Given that tertiary alkyl halides R 3 CX, undergo spontaneous dissociation to yield R 3 X + plus halide ion. Among (CH 3 ) 3 CBr and H 2 C=CH C(CH 3 ) 2 Br which will react faster is to be explained. Concept introduction: The stability of free radicals is in the order benzyl > allyl > tertiary > secondary > primary > mehyl. The benzyl and allyl radicals are stabilized by resonance while tertiary, secondary and primary radicals are stabilized by hyperconjucation. To explain: Among (CH 3 ) 3 CBr and H 2 C=CH C(CH 3 ) 2 Br which will react faster.

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Answer 1

Question

Chapter 10.SE, Problem 45AP

Interpretation Introduction

Interpretation:

Given that tertiary alkyl halides R₃CX, undergo spontaneous dissociation to yield R₃X⁺ plus halide ion. Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster is to be explained.

Concept introduction:

The stability of free radicals is in the order benzyl > allyl > tertiary > secondary > primary > mehyl. The benzyl and allyl radicals are stabilized by resonance while tertiary, secondary and primary radicals are stabilized by hyperconjucation.

To explain:

Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster.

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Answer 2

Question

Chapter 10.SE, Problem 45AP

Interpretation Introduction

Interpretation:

Given that tertiary alkyl halides R₃CX, undergo spontaneous dissociation to yield R₃X⁺ plus halide ion. Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster is to be explained.

Concept introduction:

The stability of free radicals is in the order benzyl > allyl > tertiary > secondary > primary > mehyl. The benzyl and allyl radicals are stabilized by resonance while tertiary, secondary and primary radicals are stabilized by hyperconjucation.

To explain:

Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 10.SE, Problem 45AP

Interpretation Introduction

Interpretation:

Given that tertiary alkyl halides R₃CX, undergo spontaneous dissociation to yield R₃X⁺ plus halide ion. Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster is to be explained.

Concept introduction:

The stability of free radicals is in the order benzyl > allyl > tertiary > secondary > primary > mehyl. The benzyl and allyl radicals are stabilized by resonance while tertiary, secondary and primary radicals are stabilized by hyperconjucation.

To explain:

Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 10.SE, Problem 45AP

Interpretation Introduction

Interpretation:

Given that tertiary alkyl halides R₃CX, undergo spontaneous dissociation to yield R₃X⁺ plus halide ion. Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster is to be explained.

Concept introduction:

The stability of free radicals is in the order benzyl > allyl > tertiary > secondary > primary > mehyl. The benzyl and allyl radicals are stabilized by resonance while tertiary, secondary and primary radicals are stabilized by hyperconjucation.

To explain:

Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Chapter 10.SE, Problem 45AP

Interpretation Introduction

Interpretation:

Given that tertiary alkyl halides R₃CX, undergo spontaneous dissociation to yield R₃X⁺ plus halide ion. Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster is to be explained.

Concept introduction:

The stability of free radicals is in the order benzyl > allyl > tertiary > secondary > primary > mehyl. The benzyl and allyl radicals are stabilized by resonance while tertiary, secondary and primary radicals are stabilized by hyperconjucation.

To explain:

Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster.

Answer 6

Chapter 10.SE, Problem 45AP

Interpretation Introduction

Interpretation:

Given that tertiary alkyl halides R₃CX, undergo spontaneous dissociation to yield R₃X⁺ plus halide ion. Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster is to be explained.

Concept introduction:

The stability of free radicals is in the order benzyl > allyl > tertiary > secondary > primary > mehyl. The benzyl and allyl radicals are stabilized by resonance while tertiary, secondary and primary radicals are stabilized by hyperconjucation.

To explain:

Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster.

Answer 7

Chapter 10.SE, Problem 45AP

Interpretation Introduction

Interpretation:

Given that tertiary alkyl halides R₃CX, undergo spontaneous dissociation to yield R₃X⁺ plus halide ion. Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster is to be explained.

Concept introduction:

The stability of free radicals is in the order benzyl > allyl > tertiary > secondary > primary > mehyl. The benzyl and allyl radicals are stabilized by resonance while tertiary, secondary and primary radicals are stabilized by hyperconjucation.

To explain:

Among (CH₃)₃CBr and H₂C=CH C(CH₃)₂Br which will react faster.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 10.1 - Prob. 1P Ch. 10.1 - Draw structures corresponding to the following...Ch. 10.2 - Prob. 3P Ch. 10.2 - Taking the relative reactivities of 1Â°, 2Â°, and...Ch. 10.4 - Prob. 5P Ch. 10.4 - The major product of the reaction of...Ch. 10.4 - Prob. 7P Ch. 10.5 - Prob. 8P Ch. 10.6 - Prob. 9P Ch. 10.6 - How might you replace a halogen substituent by a...

Solution Summary: The author explains that the tertiary alkyl halides R 3 CX undergo spontaneous dissociation to yield R3 X + plus

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