Organic Chemistry
Organic Chemistry
8th Edition
ISBN: 9781337516402
Author: Brown
Publisher: Cengage
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Chapter 11, Problem 11.31P

Show reagents and experimental conditions to synthesize the following compounds from 1-propanol (any derivative of 1-propanol prepared in one part of this problem may be used for the synthesis of another part of the problem).

  1. (a) Propanal
  2. (b) Propanoic acid
  3. (c) Propene
  4. (d) 2-Propanol
  5. (e) 2-Bromopropane
  6. (f) 1-Chloropropane
  7. (g) 1,2-Dibromopropane
  8. (h) Propyne
  9. (i) 2-Propanone
  10. (j) 1-Chloro-2-propanol
  11. (k) Methyloxirane
  12. (l) Dipropyl ether
  13. (m) Isopropyl propyl ether
  14. (n) 1-Mercapto-2-propanol
  15. (o) 1-Amino-2-propanol
  16. (p) 1,2-Propanediol

(a)

Expert Solution
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Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept introduction:

Oxidation of primary alcohol to aldehyde: A Primary alcohol is oxidized to an aldehyde by PCC.

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  1

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  2

The oxidation of primary alcohol to aldehyde is performed using PCC gives rise to the desired aldehyde (Propanal) product as shown above.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept introduction:

Hydroboration-oxidation: Generally alkenes react with boron followed by hydrogen peroxide and in presence of base like sodium hydroxide it gives corresponding alcohol and it follows anti-Markovnikov rule.

Oxidation of primary alcohol to aldehyde: A Primary alcohol is oxidized to an aldehyde by PCC.

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  3

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  4

The oxidation of primary alcohol (A) to carboxylic acid (C) is performed using chromic acid on heating; the aldehyde with water in the solution g9ives hydrated aldehyde intermediate leads to the desired Propanoic acid product as shown above.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept introduction:

Acid-catalyzed dehydration of alcohols: An alcohol can be converted into an alkene by dehydration (with elimination of water from adjacent carbon). By heating alcohol with either 85% phosphoric acid or concentrated sulfuric acid. Primary alcohol require high temperature as high as 180oC whereas secondary alcohol require somewhat lower temperature. Acid-catalyzed dehydration of tertiary alcohols often requires temperature only slightly above room temperature.

Ease of dehydration of alcohols:

The ease of acid-catalyzed dehydration of alcohols is in the order as follows,

1oalcohol<2oalcohol < 3oalcohol

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  5

The alcohol (A) compound undergoes acid-catalyzed dehydration

Acid-catalyzed dehydration reaction of alcohol with H3PO4 leads to the formation of alkene; the reaction follows Zaitsev’s rule (more substituted product is predominant). Rearrangement is possible for the extensive β branching.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept introduction:

Acid-catalyzed dehydration of alcohols: An alcohol can be converted into an alkene by dehydration (with elimination of water from adjacent carbon). By heating alcohol with either 85% phosphoric acid or concentrated sulfuric acid. Primary alcohol require high temperature as high as 180oC whereas secondary alcohol require somewhat lower temperature. Acid-catalyzed dehydration of tertiary alcohols often requires temperature only slightly above room temperature.

Ease of dehydration of alcohols:

The ease of acid-catalyzed dehydration of alcohols is in the order as follows,

1oalcohol<2oalcohol < 3oalcohol

Acid Catalyzed Hydration Reaction: The reaction involves breaking of pi bonds between carbon-carbon multiple bonds and addition of alcohol to more substituted position of carbon in the molecule.

First step is the acid donates proton to the alkene which leads to the formation of more stable carbo cation.

Then, the water is added to the given alkene through acid catalyzed reaction where the water gets added to the carbo cation finally, the removal of one proton from oxonium ion (oxygen with one positive charge) using water results in the formation of product.

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  6

In the above reaction, A to B involves acid-catalyzed dehydration process; followed by acid-catalyzed hydration process (B to C).

Acid-catalyzed hydration:

Initially H+ ions from H2SO4 attacks the carbon with more number of H in C=C of given reactant and forms stable secondary carbocation intermediate. The lone pair of electrons in H2O attacks the carbocation and forms compound with oxonium ion (oxygen with positive charge).

Finally, the oxonium ion removes one proton gives the secondary alcohol as the major product.

