Chapter 8, Problem 8.43P

Interpretation Introduction

(a)

Interpretation:

Complete ${\text{S}}_{\text{N}}\text{2}$ mechanism with appropriate arrows for the given reaction is to be drawn.

Concept introduction:

A bimolecular nucleophilic substitution (${\text{S}}_{\text{N}}2$)reaction is a one step reaction. In ${\text{S}}_{\text{N}}\text{2}$ reaction, the nucleophile attacks the substrate only from the side opposite to the leaving group (backside attack). The substituents that remain on the atom being attacked undergo Walden inversion, and if the atom is a chiral center, a single stereoisomer is produced, making the ${\text{S}}_{\text{N}}\text{2}$ reaction stereospecific.

Answer to Problem 8.43P

Complete ${\text{S}}_{\text{N}}\text{2}$ mechanisms with appropriate arrows for the given reactions are:

(i)

(ii)

(iii)

(iv)

(v)

Explanation of Solution

(i)

The given reaction is

In the above reaction, $\text{NaOH}$ acts as ${\text{OH}}^{\text{-}}$ (nucleophile). In ${\text{S}}_{\text{N}}\text{2}$ reaction, the nucleophile should attack from the opposite side, but the stereochemistry for substrate is not mentioned. Thus, the nucleophile attacks and the leaving group leaves from the same carbon atom to give the product shown below:

(ii)

The given reaction is

In the above reaction, ${\text{I}}^{\text{-}}$ is a leaving group and ${\text{OH}}^{\text{-}}$ act as nucleophile from $\text{NaOH}$. In ${\text{S}}_{\text{N}}\text{2}$ reaction, the nucleophile should attack from the side opposite the $\text{C-I}$ bond; in this case, from behind the plane of the page. Thus, the new $\text{C-OH}$ bond remains behind the plane of the page.

The stereochemistry of another group ${\text{(CH}}_{\text{3}}\text{)}$ attached remains the same as no bonds, attached to the carbon, were broken or formed.

(iii)

The given reaction

In the above reaction, ${\text{I}}^{\text{-}}$ is a leaving group and ${\text{OH}}^{\text{-}}$ act as nucleophile from $\text{NaOH}$. In ${\text{S}}_{\text{N}}\text{2}$ reaction, the nucleophile should attack from the side opposite the $\text{C-I}$ bind, in this case, from the front of the plane of the page. Thus, the new $\text{C-OH}$ bond remains above the plane of the page.

The stereochemistry of another group ${\text{(CH}}_{\text{3}}\text{)}$ attached remains the same as no bonds, attached to the carbon, were broken or formed.

(iv)

The given reaction

In the above reaction, ${\text{I}}^{\text{-}}$ is a leaving group and ${\text{Br}}^{\text{-}}$ acts as a nucleophile from $\text{KBr}$. In ${\text{S}}_{\text{N}}\text{2}$ reaction, the nucleophile should attack from the side opposite to the $\text{C-I}$ bond, in this case, from behind the plane of the page. Thus, the new $\text{C-Br}$ bond remains behind the plane of the page.

The stereochemistry of another group ${\text{(CH}}_{\text{3}}\text{)}$ attached remains the same as no bonds, attached to the carbon, were broken or formed.

(v)

The given reaction

In the above reaction, ${\text{Cl}}^{\text{-}}$ is a leaving group and ${\text{CH}}_{\text{3}}{\text{O}}^{\text{-}}$ acts as a nucleophile from ${\text{NaOCH}}_{\text{3}}$. In ${\text{S}}_{\text{N}}\text{2}$ reaction, the nucleophile should attack from the side opposite to the $\text{C-Cl}$ bond, in this case, on the plane of the page as the leaving group is present at the plane of paper. Thus, the new ${\text{C-OCH}}_3$ bond remains on the plane of the page.

Another group ${\text{(CH}}_{\text{3}}\text{)}$ attached at the carbon remains unchanged, above the plane of the paper.

Conclusion

The Product of ${\text{S}}_{\text{N}}\text{2}$ reaction is obtained with inversion in configuration.

Interpretation Introduction

(b)

Interpretation:

Complete ${\text{S}}_{\text{N}}1$ mechanism with appropriate arrows for the given reaction is to be drawn.

