Organic Chemistry - Standalone book
Organic Chemistry - Standalone book
10th Edition
ISBN: 9780073511214
Author: Francis A Carey Dr., Robert M. Giuliano
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 11, Problem 52P

Suggest reasonable explanations for each of the following observations:

The first-order rate constant for solvolysis of (CH 3 ) 2 C=CHCH 2 Cl in ethanol is over 6000 times greater than that of allyl chloride (25 o C) .

After a solution of 3-buten-2-ol

in aqueous sulfuric acid had been allowed to stand for 1 week, it was found to contain both 3-buten-2-ol

and 2-buten-1-ol .

Treatment of CH 3 CH=CHCH 2 OH

with hydrogen bromide gave the same mixture of 1-bromo-2-butene and 3-bromo-1-butene .

Treatment of 3-buten-2-ol with hydrogen bromide gave the same mixture of bromides as in part (c).

The major product in parts (c) and (d) was 1-bromo-2-butene .

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation:

The reasonable explanations for each of the given observations are to be suggested.

Concept introduction:

The allyl group contains the unit CH=CH-CH2.

In substitution reactions, allylic halides react faster than the corresponding alkyl halides.

The primary or secondary allylic carbocations are less stable than tertiary allylic carbocations.

In allylic carbocations that are not symmetrically substituted, the two resonance structures are not equivalent and do not contribute equally to the resonance hybrid. A more stable resonance form contributes more to the resonance hybrid.

Two non-equivalent resonance forms of the allylic carbocation yield two different products.

Reactions that occur via SN1 mechanism may produce more than one product due to non-equivalent resonance forms of allylic carbocations.

The regioselectivity of an SN1 reaction depends on the stability of the carbocation intermediate produced in the transition state.

A more substituted alkene is stable and forms more readily than a less substituted alkene.

Answer to Problem 52P

Solution:

a) Resonance stabilization of the carbocation formed by the removal of chloride ion from (CH3)2C=CH-CH2-Cl gives tertiary carbocation which is more stable as compared to the primary and secondary carbocation formed in case of allyl chloride due to which its rate of solvolysis is faster.

b) The resonance stabilization of the carbocation formed by the reaction of 3-buten-2-ol cause the formation of carbocation at first and third position which leads to the formation of corresponding alcohols.

c) The resonance stabilization of the carbocation formed by the reaction of 2-buten-1-ol cause the formation of carbocation at first and third position which leads to the formation of corresponding alcohols.

d) The resonance stabilization of the carbocation formed by the reaction of both 3-buten-2-ol and 2-buten-1-ol cause the formation of carbocation at first and third position which leads to the formation of same products.

e) 1-bromo-2-butene is a major product in both the reactions because it is a more substituted alkene and hence, more stable.

Explanation of Solution

a) The comparison of rate constant of solvolysis of the given allylic chlorides.

The expanded structures for the two given allylic chlorides are shown below:

Organic Chemistry - Standalone book, Chapter 11, Problem 52P , additional homework tip  1

Both the allylic chlorides follow the SN1 reaction mechanism. Under the SN1 conditions such as solvolysis in ethanol, the (CH3)2C=CH-CH2-Cl reacts over 6000 times faster than allyl chloride. Both reactions follow the first-order rate law, and their relative rates reflect the greater stability of tertiary allylic carbocation compared with the primary allylic carbocation. In the (CH3)2C=CH-CH2-Cl, the primary allylic carbocation formed upon dissociation of chlorine can be resonance stabilized to a more stable tertiary allylic carbocation. The allylic carbocation formed in allyl chloride cannot be resonance stabilized into a more stable carbocation intermediate. Thus, under the SN1 conditions such as solvolysis in ethanol, the (CH3)2C=CH-CH2-Cl reacts over 6000 times faster than allyl chloride.

Organic Chemistry - Standalone book, Chapter 11, Problem 52P , additional homework tip  2

b) Reaction of 3-buten-2-ol with aqueous sulfuric acid.

The expanded structure for 3-buten-2-ol is as shown below:

Organic Chemistry - Standalone book, Chapter 11, Problem 52P , additional homework tip  3

When a solution of 3-buten-2-ol in aqueous sulfuric acid is allowed to stand for one week, it is found to contain both 3-buten-2-ol and 3-butene-1-ol. This is because sulfuric acid is a strong acid and will protonate the hydroxyl group in 3-buten-2-ol. This will produce an allylic carbocation which can be resonance stabilized. However, the two resonance forms for this allylic carbocation are not equivalent. Each allylic carbocation will undergo nucleophilic substitution reaction to yield two different products as shown below.

Organic Chemistry - Standalone book, Chapter 11, Problem 52P , additional homework tip  4

c) Treatment of the given compound with hydrogen bromide.

