Organic Chemistry
Organic Chemistry
2nd Edition
ISBN: 9781118452288
Author: David R. Klein
Publisher: WILEY
Question
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Chapter 8, Problem 78IP

(a)

Interpretation Introduction

Interpretation:

The mechanism, rate, changes in rate when concentration of base is doubled, energy diagram and transition state of the given reaction of 2-bromo-2-methylhexane with sodium ethoxide are needed to be find out.

Concept introduction:

E2 reaction is a bimolecular elimination reaction in which alkene compounds formed in a single step. Alkenes are formed when alkyl halides are treated with bases via eliminating one β-proton and one α-halo group of the alkyl halide.

Curved arrows are used for drawing the mechanism of reaction.

Rate of E2 reaction is depends upon the concentration of substrate and concentration of base. Because in a bimolecular reaction, there should involves two chemical entities.

Equation for the rate of E2 reaction is,

Rate=k[substrate][base]

Energy diagram of an E2 reaction has only one hump because only one step is involved in the E2 reaction.

Transition state is a state in between the reactant and product.

 In the transition state of E2 reaction: the abstraction of β-proton by the base, removal of leaving group (halo-group) and formation of double bond are taking place.

(b)

Interpretation Introduction

Interpretation:

The mechanism, rate, changes in rate when concentration of base is doubled, energy diagram and transition state of the given reaction of 2-bromo-2-methylhexane with sodium ethoxide are needed to be find out.

Concept introduction:

E2 reaction is a bimolecular elimination reaction in which alkene compounds formed in a single step. Alkenes are formed when alkyl halides are treated with bases via eliminating one β-proton and one α-halo group of the alkyl halide.

Curved arrows are used for drawing the mechanism of reaction.

Rate of E2 reaction is depends upon the concentration of substrate and concentration of base. Because in a bimolecular reaction, there should involves two chemical entities.

Equation for the rate of E2 reaction is,

Rate=k[substrate][base]

Energy diagram of an E2 reaction has only one hump because only one step is involved in the E2 reaction.

Transition state is a state in between the reactant and product.

 In the transition state of E2 reaction: the abstraction of β-proton by the base, removal of leaving group (halo-group) and formation of double bond are taking place.

(c)

Interpretation Introduction

Interpretation:

The mechanism, rate, changes in rate when concentration of base is doubled, energy diagram and transition state of the given reaction of 2-bromo-2-methylhexane with sodium ethoxide are needed to be find out.

Concept introduction:

E2 reaction is a bimolecular elimination reaction in which alkene compounds formed in a single step. Alkenes are formed when alkyl halides are treated with bases via eliminating one β-proton and one α-halo group of the alkyl halide.

Curved arrows are used for drawing the mechanism of reaction.

Rate of E2 reaction is depends upon the concentration of substrate and concentration of base. Because in a bimolecular reaction, there should involves two chemical entities.

Equation for the rate of E2 reaction is,

Rate=k[substrate][base]

Energy diagram of an E2 reaction has only one hump because only one step is involved in the E2 reaction.

Transition state is a state in between the reactant and product.

 In the transition state of E2 reaction: the abstraction of β-proton by the base, removal of leaving group (halo-group) and formation of double bond are taking place.

(d)

Interpretation Introduction

Interpretation:

The mechanism, rate, changes in rate when concentration of base is doubled, energy diagram and transition state of the given reaction of 2-bromo-2-methylhexane with sodium ethoxide are needed to be find out.

Concept introduction:

E2 reaction is a bimolecular elimination reaction in which alkene compounds formed in a single step. Alkenes are formed when alkyl halides are treated with bases via eliminating one β-proton and one α-halo group of the alkyl halide.

Curved arrows are used for drawing the mechanism of reaction.

Rate of E2 reaction is depends upon the concentration of substrate and concentration of base. Because in a bimolecular reaction, there should involves two chemical entities.

Equation for the rate of E2 reaction is,

Rate=k[substrate][base]

Energy diagram of an E2 reaction has only one hump because only one step is involved in the E2 reaction.

Transition state is a state in between the reactant and product.

 In the transition state of E2 reaction: the abstraction of β-proton by the base, removal of leaving group (halo-group) and formation of double bond are taking place.

