Chapter 12, Problem 26P

(a)

Interpretation Introduction

Interpretation:

The product of the given reaction should be given.

Concept introduction:

Bromination of Allylic Carbons:

N-bromosuccinimide (NBS) is used for the allylic bromination through radical reaction. bromination of  allylicc carbon requires low concentration of bromine and low concentration of hydrobromic acid. If high concentration of bromine and high concentration of hydrobromic acid which leads to the formation of bromination in the double bond.

Bromination reaction starts with the homolytic cleavage of $\text{N-Br}$ bond in N-bromosuccinimide (NBS) which creates bromine radical to initiate the radical bromination reaction.

NBS bromine radical removes the allylic hydrogen which forms hydrogen bromide and allylic radical in the first propagation step, the allylic radical is stabilized by the double bond in ring. This allylic radical reaction with bromine molecule and forms allylic bromide in the second propagation step which are shown above.

Answer to Problem 26P

1-pentene undergoes bromination using N-bromosuccinamide and yields brominated compound A and B which is shown below.

Explanation of Solution

N-bromosuccinimide (NBS) is used for the allylic bromination through radical reaction.  Bromination of allylic carbon requires low concentration of bromine and low concentration of hydrobromic acid

Bromination reaction starts with the homolytic cleavage of $\text{N-Br}$ bond in N-bromosuccinimide (NBS) which creates bromine radical to initiate the radical bromination reaction.  1-pentene has one allylic position and it undergoes resonance which forms two types of radical.  Therefore, it undergoes bromination using N-bromo succinamide and yields brominated compound A and B which is shown.

(b)

Interpretation Introduction

Interpretation:

The product of the given reaction should be given.

Concept introduction:

Radical or free radical: unpaired valence electron of an atom, molecule, or ion is called as radical.

Bromination of Allylic Carbons:

N-bromosuccinimide (NBS) is used for the allylic bromination through radical reaction. bromination of  allylic carbon requires low concentration of bromine and low concentration of hydrobromic acid. If high concentration of bromine and high concentration of hydrobromic acid which leads to the formation of bromination in the double bond.

Bromination reaction starts with the homolytic cleavage of $\text{N-Br}$ bond in N-bromosuccinimide (NBS) which creates bromine radical to initiate the radical bromination reaction.

NBS bromine radical removes the allylic hydrogen which forms hydrogen bromide and allylic radical in the first propagation step, the allylic radical is stabilized by the double bond in ring. This allylic radical reaction with bromine molecule forms allylic bromide in the second propagation step which are shown above.

Answer to Problem 26P

2-methyl-2-pentene undergoes bromination using N-bromo succinamide and yields brominated compound A, B as a major product and C, D as minor product which is shown below.

Explanation of Solution

N-bromosuccinimide (NBS) is used for the allylic bromination through radical reaction. bromination of  allylicc carbon requires low concentration of bromine and low concentration of hydrobromic acid

Bromination reaction starts with the homolytic cleavage of $\text{N-Br}$ bond in N-bromosuccinimide (NBS) which creates bromine radical to initiate the radical bromination reaction. 2-methyl-2-pentene has three allylic position and it undergoes resonance which forms two types of radical.  Therefore, it undergoes bromination using N-bromo succinamide and yields A, B as a major product and C, D as minor product which is shown below.

(c)

Interpretation Introduction

Interpretation:

The product of the given reaction should be given

Concept introduction:

Radical or free radical: unpaired valence electron of an atom, molecule, or ion is called as radical.

Bromination:

2-methyl propane undergoes radical bromination which yields the 2-bromo-2-methylpropane.because bromination will occur where the tertiary radical is present. (bromination reactions are more selective reaction).

Bromination will occur on tertiary radical than the secondary than primary radical, tertiary radical is more stable radical than the other radicals.

Answer to Problem 26P

3-methyl hexane undergoes radical bromination and yields the 3-bromo-3-methylhexane which is shown below

Explanation of Solution

3-methyl hexane undergoes radical bromination and yields the 3-bromo-3-methylhexane which is shown below

(d)

Interpretation Introduction

Interpretation:

The product of the given reaction should be given.

Concept introduction:

Radical or free radical: unpaired valence electron of an atom, molecule, or ion is called as radical.

Chlorination:

2-methyl propane undergoes radical chlorination and yields the 2-bromo-2-methylpropane and 1-bromo-2-methyl propane.

Answer to Problem 26P

Cyclohexane undergoes radical chlorination and yields the 1-chloro cyclohexane which is shown below

Explanation of Solution

Cyclohexane undergoes radical chlorination, all the carbons in cyclohexane are secondary.  Therefore, it yields the 1-chloro cyclohexane which is shown above.

(e)

Interpretation Introduction

Interpretation:

The product of the given reaction should be given.

Concept introduction:

Radical or free radical: unpaired valence electron of an atom, molecule, or ion is called as radical.

Chlorination:

2-methyl propane undergoes radical chlorination and yields the 2-bromo-2-methylpropane and 1-bromo-2-methyl propane.

Answer to Problem 26P

Cyclopentane has no reaction with chlorine in dichloromethane which is shown below

Explanation of Solution

Cyclopentane has no reaction with chlorine in dichloromethane, because the reaction will not go without light or heat which is shown below

(f)

Interpretation Introduction

Interpretation:

The product of the given reaction should be given.

Concept introduction:

Radical or free radical: unpaired valence electron of an atom, molecule, or ion is called as radical.

Chlorination:

2-methyl propane undergoes radical chlorination and yields the 2-bromo-2-methylpropane and 1-bromo-2-methyl propane.

Answer to Problem 26P

Explanation of Solution

Methyl cyclopentane undergoes radical chlorination, the carbons in cyclopentane are secondary and primary.  Therefore, it yields the four types of chlorocyclopentane which is shown below.