Chapter_2 - P1 PBU/OVY JWDD052-02 P2 PBU/OVY QC PBU/OVY T1...

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-02 JWDD052-Solomons-v2 May 23, 2007 14:54 2 REPRESENTATIVE CARBON COMPOUNDS: FUNCTIONAL GROUPS, INTERMOLECULAR FORCES, AND INFRARED (IR) SPECTROSCOPY SOLUTIONS TO PROBLEMS 2.1 F H (a) or F H δ+ δ− Br I (b) or Br I Br Br (c) μ = 0 D F F (d) μ = 0 D 2.2 VSEPR theory predicts a planar structure for BF 3 . B F F F μ = 0 D The vector sum of the bond moments of a trigonal planar structure would be zero, resulting in a prediction of μ = 0 D for BF 3 . This correlates with the experimental observation and con±rms the prediction of VSEPR theory. 2.3 The shape of CCl 2 CCl 2 (below) is such that the vector sum of all of the C—Cl bond moments is zero. C C Cl Cl Cl Cl 2.4 The fact that SO 2 has a dipole moment indicates that the molecule is angular, not linear. O O S μ = 1.63 D μ = 0 D SO O not An angular shape is what we would expect from VSEPR theory, too. 17
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-02 JWDD052-Solomons-v2 May 23, 2007 14:54 18 REPRESENTATIVE CARBON COMPOUNDS 2.5 Again, this is what VSEPR theory predicts. net dipole O H H 3 C 2.6 In CFCl 3 the large C—F bond moment opposes the C—Cl moments, leading to a net dipole moment in the direction of the ±uorine. Because hydrogen is much less electronegative than ±uorine, no such opposing effect occurs in CHCl 3 ; therefore, it has a net dipole moment that is larger and in the direction of the chlorine atoms. Smaller net dipole moment F Cl Cl Larger net dipole moment H Cl Cl C Cl C Cl 2.7 (a) C H H FF C net dipole moment (c) (d) net dipole moment C H H F F C (b) C H F H F C μ = 0 D C F F F F C μ = 0 D 2.8 (a) μ = 0 D net dipole moment C H HB r Br C Cis-trans isomers net dipole moment cis C HH Br Br C trans C H H Br Br C (b) μ = 0 D net dipole moment C Cl Cl Br Br C net dipole moment cis C Cl Cl Br Br C trans Cl Cl C Br Br C Cis-trans isomers
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-02 JWDD052-Solomons-v2 May 23, 2007 14:54 REPRESENTATIVE CARBON COMPOUNDS 19 2.9 and (a) (b) (c) Br Br Br Br 2.10 (a) F (b) Cl (c) Propyl bromide (d) Isopropyl fuoride (e) Phenyl iodide 2.11 and (a) (b) OH OH (c) OH OH 2.12 (a) (b) OH OH 2.13 (a) O (b) O (c) O (d) Methyl propyl ether (e) Diisopropyl ether (±) Methyl phenyl ether 2.14 (a) Isopropylpropylamine (b) Tripropylamine (c) Methylphenylamine (d) Dimethylphenylamine (e) NH 2 (±) CH 3 CH 3 CH 3 (CH 3 ) 3 N or N (g) CH 3 N 2.15 (a) (e) only (b) (a,c) (c) (b, d, ±, g) 2.16 (a) CH 3 H Cl CH 3 CH 3 N CH 3 CH 3 CH 3 NH ++ + Cl - (b) sp 3
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-02 JWDD052-Solomons-v2 May 23, 2007 14:54 20 REPRESENTATIVE CARBON COMPOUNDS 2.17 (a) CH 3 CH 2 CH 2 CH 2 OH would boil higher because its molecules can form hydrogen bonds to each other through the H O group. (b) CH 3 CH 2 NHCH 3 would boil higher because its molecules can form hydrogen bonds to each other through the H N group.
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This note was uploaded on 09/10/2008 for the course CHE 321 taught by Professor Fowler/sampson during the Spring '08 term at SUNY Stony Brook.

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Chapter_2 - P1 PBU/OVY JWDD052-02 P2 PBU/OVY QC PBU/OVY T1...

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