Chapter 18, Problem 18.1P

Interpretation Introduction

(a)

Interpretation:

The increasing order of ${\text{S}}_{\text{N}}\text{2}$ reaction of the given compounds along with the reason of order is to be stated.

Concept introduction:

The nucleophilic substitution reactions are the reactions in which one nucleophile is substituted by another nucleophile. These reactions depend upon the nucleophilicity and concentration of the nucleophile. It is of two types, ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}\text{2}$.

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution bimolecular single-step reaction in which the addition of nucleophile and removal of leaving group takes place simultaneously.

The ${\text{S}}_{\text{N}}\text{1}$ reaction is a nucleophilic substitution reaction in which the substitution of nucleophile takes place. This reaction takes place in two steps. This generates a carbocation intermediate in the reaction.

Answer to Problem 18.1P

The given compounds with the increasing order of ${\text{S}}_{\text{N}}\text{2}$ reaction are arranged as shown below. The order is obtained due to the steric hindrance of the groups attached.

$\left(3-\text{bromopropyl}\right)\text{benzene}>\text{benzyl bromide}>p-\text{bromotoluene}$

Explanation of Solution

The structure of given compounds is shown below

Figure 1

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution bimolecular single-step reaction in which the addition of nucleophile and removal of leaving group takes place simultaneously. It depends on the concentration of nucleophile and substrate. It also depends on how easily a leaving group departs when nucleophile attack from the backside. It is written as shown below.

$\text{Rate}=k\left(\text{Substrate}\right)\left(\text{nucleophile}\right)$ …(1)

In $p-\text{bromotoluene}$, bromide group is directly attached to the benzene ring. When nucleophile attacks on the benzene ring. The double bond undergoes delocalization which will destroy the aromaticity of the ring. It means if the leaving group departs the conjugation is lost and a non-aromatic compound is formed which is less stable as compared to aromatic. In benzyl bromide, attack of nucleophile is difficult from backside because of the presence of steric hindrance of the phenyl ring. It is also repelled by $\text{π}$ electrons of the phenyl ring. Therefore, its rate is slower than $\left(3-\text{bromopropyl}\right)\text{benzene}$ where steric hindrance is less.

Therefore, the given compounds with the increasing order of ${\text{S}}_{\text{N}}\text{2}$ reaction are arranged as shown below.

$\left(3-\text{bromopropyl}\right)\text{benzene}>\text{benzyl bromide}>p-\text{bromotoluene}$

Conclusion

The increasing order of ${\text{S}}_{\text{N}}\text{2}$ reaction is shown below.

$\left(3-\text{bromopropyl}\right)\text{benzene}>\text{benzyl bromide}>p-\text{bromotoluene}$

Interpretation Introduction

(b)

Interpretation:

The increasing order of ${\text{S}}_{\text{N}}\text{2}$ reaction of the given compounds along with the reason of order is to be stated.

Concept introduction:

The nucleophilic substitution reactions are the reactions in which one nucleophile is substituted by another nucleophile. These reactions depend upon the nucleophilicity and concentration of the nucleophile. It is of two types, ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}\text{2}$.

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution bimolecular single-step reaction in which the addition of nucleophile and removal of leaving group takes place simultaneously.

The ${\text{S}}_{\text{N}}\text{1}$ reaction is a nucleophilic substitution reaction in which the substitution of nucleophile takes place. This reaction takes place in two steps. This generates a carbocation intermediate in the reaction.

Answer to Problem 18.1P

The given compounds with the increasing order of ${\text{S}}_{\text{N}}\text{2}$ reaction are arranged as shown below. The order is obtained due to the stability order of the groups attached.

$1-\text{bromocyclohexene}<\text{bromocyclohexane}<1-\left(\text{bromomethyl}\right)\text{cyclohexene}$

Explanation of Solution

The structure of given compounds is shown below.

Figure 2

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution bimolecular single-step reaction in which the addition of nucleophile and removal of leaving group takes place simultaneously. It depends on the concentration of nucleophile and substrate. It also depends on how easily a leaving group departs when nucleophile attack from the backside. It is written as shown below.

$\text{Rate}=k\left(\text{Substrate}\right)\left(\text{nucleophile}\right)$ …(1)

In $1-\text{bromocyclohexene}$, it is least reactive because of the presence of $\text{π}$ electrons cloud of the double bond present. In ${\text{S}}_{\text{N}}\text{2}$ reaction, primary alkyl halide is more reactive than secondary. The compound, $1-\left(\text{bromomethyl}\right)\text{cyclohexene}$ is a primary alkyl halide species and $\text{bromocyclohexane}$ is a secondary alkyl halide species. It means $1-\left(\text{bromomethyl}\right)\text{cyclohexene}$ is more reactive than $\text{bromocyclohexane}$ because of their stability as alkyl halides. Their overall increasing order of ${\text{S}}_{\text{N}}\text{2}$ reaction is written in the order as shown below.

$1-\text{bromocyclohexene}<\text{bromocyclohexane}<1-\left(\text{bromomethyl}\right)\text{cyclohexene}$.

Conclusion

The increasing order of ${\text{S}}_{\text{N}}\text{2}$ reaction is shown below.

$1-\text{bromocyclohexene}<\text{bromocyclohexane}<1-\left(\text{bromomethyl}\right)\text{cyclohexene}$