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept introduction:

Oxidation of secondary alcohol to aldehyde: A secondary alcohol is oxidized to a aldehyde by chromic acid. The mechanism involves initial formation of an alkyl chromate intermediate, followed by reaction with base to remove a proton, generating the carbonyl group of an aldehyde reducing the chromium (VI) to chromium (IV).

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  7

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  8

The Propanol undergoes acid-catalyzed dehydration; further acid-catalyzed hydration forming secondary alcohol (C), reaction of secondary alcohol with HBr leads to desired compound (D)

(or)

Reaction of Compound (B) reacts with HBr gives desired compound (D).

(f)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept introduction:

Conversion of alcohol into haloalkane:

Conversion of primary and secondary alcohols to Chloroalkane is thionyl chloride. The reaction is carried out in the presence of base such as pyridine or trimethylamine.

The reaction of an alcohol with thionyl chloride is the formation of an alkyl Chlorosulfite that converts hydroxide ion (poor leaving group) into Chlorosulfite (good leaving group).  Nucleophilic displacement of this leaving group gives product.

Leaving group: Leaving group can be any groups or atoms that get detached from either neutral or charged organic compounds. The stability of the leaving group is to stabilize the electron density that results from heterolysis cleavage of bond.

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  9

Reaction of alcohol (A) with SOCl2 in presence of base (pyridine) leads to the formation of chloroalkane (B) with an inversion of configuration due to the backside attack of chloride ion.

(g)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept introduction:

Acid-catalyzed dehydration of primary and secondary alcohols: Dehydration of primary and secondary alcohols is often accompanied by rearrangement process (shift of hydride or alkyl from β- carbon to α- carbon). Primary alcohol with little or no β- branching undergoes acid-catalyzed dehydration to give a terminal alkene and rearranged alkene. Always the more substituted alkene product predominates.

Addition of halogen across double bond: The reaction of alkene with halogen molecules undergoes electrophilic addition reaction forming 1,2-dihaloalkane compound.

Explanation of Solution

The reaction is shown below,

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  10

Acid-catalyzed dehydration reaction of alcohol with H2SO4 leads to the formation of alkene; the reaction follows Zaitsev’s rule (more substituted product is predominant). Rearrangement is possible for the extensive β branching.

Further reaction of alkene by addition of bromine across double bond gives 1,2-dibromopropane.

(h)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept introduction:

Acid-catalyzed dehydration of primary and secondary alcohols: Dehydration of primary and secondary alcohols is often accompanied by rearrangement process (shift of hydride or alkyl from β- carbon to α- carbon). Primary alcohol with little or no β- branching undergoes acid-catalyzed dehydration to give a terminal alkene and rearranged alkene. Always the more substituted alkene product predominates.

Addition of halogen across double bond: The reaction of alkene with halogen molecules undergoes electrophilic addition reaction forming 1,2-dihaloalkane compound.

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  11

Acid-catalyzed dehydration reaction of alcohol with H2SO4 leads to the formation of alkene; the reaction follows Zaitsev’s rule (more substituted product is predominant). Rearrangement is possible for the extensive β branching.

Further reaction of alkene by addition of bromine across double bond gives 1,2-dibromopropane.

Addition of strong base leads to the elimination of two moles of HBr leads to the desired alkyne product (D).

(i)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept introduction:

Acid-catalyzed dehydration of primary and secondary alcohols: Dehydration of primary and secondary alcohols is often accompanied by rearrangement process (shift of hydride or alkyl from β- carbon to α- carbon). Primary alcohol with little or no β- branching undergoes acid-catalyzed dehydration to give a terminal alkene and rearranged alkene. Always the more substituted alkene product predominates.

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  12

Propanol (A) undergoes dehydration followed by hydration gives secondary alcohol; the oxidation of secondary alcohol to ketone (D) the desired product.

(j)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept introduction:

Acid-catalyzed dehydration of primary and secondary alcohols: Dehydration of primary and secondary alcohols is often accompanied by rearrangement process (shift of hydride or alkyl from β- carbon to α- carbon). Primary alcohol with little or no β- branching undergoes acid-catalyzed dehydration to give a terminal alkene and rearranged alkene. Always the more substituted alkene product predominates.

Addition of halogen across double bond: The reaction of alkene with halogen molecules undergoes electrophilic addition reaction forming 1,2-dihaloalkane compound.

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  13

Propanol (A) reacts with phosphoric acid undergoes acid-catalyzed dehydration forming compound (B); further reaction with chlorine molecule leads to the compound (B); further reaction of compound (C).

