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There are two main types of covalent bond breakage. In
Chapter 9, Problem 65P(choose chapter or problem)
There are two main types of covalent bond breakage. In homolytic breakage (as in Table 9.2, p. 353), each atom in the bond gets one of the shared electrons. In some cases, the electronegativity of adjacent atoms affects the bond energy. In heterolytic breakage, one atom gets both electrons and the other gets none; thus, a cation and an anion form.
(a) Why is the C–C bond in H3C–CF3 (423 kJ/mol) stronger than that in H3C–CH3 (376 kJ/mol)?
(b) Use bond energy and any other data to calculate the heat of reaction for the heterolytic cleavage of O2.
Questions & Answers
QUESTION: Problem 65P
There are two main types of covalent bond breakage. In homolytic breakage (as in Table 9.2, p. 353), each atom in the bond gets one of the shared electrons. In some cases, the electronegativity of adjacent atoms affects the bond energy. In heterolytic breakage, one atom gets both electrons and the other gets none; thus, a cation and an anion form.
(a) Why is the C–C bond in H3C–CF3 (423 kJ/mol) stronger than that in H3C–CH3 (376 kJ/mol)?
(b) Use bond energy and any other data to calculate the heat of reaction for the heterolytic cleavage of O2.
ANSWER:
Problem 65P
Step 1
(a) Here we have to explain why is the C–C bond in H3C–CF3 (423 kJ/mol) stronger than that in H3C–CH3 (376 kJ/mol).
The C–C bond in H3C–CF3 (423 kJ/mol) stronger than that in H3C–CH3 (376 kJ/mol) due to presence of very highly electronegative fluorine atoms bonded to one of the carbon atoms in H3C–CF3. The presence of the electronegative fluorine atom is making the C-C more polar as compared to C-C in H3C–CH3. The polar bond will tend to undergo heterolytic cleavage rather than homolytic cleavage. Hence more energy is required for heterolytic cleavage.
Thus presence of C-C polar bond in H3C–CF3 due to the highly electronegative fluorine atom is more stronger than H3C–CH3.