10_ - P1 PCX/PBR JWDD052-10 P2 PBU JWDD052-Solomons-v2 10...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
P1: PCX/PBR P2: PBU Printer: Hamilton JWDD052-10 JWDD052-Solomons-v2 April 12, 2007 16:36 10 RADICAL REACTIONS SOLUTIONS TO PROBLEMS 10.1 (a) HH ( DH ° = 436) 2 H F 2( DH ° = 570) FF ( DH ° = 159) + + 595 kJ mol 1 is required for bond cleavage 1140 kJ mol 1 is evolved in formation of the bonds in 2 mol of HF 1 H =+ 595 1140 =− 545 kJ mol 1 (b) CH 3 H CH 3 F ( DH ° = 440) ( DH ° = 461) ( DH ° = 159) + HF ( DH ° = 570) + + 599 kJ mol 1 is required for bond cleavage 1031 kJ mol 1 is evolved in bond formation 1 H 599 1031 432 kJ mol 1 (c) CH 3 H CH 3 Cl ( DH ° = 440) ( DH ° = 352) Cl Cl ( DH ° = 243) + HC l ( DH ° = 432) + + 683 kJ mol 1 is required for bond cleavage 784 kJ mol 1 is evolved in bond formation 1 H 683 784 101 kJ mol 1 (d) CH 3 H CH 3 Br ( DH ° = 440) ( DH ° = 293) Br Br ( DH ° = 193) + HB r ( DH ° = 366) + + 633 kJ mol 1 is required for bond cleavage 659 kJ mol 1 is evolved in bond formation 1 H 633 659 26 kJ mol 1 169
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
P1: PCX/PBR P2: PBU Printer: Hamilton JWDD052-10 JWDD052-Solomons-v2 April 12, 2007 16:36 170 RADICAL REACTIONS (e) CH 3 H ( DH ° = 440) CH 3 I ( DH ° = 240) II ( DH ° = 151) + HI ( DH ° = 298) + + 591 kJ mol 1 is required for bond cleavage 538 kJ mol 1 is evolved in bond formation 1 H =+ 591 538 53 kJ mol 1 (f) CH 3 CH 2 H ( DH ° = 421) CH 3 CH 2 Cl ( DH ° = 353) Cl Cl ( DH ° = 243) + HC l ( DH ° = 432) + + 664 kJ mol 1 is required for bond cleavage 785 kJ mol 1 is evolved in bond formation 1 H 664 785 =− 121 kJ mol 1 (g) (CH 3 ) 2 CH H ( DH ° = 413) (CH 3 ) 2 CH Cl ( DH ° = 355) Cl Cl ( DH °= 243) + l ( DH ° = 432) + + 656 kJ mol 1 is required for bond cleavage 787 kJ mol 1 is evolved in bond formation 1 H 656 787 131 kJ mol 1 (h) (CH 3 ) 3 CH ( DH ° = 400) (CH 3 ) 3 CC l ( DH ° = 349) Cl Cl ( DH ° = 243) + l ( DH ° = 432) + + 643 kJ mol 1 is required for bond cleavage 781 kJ mol 1 is evolved in bond formation 1 H 643 781 138 kJ mol 1 10.2 3 ° 2 ° 1 ° Methyl > CH 3 CH 2 CH 3 CH 3 C > CH 3 (CH 3 ) 2 CHCH 2 > >> > CH 3 CH 2 CH CH 3 10.3 The compounds all have different boiling points. They could, therefore, be separated by careful fractional distillation. Or, because the compounds have different vapor pressures, they could easily be separated by gas chromatography. GC/MS (gas chromatography/mass spectrometry) could be used to separate the compounds as well as provide structural in- formation from their mass spectra. Their mass spectra would show contributions from the naturally occurring 35 Cl and 37 Cl isotopes. The natural abundance of 35 Cl is approximately
Background image of page 2
P1: PCX/PBR P2: PBU Printer: Hamilton JWDD052-10 JWDD052-Solomons-v2 April 12, 2007 16:36 RADICAL REACTIONS 171 75% and that of 37 Cl is approximately 25%. Thus, for CH 3 C1, containing only one chlorine atom, there will be an M . + peak and an M . + + 2 peak in roughly a 3 : 1 (0.75 : 0.25) ratio of intensities. For CH 2 Cl 2 there will be M . + ,M . + + 2, and M . + + 4 peaks in roughly a 9 : 6 : 1 ratio, respectively. [The probability of a molecular ion M . + with both chlorine atoms as 35 Cl is (.75)(.75) = .56, the probability of an M .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/20/2011 for the course CHEMISTRY ES110 taught by Professor Te.li during the Spring '11 term at Central Texas College.

Page1 / 18

10_ - P1 PCX/PBR JWDD052-10 P2 PBU JWDD052-Solomons-v2 10...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online