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

How would you prepare 1 liter of a 50 mM Phosphate buffer at pH 7.5 beginning with K3PO4 and 1 M HCl or 1 M NaOH? Please help and show calculations. Thank you

Draw the four most importantcontributing structures of the cation intermediate thatforms in the electrophilic chlorination of phenol,(C6H5OH) to form p-chlorophenol. Put a circle aroundthe best one. Can you please each step and also how you would approach a similar problem. Thank you!

A 100mM lactic acid/lactate buffer was found to have a lactate to lactic acid ratio of 2 and a pH of 4.2. What is the pKa of lactic acid? Can you please help show the calculations?

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