The structure for CH3CH=CHCH2OH is shown below:

Organic Chemistry - Standalone book, Chapter 11, Problem 52P , additional homework tip  5

When CH3-CH=CH-CH2-OH is treated with hydrogen bromide, it yields a mixture of 1-bromo-2-butene and 3-bromo-1-butene. This is because hydrogen bromide protonates the hydroxyl group in CH3-CH=CH-CH2-OH. This will produce an allylic carbocation which can be

resonance stabilized. However, the two resonance forms for this allylic carbocation are not equivalent. Each allylic carbocation will undergo nucleophilic substitution reaction to form two different allylic bromides as shown below.

Organic Chemistry - Standalone book, Chapter 11, Problem 52P , additional homework tip  6

d) Treatment of 3-butene-2-ol with hydrogen bromide.

The structure for 3-butene-2-ol is as shown below:

Organic Chemistry - Standalone book, Chapter 11, Problem 52P , additional homework tip  7

When 3-buten-2-ol is treated with hydrogen bromide, it yields a mixture of 1-bromo-2-butene and 3-bromo-1-butene. This is because hydrogen bromide protonates the hydroxyl group in (CH3)2C=CH-CH2-OH. This will produce an allylic carbocation which can be resonance stabilized. However, the two resonance forms for this allylic carbocation are not equivalent. Each allylic carbocation will undergo substitution reaction to form two different allylic bromides as shown below:

Organic Chemistry - Standalone book, Chapter 11, Problem 52P , additional homework tip  8

e) Major products obtained in part (c) and (d).

The products obtained in part (c) and (d) are as follows:

Organic Chemistry - Standalone book, Chapter 11, Problem 52P , additional homework tip  9

3-butene-1-ol and 2-buten-1-ol, when treated with hydrogen bromide, yield a mixture of 1-bromo-2-butene and 3-bromo-1-butene. The major product is 1-bromo-2-butene.

This is because this reaction follows SN1 mechanism pathway. The first step in an SN1 mechanism is the formation of a carbocation intermediate. Its key features are carbocation formation in step 1 and bonding of the nucleophile (bromide) to the carbocation in step 2. The protonation of the hydroxyl group in both the starting alcohols will produce identical allylic carbocation which can be resonance stabilized.

The major product formed in the reaction is governed by the fact that a more substituted alkene is stable and forms readily than a less substituted alkene. The bromine atom bonds to the carbon that carries the positive charge, giving 1-bromo-2-butene as the major product and 3-bromo-1-butene as the minor product because 1-bromo-2-butene is a more substituted alkene while 3-bromo-1-butene is a less substituted alkene.

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Chapter 11 Solutions

Organic Chemistry - Standalone book

Ch. 11.5 - Prob. 11PCh. 11.6 - Prob. 12PCh. 11.8 - Prob. 13PCh. 11.9 - What dienes containing isolated double bonds are...Ch. 11.10 - Prob. 15PCh. 11.10 - Prob. 16PCh. 11.11 - Prob. 17PCh. 11.12 - Dicarbonyl compounds such as quinones are reactive...Ch. 11.12 - 2,3-Di-tert-butyl-1,3-butadiene is extremely...Ch. 11.12 - Methyl acrylate (H2C=CHCO2CH3) reacts with...Ch. 11.13 - Prob. 21PCh. 11.14 - What diene and dienophile could you use to prepare...Ch. 11.14 - Write equations in the synthetic direction for the...Ch. 11.16 - Prob. 24PCh. 11.16 - Prob. 25PCh. 11 - Write structural formulas for each of the...Ch. 11 - Give an acceptable IUPAC name for each of the...Ch. 11 - A certain species of grasshopper secretes an...Ch. 11 - Which of the following are chiral?...Ch. 11 - Describe the molecular geometry expected for...Ch. 11 - Prob. 31PCh. 11 - What compound of molecular formula C6H10 gives...Ch. 11 - Prob. 33PCh. 11 - Prob. 34PCh. 11 - Prob. 35PCh. 11 - Prob. 36PCh. 11 - Identify the more reactive dienophile in each of...Ch. 11 - Prob. 38PCh. 11 - Prob. 39PCh. 11 - Prob. 40PCh. 11 - Prob. 41PCh. 11 - Prob. 42PCh. 11 - Prob. 43PCh. 11 - Prob. 44PCh. 11 - Prob. 45PCh. 11 - Prob. 46PCh. 11 - Show how to prepare each of the following...Ch. 11 - Prob. 48PCh. 11 - Prob. 49PCh. 11 - Prob. 50PCh. 11 - Compound A was converted to compound B by the...Ch. 11 - Suggest reasonable explanations for each of the...Ch. 11 - Prob. 53PCh. 11 - Prob. 54PCh. 11 - Prob. 55DSPCh. 11 - Prob. 56DSPCh. 11 - Prob. 57DSPCh. 11 - Prob. 58DSPCh. 11 - Prob. 59DSP
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