Concept introduction:

A unimolecular nucleophilic substitution (${\text{S}}_{\text{N}}\text{1}$) reaction consists of two steps. First, the leaving group leaves in a heterolysis step, yielding a carbocation intermediate, then a nucleophile attacks the carbocation in the coordination step. Intermediates are not included in the overall mechanism.

If ${\text{S}}_{\text{N}}\text{1}$ reaction takes place at a chiral center, the products contain a mixture of both stereochemical configurations (retention as well as inversion in configuration).

Answer to Problem 8.43P

Complete ${\text{S}}_{\text{N}}1$ mechanisms with appropriate arrows for the given reactions are as:

(i)

(ii)

(iii)

(iv)

(v)

Explanation of Solution

(i)

The given reaction is

$\text{Br}$ acts as a leaving group coming up as ${\text{Br}}^{\text{-}}$. The ${\text{OH}}^{\text{-}}$ ion can act as a nucleophile. The ${\text{S}}_{\text{N}}\text{1}$ mechanism takes places in two steps. First, the leaving group leaves via heterolysis, yielding a planar carbocation, then ${\text{OH}}^{\text{-}}$ attacks ${\text{C}}^{\text{+}}$ via a coordination step, yielding the overall product.

The leaving group in the above reaction is present on the plane of the planar, no stereochemistry is mentioned. Therefore, stereochemistry is not concerned at C, and the only isomer formed is the one shown in the above diagram.

(ii)

The given reaction is

$\text{I}$ acts as a leaving group coming up as ${\text{I}}^{\text{-}}$. The ${\text{OH}}^{\text{-}}$ ion can act as a nucleophile. The ${\text{S}}_{\text{N}}\text{1}$ mechanism takes places in two steps. First, the leaving group leaves via heterolysis, yielding a planar carbocation, then ${\text{OH}}^{\text{-}}$ attacks ${\text{C}}^{\text{+}}$ via a coordination step, yielding the overall product.

In the above reaction, leaving group (I) is present above the plane of the paper, thus the nucleophile attacks from both sides of the paper- behind the paper and in front of the paper resulting in the formation of two stereoisomers as shown above.

(iii)

The given reaction

$\text{I}$ acts as a leaving group coming up as ${\text{I}}^{\text{-}}$. The ${\text{OH}}^{\text{-}}$ ion can act as a nucleophile. The ${\text{S}}_{\text{N}}\text{1}$ mechanism takes places in two steps. First, the leaving group leaves via heterolysis, yielding a planar carbocation, then ${\text{OH}}^{\text{-}}$ attacks ${\text{C}}^{\text{+}}$ via a coordination step, yielding the overall product.

In the above reaction, leaving group (I) is present behind the plane of the paper, thus the nucleophile attacks from both sides of the paper- behind the paper and in front of the paper resulting in the formation of two stereoisomers as shown above.

(iv)

The given reaction

$\text{I}$ acts as a leaving group coming up as ${\text{I}}^{\text{-}}$. The ${\text{Br}}^{\text{-}}$ ion can act as a nucleophile. The ${\text{S}}_{\text{N}}\text{1}$ mechanism takes places in two steps. First, the leaving group leaves via heterolysis, yielding a planar carbocation, then ${\text{Br}}^{\text{-}}$ attacks ${\text{C}}^{\text{+}}$ via a coordination step, yielding the overall product.

In the above reaction, the carbocation is stabilized by proton transfer. The new carbocation formed at the adjacent carbon from the leaving group is a more stable carbocation. The stereochemistry at the carbon (more stable carbocation) is of no concern. Therefore, the only isomer formed is as shown above.

(v)

The given reaction

$\text{Cl}$ acts as a leaving group coming up as ${\text{Cl}}^{\text{-}}$. The ${\text{CH}}_{\text{3}}{\text{O}}^{\text{-}}$ ion can act as a nucleophile. The ${\text{S}}_{\text{N}}\text{1}$ mechanism takes places in two steps. First, the leaving group leaves via heterolysis, yielding a planar carbocation, then ${\text{CH}}_{\text{3}}{\text{O}}^{\text{-}}$ attacks ${\text{C}}^{\text{+}}$ via a coordination step, yielding the overall products.

In the above reaction, chiral carbocation is formed once the leaving group leaves. Therefore, the nucleophile can attack from either side- from behind the paper and in front of the paper resulting in the formation of two stereoisomers as shown above.

Conclusion

Product of ${\text{S}}_{\text{N}}\text{1}$ reaction is a mixture of both stereochemical configurations.