The reason as to why the bond dissociation energy for the C − C bond in ethane is much higher than the bond dissociation energy for the labeled C − C bond in but − 1 − ene is to be explained. Concept introduction: In organic chemistry , the formation of carbocation or carbanion or radical occurs due to the heterolysis or homolysis process Heterolysis is the process in which unequal sharing of electrons results in the breaking of the bond, whereas homolysis is the process in which equal sharing of electrons results in the breaking of the bond.

Question

Want to see the full answer?

Answer 1

Question

Formula Formula Bond dissociation energy (BDE) is the energy required to break a bond, making it an endothermic process. BDE is calculated for a particular bond and therefore consists of fragments such as radicals since it undergoes homolytic bond cleavage. For the homolysis of a X-Y molecule, the energy of bond dissociation is calculated as the difference in the total enthalpy of formation for the reactants and products. X-Y → X + Y BDE = Δ H f X + Δ H f Y – Δ H f X-Y where, ΔHf is the heat of formation.

Chapter 14, Problem 38P

Interpretation Introduction

Interpretation: The reason as to why the bond dissociation energy for the C−C bond in ethane is much higher than the bond dissociation energy for the labeled C−C bond in but−1−ene is to be explained.

Concept introduction: In organic chemistry, the formation of carbocation or carbanion or radical occurs due to the heterolysis or homolysis process Heterolysis is the process in which unequal sharing of electrons results in the breaking of the bond, whereas homolysis is the process in which equal sharing of electrons results in the breaking of the bond.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

7 The pentadienyl radical, H2C“CH¬CH“CH¬CH2#, has its unpaired electron delocalized over three carbon atoms.(a) Use resonance forms to show which three carbon atoms bear the unpaired electron.

Draw the complete Lewis structure of (CH3)2CHCH(NH2)CO2H and identify all H-bond donors and all H-bond acceptors.

The pentadienyl radical, H2C“CH¬CH“CH¬CH2#, has its unpaired electron delocalized over three carbon atoms.(a) Use resonance forms to show which three carbon atoms bear the unpaired electron.(b) How many MOs are there in the molecular orbital picture of the pentadienyl radical?

Answer 2

Question

Formula Formula Bond dissociation energy (BDE) is the energy required to break a bond, making it an endothermic process. BDE is calculated for a particular bond and therefore consists of fragments such as radicals since it undergoes homolytic bond cleavage. For the homolysis of a X-Y molecule, the energy of bond dissociation is calculated as the difference in the total enthalpy of formation for the reactants and products. X-Y → X + Y BDE = Δ H f X + Δ H f Y – Δ H f X-Y where, ΔHf is the heat of formation.

Chapter 14, Problem 38P

Interpretation Introduction

Interpretation: The reason as to why the bond dissociation energy for the C−C bond in ethane is much higher than the bond dissociation energy for the labeled C−C bond in but−1−ene is to be explained.

Concept introduction: In organic chemistry, the formation of carbocation or carbanion or radical occurs due to the heterolysis or homolysis process Heterolysis is the process in which unequal sharing of electrons results in the breaking of the bond, whereas homolysis is the process in which equal sharing of electrons results in the breaking of the bond.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Formula Formula Bond dissociation energy (BDE) is the energy required to break a bond, making it an endothermic process. BDE is calculated for a particular bond and therefore consists of fragments such as radicals since it undergoes homolytic bond cleavage. For the homolysis of a X-Y molecule, the energy of bond dissociation is calculated as the difference in the total enthalpy of formation for the reactants and products. X-Y → X + Y BDE = Δ H f X + Δ H f Y – Δ H f X-Y where, ΔHf is the heat of formation.

Chapter 14, Problem 38P

Interpretation Introduction

Interpretation: The reason as to why the bond dissociation energy for the C−C bond in ethane is much higher than the bond dissociation energy for the labeled C−C bond in but−1−ene is to be explained.

Concept introduction: In organic chemistry, the formation of carbocation or carbanion or radical occurs due to the heterolysis or homolysis process Heterolysis is the process in which unequal sharing of electrons results in the breaking of the bond, whereas homolysis is the process in which equal sharing of electrons results in the breaking of the bond.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Formula Formula Bond dissociation energy (BDE) is the energy required to break a bond, making it an endothermic process. BDE is calculated for a particular bond and therefore consists of fragments such as radicals since it undergoes homolytic bond cleavage. For the homolysis of a X-Y molecule, the energy of bond dissociation is calculated as the difference in the total enthalpy of formation for the reactants and products. X-Y → X + Y BDE = Δ H f X + Δ H f Y – Δ H f X-Y where, ΔHf is the heat of formation.

Chapter 14, Problem 38P

Interpretation Introduction