1.7 Structural Formulas of Organic Molecules

1. 1.7Structural Formulas of Organic Molecules

2. Constitution • The order in which the atoms of a molecule are connected is called its constitution or connectivity. • The constitution of a molecule must be determined in order to write a Lewis structure.

3. H H H C H C C H : H : H O (CH3)2CHOH or CH3CHCH3 H OH Condensed structural formulas • Lewis structures in which many (or all) covalent bonds and electron pairs are omitted. can be condensed to:

4. CH3CH2CH2CH3 is shown as CH3CH2CH2CH2OHis shown as OH Bond-line formulas • Omit atom symbols. Represent structure by showing bonds between carbons and atoms other than hydrogen. • Atoms other than carbon and hydrogen are called heteroatoms.

5. H Cl Cl C H2C CH2 H2C CH2 C H H Bond-line formulas • Omit atom symbols. Represent structure by showing bonds between carbons and atoms other than hydrogen. • Atoms other than carbon and hydrogen are called heteroatoms. is shown as

6. 1.8Constitutional Isomers

7. Constitutional isomers • Isomers are different compounds that have the same molecular formula. • Constitutional isomers are isomers that differ in the order in which the atoms are connected. • An older term for constitutional isomers is “structural isomers.”

8. O H2NCNH2 A Historical Note NH4OCN Ammonium cyanate Urea • In 1823 Friedrich Wöhler discovered that when ammonium cyanate was dissolved in hot water, it was converted to urea. • Ammonium cyanate and urea are constitutional isomers of CH4N2O. • Ammonium cyanate is “inorganic.” Urea is “organic.” Wöhler is credited with an important early contribution that helped overturn the theory of “vitalism.”

9. H .. .. : H C O N O .. .. H Examples of constitutional isomers .. • Both have the molecular formula CH3NO2 but the atoms are connected in a different order. H : O + N H C – : : O H .. Nitromethane Methyl nitrite

10. 1.9Resonance

11. Resonance • two or more Lewis structures may be written for certain compounds (or ions) • Recall from Table 1.5

12. H .. .. : H C O N O .. .. .. H Table 1.5 How to Write Lewis Structures • If an atom lacks an octet, use electron pairs on an adjacent atom to form a double or triple bond. • Example:Nitrogen has only 6 electrons in the structure shown.

13. H .. .. : H C O N O .. .. H Table 1.5 How to Write Lewis Structures • If an atom lacks an octet, use electron pairs on an adjacent atom to form a double or triple bond. • Example:All the atoms have octets in this Lewis structure.

14. H .. .. : H C O N O .. .. H Table 1.5 How to Write Lewis Structures • Calculate formal charges. • Example:None of the atoms possess a formal charge in this Lewis structure.

15. H .. + – : H C O N O .. .. .. H Table 1.5 How to Write Lewis Structures • Calculate formal charges. • Example:This structure has formal charges; is less stable Lewis structure.

16. H H .. .. .. + – : : H C O N O H C O N O .. .. .. .. .. H H Resonance Structures of Methyl Nitrite • same atomic positions • differ in electron positions more stable Lewis structure less stable Lewis structure

17. H H .. .. .. + – : : H C O N O H C O N O .. .. .. .. .. H H Resonance Structures of Methyl Nitrite • same atomic positions • differ in electron positions more stable Lewis structure less stable Lewis structure

18. Why Write Resonance Structures? • Electrons in molecules are often delocalizedbetween two or more atoms. • Electrons in a single Lewis structure are assigned to specific atoms-a single Lewis structure is insufficient to show electron delocalization. • Composite of resonance forms more accurately depicts electron distribution.

19. + – •• •• O O O •• •• •• •• Example • Ozone (O3) • Lewis structure of ozone shows one double bond and one single bond Expect: one short bond and one long bond Reality: bonds are of equal length (128 pm)

20. + – •• •• O O O •• •• •• •• – + + – •• •• •• O O O O O O •• •• •• •• •• •• •• •• •• Example • Ozone (O3) • Lewis structure of ozone shows one double bond and one single bond Resonance:

21. – + + – •• •• •• O O O O O O •• •• •• •• •• •• •• •• •• Example • Ozone (O3) • Electrostatic potentialmap shows both endcarbons are equivalentwith respect to negativecharge. Middle atomis positive.

22. 1.10The Shapes of Some Simple Molecules

23. Valence Shell Electron Pair Repulsions • The most stable arrangement of groups attached to a central atom is the one that has the maximum separation of electron pairs(bonded or nonbonded).

