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13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. cleavage Bond A •CH3 + 26.← Cleavage 2°C. + Bond C +3°C• CH3 2C Cleavage E 2°C. 26. weakest bond Intact molecule Strongest 3°C 20. Gund Largest argest a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. C Weakest bond A Produces Most Bond Strongest Bond Strongest Gund produces least stable radicals Weakest Stable radical b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 13°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. • CH3 methyl radical Formed in Gund A Cleavage c. (5pts) Use principles discussed in lecture, supported by relevant structures, to succinctly explain the why your part b (i) radical is more stable than your part b(ii) radical. Written explanation can be no more than one-two succinct sentence(s)! E H .C. -H methyl radical higher in E since it M 1 H₂C це H has zero hyperconjugation - No 5 bonds off of carbons adjacent to half-filled Lower by P-orbital 3 radical stabilized hyperconjugation CH₂ from three different directions See "interaction" = lines between half-filled P bond orbitals. orbital and adjacent D

13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. cleavage Bond A •CH3 + 26.← Cleavage 2°C. + Bond C +3°C• CH3 2C Cleavage E 2°C. 26. weakest bond Intact molecule Strongest 3°C 20. Gund Largest argest a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. C Weakest bond A Produces Most Bond Strongest Bond Strongest Gund produces least stable radicals Weakest Stable radical b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 13°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. • CH3 methyl radical Formed in Gund A Cleavage c. (5pts) Use principles discussed in lecture, supported by relevant structures, to succinctly explain the why your part b (i) radical is more stable than your part b(ii) radical. Written explanation can be no more than one-two succinct sentence(s)! E H .C. -H methyl radical higher in E since it M 1 H₂C це H has zero hyperconjugation - No 5 bonds off of carbons adjacent to half-filled Lower by P-orbital 3 radical stabilized hyperconjugation CH₂ from three different directions See "interaction" = lines between half-filled P bond orbitals. orbital and adjacent D

Organic Chemistry: A Guided Inquiry
Organic Chemistry: A Guided Inquiry
2nd Edition
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Andrei Straumanis
Chapter5: Resonance
Section: Chapter Questions
Problem 6CTQ
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13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. cleavage Bond A •CH3 + 26.← Cleavage 2°C. + Bond C +3°C• CH3 2C Cleavage E 2°C. 26. weakest bond Intact molecule Strongest 3°C 20. Gund Largest argest a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. C Weakest bond A Produces Most Bond Strongest Bond Strongest Gund produces least stable radicals Weakest Stable radical b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 13°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. • CH3 methyl radical Formed in Gund A Cleavage c. (5pts) Use principles discussed in lecture, supported by relevant structures, to succinctly explain the why your part b (i) radical is more stable than your part b(ii) radical. Written explanation can be no more than one-two succinct sentence(s)! E H .C. -H methyl radical higher in E since it M 1 H₂C це H has zero hyperconjugation - No 5 bonds off of carbons adjacent to half-filled Lower by P-orbital 3 radical stabilized hyperconjugation CH₂ from three different directions See "interaction" = lines between half-filled P bond orbitals. orbital and adjacent D
Transcribed Image Text:13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. cleavage Bond A •CH3 + 26.← Cleavage 2°C. + Bond C +3°C• CH3 2C Cleavage E 2°C. 26. weakest bond Intact molecule Strongest 3°C 20. Gund Largest argest a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. C Weakest bond A Produces Most Bond Strongest Bond Strongest Gund produces least stable radicals Weakest Stable radical b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 13°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. • CH3 methyl radical Formed in Gund A Cleavage c. (5pts) Use principles discussed in lecture, supported by relevant structures, to succinctly explain the why your part b (i) radical is more stable than your part b(ii) radical. Written explanation can be no more than one-two succinct sentence(s)! E H .C. -H methyl radical higher in E since it M 1 H₂C це H has zero hyperconjugation - No 5 bonds off of carbons adjacent to half-filled Lower by P-orbital 3 radical stabilized hyperconjugation CH₂ from three different directions See "interaction" = lines between half-filled P bond orbitals. orbital and adjacent D
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