(k)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept-Introduction:

Acid-catalyzed dehydration of primary and secondary alcohols: Dehydration of primary and secondary alcohols is often accompanied by rearrangement process (shift of hydride or alkyl from β- carbon to α- carbon). Primary alcohol with little or no β- branching undergoes acid-catalyzed dehydration to give a terminal alkene and rearranged alkene. Always the more substituted alkene product predominates.

Formation of epoxide: The alkene can be converted to epoxide when alkene is treated with MCPBA (m-Chloro per benzoic acid)

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  14

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  15

Propanol (A) with phosphoric acid on heating gives alkene compound (B) and further reaction with peroxyacid leads to the formation of epoxide compound (C).

(l)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept-Introduction:

Williamson ether synthesis: A general method for the synthesis of dialkyl ethers by a SN2 reaction between haloalkane and an alkoxide ion. The reaction involves a nucleophilic displacement of a halide ion or another good leaving group by an alkoxide ion.

Yield of ether becomes highest when the halide to be displaced is on primary carbon; yields are low in the displacement from secondary halides (because of β- elimination), and the Williamson ether synthesis fails altogether with tertiary halides (because of β- elimination via E2 mechanism).

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  16

Propanol (A) reacts with SOCl2 to give Chloropropane which further reacts with alkoxide (obtained by deprotonation of propanol with strong base) leads to the formation ether, known as Williamson’s ether synthesis.

(m)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept-Introduction:

Williamson ether synthesis: A general method for the synthesis of dialkyl ethers by a SN2 reaction between haloalkane and an alkoxide ion. The reaction involves a nucleophilic displacement of a halide ion or another good leaving group by an alkoxide ion.

Yield of ether becomes highest when the halide to be displaced is on primary carbon; yields are low in the displacement from secondary halides (because of β- elimination), and the Williamson ether synthesis fails altogether with tertiary halides (because of β- elimination via E2 mechanism).

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  17

Propanol (A) reacts with SOCl2 to give Chloropropane (B); and conversion of primary alcohol (A) to secondary alkoxide (C) shown above.

The reaction of compound (B) and compound (C) undergoes Williamson’s ether synthesis forming ether the desired product.

(n)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept-Introduction:

Acid-catalyzed dehydration of primary and secondary alcohols: Dehydration of primary and secondary alcohols is often accompanied by rearrangement process (shift of hydride or alkyl from β- carbon to α- carbon). Primary alcohol with little or no β- branching undergoes acid-catalyzed dehydration to give a terminal alkene and rearranged alkene. Always the more substituted alkene product predominates.

Formation of epoxide: The alkene can be converted to epoxide when alkene is treated with MCPBA (m-Chloro per benzoic acid)

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  18

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  19

Propanol (A) with phosphoric acid on heating gives alkene compound (B) and further reaction with peroxyacid leads to the formation of epoxide compound (C).

The epoxide ring opening using the nucleophile (SH), results in the racemic mixture of desired compound (D) due to the chiral center.

(o)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept Introduction:

Acid-catalyzed dehydration of primary and secondary alcohols: Dehydration of primary and secondary alcohols is often accompanied by rearrangement process (shift of hydride or alkyl from β- carbon to α- carbon). Primary alcohol with little or no β- branching undergoes acid-catalyzed dehydration to give a terminal alkene and rearranged alkene. Always the more substituted alkene product predominates.

Formation of epoxide: The alkene can be converted to epoxide when alkene is treated with MCPBA (m-Chloro per benzoic acid)

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  20

Explanation of Solution

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  21

Propanol (A) with phosphoric acid on heating gives alkene compound (B) and further reaction with peroxyacid leads to the formation of epoxide compound (C).

The epoxide ring opening using the nucleophile (NH3), results in the racemic mixture of desired compound (D) due to the chiral center.

(p)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Conversion of Propanol to given desired product has to be shown.

Concept introduction:

An alkene undergoes an oxidation reaction with Osmium tetra oxide and followed by hydrolysis to give a cis 1, 2 diol for example,

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  22

Explanation of Solution

The reaction is shown below,

Organic Chemistry, Chapter 11, Problem 11.31P , additional homework tip  23

Acid-catalyzed dehydration gives the alkene compound (B) and the alkene (B) undergoes oxidation using osmium tetra oxide gives cyclic osmate intermediate, this intermediate further reaction with hydrogen peroxide hydrolysis that leads to product 1,2 diol (C). The racemic mixture obtained.

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Chapter 11 Solutions

Organic Chemistry

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