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11 Intermolecular Forces Solutions to Exercises 11.89 (a) 24 valence pairs H H H H The geometry around the central C atom is trigonal planar, and around the two terminal C atoms, tetrahedral. (b) Polar. The C=O bond is quite polar and the dipoles in the trigonal plane around the central C atom do not cancel. (c) Dipole-dipole and London-dispersion forces (d) Since the molecular weights of acetone and 1-propanol are similar, the strength of the London-dispersion forces in the two compounds is also similar. The big difference is that 1-propanol has hydrogen bonding, while acetone does not. These relatively strong attractive forces lead to the higher boiling point for 1-propanol. Br 11.90 CH₂ CH₂ CH₂ CH CH₃ CH₃ CH₃ CH₂ CH₃ CH₃ CH₃ (i) MM = 44 (ii) MM = 72 (iii) MM = 123 CH₂ Br CH₂ OH CH₃ CH₃ CH₃ CH₂ CH₃ CH₂ (iv) MM = 58 (v) MM = 123 (vi) MM = 60 It is useful to draw the structural formulas because intermolecular forces are determined by the size and shape (structure) of molecules. (a) Molar mass: compounds (i) and (ii) have similar rod-like structures; (ii) has a longer rod. The longer chain leads to greater molar mass, stronger London- dispersion forces and higher heat of vaporization. (b) Molecular shape: compounds (iii) and (v) have the same chemical formula and molar mass but different molecular shapes (they are structural isomers). The more rod-like shape of (v) leads to more contact between molecules, stronger dispersion forces and higher heat of vaporization. (c) Molecular polarity: rod-like hydrocarbons (i) and (ii) are essentially nonpolar, owing to free rotation about C-C σ bonds, while (iv) is quite polar, owing to the C=O group. (iv) has a smaller molar mass than (ii) but a larger heat of vaporization, which must be due to the presence of dipole-dipole forces in (iv). [Note that (iii) and (iv), with similar shape and molecular polarity, have very similar heats of vaporization.] (d) Hydrogen-bonding interactions: molecules (v) and (vi) have similar structures, but (vi) has hydrogen bonding and (v) does not. Even though molar mass and thus dispersion forces are larger for (v), (vi) has the higher heat of vaporization. This must be due to hydrogen bonding interactions. 331