24. Table 1.6 Methane • tetrahedral geometry • H—C—H angle = 109.5°

25. Table 1.6 Methane • tetrahedral geometry • each H—C—H angle = 109.5°

26. Table 1.6 Water • bent geometry • H—O—H angle = 105° H H : O .. but notice the tetrahedral arrangement of electron pairs

27. Table 1.6 Ammonia • trigonal pyramidal geometry • H—N—H angle = 107° H H : N H but notice the tetrahedral arrangement of electron pairs

28. Table 1.6 Boron Trifluoride • F—B—F angle = 120° • trigonal planar geometry allows for maximum separationof three electron pairs

29. Multiple Bonds • Four-electron double bonds and six-electron triple bonds are considered to be similar to a two-electron single bond in terms of their spatialrequirements.

30. H O C H Table 1.6: Formaldehyde • H—C—H and H—C—Oangles are close to 120° • trigonal planar geometry

31. O O C Table 1.6 Carbon Dioxide • O—C—O angle = 180° • linear geometry

32. 1.11Molecular Dipole Moments

33. +— Dipole Moment • A substance possesses a dipole moment if its centers of positive and negative charge do not coincide. •  = e x d • (expressed in Debye units) not polar

34. — + Dipole Moment • A substance possesses a dipole moment if its centers of positive and negative charge do not coincide. •  = e x d • (expressed in Debye units) polar

35. O O C Molecular Dipole Moments + - - • molecule must have polar bonds • necessary, but not sufficient • need to know molecular shape • because individual bond dipoles can cancel

36. O O C Molecular Dipole Moments Carbon dioxide has no dipole moment;  = 0 D

37. Figure 1.7 Dichloromethane Carbon tetrachloride  = 0 D  = 1.62 D

38. Figure 1.7 Resultant of thesetwo bond dipoles is Resultant of thesetwo bond dipoles is  = 0 D Carbon tetrachloride has no dipolemoment because all of the individualbond dipoles cancel.

39. Figure 1.7 Resultant of thesetwo bond dipoles is Resultant of thesetwo bond dipoles is  = 1.62 D The individual bond dipoles do notcancel in dichloromethane; it hasa dipole moment.

40. 1.12Acids and Bases:The Arrhenius View

41. Definitions • Arrhenius • An acid ionizes in water to give protons. A base ionizes in water to give hydroxide ions. • Brønsted-Lowry • An acid is a proton donor. A base is a proton acceptor. • Lewis • An acid is an electron pair acceptor. A base is an electron pair donor.

42. . . + .. .. – – . . A + H A + OH M + OH M H .. .. Arrhenius Acids and Bases • An acid is a substance that ionizes to give protons when dissolved in water. A base is a substance that ionizes to give hydroxide ions when dissolved in water.

43. . . + .. .. – – . . A H A + OH OH M H .. .. Arrhenius Acids and Bases • Strong acids dissociate completely in water. Weak acids dissociate only partially. Strong bases dissociate completely in water. Weak bases dissociate only partially. + M +

44. [H+][A–] . Ka= . [HA] + – A H A + H Acid Strength is Measured by pKa pKa= – log10Ka

45. 1.13Acids and Bases:The Brønsted-Lowry View • Brønsted-Lowry definitionan acid is a proton donora base is a proton acceptor

46. . . A Brønsted Acid-Base Reaction • A proton is transferred from the acid to the base. + – . . + B H A H A B + base acid

47. . . A Brønsted Acid-Base Reaction • A proton is transferred from the acid to the base. + – . . + B H A H A B + base acid conjugate acid conjugate base

48. . . . . . . . . Proton Transfer from HBr to Water hydronium ion • base acid conjugate conjugate acid base H H .. + – .. . . . . O H Br H Br O + + .. .. H H

49. H H – .. . . . . . . . . . . Br .. H H Equilibrium Constant for Proton Transfer .. + . . + O + H Br H O .. • Takes the same form as for Arrhenius Ka, but H3O+ replaces H+. H3O+ and H+ are considered equivalent, and there is no difference in Ka values for Arrhenius and Brønsted acidity. [H3O+][Br–] Ka = [HBr]

50. H H – .. . . . . . . . . . . Br .. H H Equilibrium Constant for Proton Transfer .. + . . + O + H Br H O .. [H3O+][Br–] Ka = [HBr] pKa= – log10Ka