The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH 3 O - has to be drawn. Concept Introduction: Elimination reaction: Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily. E2 is single step elimination. E2 refers bimolecular elimination in which both Carbon-halogen and Carbon-hydrogen bond should broke to gives double bond.

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Answer 1

Question

Chapter 9.13, Problem 47P

(a)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3O- has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

E2 is single step elimination. E2 refers bimolecular elimination in which both Carbon-halogen and Carbon-hydrogen bond should broke to gives double bond.

(b)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3OH has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

Allylic and benzylic halides also can undergo E1 elimination because of they can form relatively stable carbocation.

E1 is dual step elimination. E1 refers unimolecular elimination in which Carbon-halogen bond should broke to gives stable carbocation.

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3. Consider the compounds below and determine if they are aromatic, antiaromatic, or non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly drawn and you should be able to tell that the bonding electrons and lone pair electrons should reside in which hybridized atomic orbital 2. You should consider ring strain- flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti- aromaticity) H H N N: NH2 N Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic TT electrons Me H Me Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic πT electrons H HH…

A chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.

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Answer 2

Question

Chapter 9.13, Problem 47P

(a)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3O- has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

E2 is single step elimination. E2 refers bimolecular elimination in which both Carbon-halogen and Carbon-hydrogen bond should broke to gives double bond.

(b)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3OH has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

Allylic and benzylic halides also can undergo E1 elimination because of they can form relatively stable carbocation.

E1 is dual step elimination. E1 refers unimolecular elimination in which Carbon-halogen bond should broke to gives stable carbocation.

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Answer 3

Question

Chapter 9.13, Problem 47P

(a)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3O- has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

E2 is single step elimination. E2 refers bimolecular elimination in which both Carbon-halogen and Carbon-hydrogen bond should broke to gives double bond.

(b)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3OH has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

Allylic and benzylic halides also can undergo E1 elimination because of they can form relatively stable carbocation.

E1 is dual step elimination. E1 refers unimolecular elimination in which Carbon-halogen bond should broke to gives stable carbocation.

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Answer 4

Question

Chapter 9.13, Problem 47P

(a)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3O- has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

E2 is single step elimination. E2 refers bimolecular elimination in which both Carbon-halogen and Carbon-hydrogen bond should broke to gives double bond.

(b)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3OH has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

Allylic and benzylic halides also can undergo E1 elimination because of they can form relatively stable carbocation.

E1 is dual step elimination. E1 refers unimolecular elimination in which Carbon-halogen bond should broke to gives stable carbocation.

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Answer 5

Chapter 9.13, Problem 47P

(a)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3O- has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

E2 is single step elimination. E2 refers bimolecular elimination in which both Carbon-halogen and Carbon-hydrogen bond should broke to gives double bond.

(b)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3OH has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

Allylic and benzylic halides also can undergo E1 elimination because of they can form relatively stable carbocation.

E1 is dual step elimination. E1 refers unimolecular elimination in which Carbon-halogen bond should broke to gives stable carbocation.

Answer 6

Chapter 9.13, Problem 47P

(a)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3O- has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

E2 is single step elimination. E2 refers bimolecular elimination in which both Carbon-halogen and Carbon-hydrogen bond should broke to gives double bond.

Answer 7

Chapter 9.13, Problem 47P

(a)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3O- has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

E2 is single step elimination. E2 refers bimolecular elimination in which both Carbon-halogen and Carbon-hydrogen bond should broke to gives double bond.

Answer 8

(b)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3OH has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

Allylic and benzylic halides also can undergo E1 elimination because of they can form relatively stable carbocation.

E1 is dual step elimination. E1 refers unimolecular elimination in which Carbon-halogen bond should broke to gives stable carbocation.

Answer 9

(b)

Interpretation Introduction

Interpretation:

The elimination product formed when 3-bromo-3-methyl-1-butene reacts with CH3OH has to be drawn.

Concept Introduction:

Elimination reaction:

Allylic and benzylic halides contains β- hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed easily.

Allylic and benzylic halides also can undergo E1 elimination because of they can form relatively stable carbocation.

E1 is dual step elimination. E1 refers unimolecular elimination in which Carbon-halogen bond should broke to gives stable carbocation.

Answer 10

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Answer 11

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Answer 12

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Answer 13

Ch. 9.1 - Prob. 3P Ch. 9.1 - Does increasing the energy barrier for an SN2...Ch. 9.1 - Rank the following alkyl bromides from most...Ch. 9.1 - Prob. 8P Ch. 9.2 - Prob. 9P Ch. 9.2 - Prob. 10P Ch. 9.2 - Prob. 11P Ch. 9.2 - Prob. 12P Ch. 9.2 - Which substitution reaction lakes place more...Ch. 9.2 - Prob. 15P

Solution Summary: The author explains that 3-bromo-3-methyl-1-butene has - hydrogens that readily undergo E2 elimination reaction because of double bond formed in the product is relatively stable and so formed

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