.2% H. Steps 1 & 2: % to Mass; Mass to Moles

  31. Calculating Empirical & Molecular Formulas Step 3: Divide by Smallest C: 6.82 = 1.00 = 1 6.82 H: 18.0 = 2.64 Step 4: Multiply ‘til Whole Since CH2.64 doesn’t make any sense, you must multiply to get a whole number ratio. 2.64 ~ 2.67 ~ 2 2/3 or 8/3. So multiply both numbers by 3. C3H8

  32. Calculating Empirical & Molecular Formulas • Molecular formulas are always some whole number multiple of the empirical formula: C2H4O (acetic acid) CH2O C6H12O6 (glucose) X 2 X 6

  33. Calculating Empirical & Molecular Formulas • Molecular weight is the same whole number multiple of the empirical formula’s formula weight (FW). C2H4O (acetic acid) CH2O C6H12O6 (glucose) X 2 MW = 30.0 amu x 2 = 60.0 amu FW = 30.0 amu X 6 MW = 30.0 amu x 6 = 180.0 amu

  34. Calculating Empirical & Molecular Formulas • Molecular weights can be determined experimentally in several ways: • colligative properties • ideal gas law • mass spectrometry • To determine the molecular formula for an unknown: • calculate the empirical formula • calculate ratio of MW/empirical FW • multiply subscripts of empirical formula by previous ratio

  35. Calculating Empirical & Molecular Formulas Example:Determine the molecular formula for a compound that contains 39.8 % C, 3.32% H, and 39.2% Cl if its molecular weight is about 181 amu. Step 1: Find Empirical Formula

  36. Calculating Empirical & Molecular Formulas Empirical Formula = C3H3ClO Step 2: Find MW/FW (empirical) MW = 181 (given) FW (empirical) = 3(12.0) + 3 (1.01) + 35.5 + 16.0 = 90.5 MW = 181 = 2 FW 90.5 Step 3: Multiply subscripts Molecular Formula = C(3x2)H(3 x 2)Cl(1x2)O(1x2) = C6H6Cl2O2

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