(e)

Interpretation Introduction

Interpretation:

The mechanism, rate, changes in rate when concentration of base is doubled, energy diagram and transition state of the given reaction of 2-bromo-2-methylhexane with sodium ethoxide are needed to be find out.

Concept introduction:

E2 reaction is a bimolecular elimination reaction in which alkene compounds formed in a single step. Alkenes are formed when alkyl halides are treated with bases via eliminating one β-proton and one α-halo group of the alkyl halide.

Curved arrows are used for drawing the mechanism of reaction.

Rate of E2 reaction is depends upon the concentration of substrate and concentration of base. Because in a bimolecular reaction, there should involves two chemical entities.

Equation for the rate of E2 reaction is,

Rate=k[substrate][base]

Energy diagram of an E2 reaction has only one hump because only one step is involved in the E2 reaction.

Transition state is a state in between the reactant and product.

 In the transition state of E2 reaction: the abstraction of β-proton by the base, removal of leaving group (halo-group) and formation of double bond are taking place.

Blurred answer

Chapter 8 Solutions

Organic Chemistry

Ch. 8.5 - Prob. 8PTSCh. 8.6 - Prob. 4LTSCh. 8.6 - Prob. 9PTSCh. 8.6 - Prob. 10PTSCh. 8.6 - Prob. 11ATSCh. 8.6 - Prob. 12ATSCh. 8.7 - Prob. 13CCCh. 8.7 - Prob. 14CCCh. 8.7 - Prob. 5LTSCh. 8.7 - Prob. 15PTSCh. 8.7 - Prob. 16ATSCh. 8.7 - Prob. 17ATSCh. 8.7 - Prob. 6LTSCh. 8.7 - Prob. 18PTSCh. 8.7 - Prob. 19ATSCh. 8.7 - Prob. 20CCCh. 8.7 - Prob. 21CCCh. 8.8 - Prob. 7LTSCh. 8.8 - Prob. 22PTSCh. 8.8 - Prob. 23ATSCh. 8.8 - Prob. 24ATSCh. 8.8 - Prob. 25ATSCh. 8.9 - Prob. 26CCCh. 8.9 - Prob. 27CCCh. 8.9 - Prob. 28CCCh. 8.9 - Prob. 8LTSCh. 8.9 - Prob. 29PTSCh. 8.9 - Prob. 31CCCh. 8.10 - Prob. 32CCCh. 8.10 - Prob. 33CCCh. 8.10 - Prob. 9LTSCh. 8.10 - Prob. 34PTSCh. 8.10 - Prob. 35ATSCh. 8.10 - Prob. 36ATSCh. 8.11 - Prob. 37CCCh. 8.11 - Prob. 38CCCh. 8.12 - Prob. 10LTSCh. 8.13 - Prob. 11LTSCh. 8.14 - Prob. 12LTSCh. 8.14 - Prob. 46PTSCh. 8.14 - Prob. 48ATSCh. 8.14 - Prob. 49ATSCh. 8 - Prob. 50PPCh. 8 - Prob. 51PPCh. 8 - Prob. 52PPCh. 8 - Prob. 53PPCh. 8 - Prob. 54PPCh. 8 - Prob. 55PPCh. 8 - Prob. 56PPCh. 8 - Prob. 57PPCh. 8 - Prob. 58PPCh. 8 - Prob. 59PPCh. 8 - Prob. 60PPCh. 8 - Prob. 61PPCh. 8 - Prob. 62PPCh. 8 - Prob. 63PPCh. 8 - Prob. 64PPCh. 8 - Prob. 65PPCh. 8 - Prob. 66PPCh. 8 - Prob. 67PPCh. 8 - Prob. 68PPCh. 8 - Prob. 69PPCh. 8 - Prob. 70PPCh. 8 - Prob. 71PPCh. 8 - Prob. 72PPCh. 8 - Prob. 73PPCh. 8 - Prob. 74PPCh. 8 - Prob. 75PPCh. 8 - Prob. 76PPCh. 8 - Prob. 77IPCh. 8 - Prob. 78IPCh. 8 - Prob. 79IPCh. 8 - Prob. 80IPCh. 8 - Prob. 81IPCh. 8 - Prob. 82IPCh. 8 - Prob. 83IPCh. 8 - Prob. 84IPCh. 8 - Prob. 85IPCh. 8 - Prob. 86IPCh. 8 - Prob. 87IP
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