Chapter 18, Problem 18.46P

Interpretation Introduction

(a)

Interpretation:

The mechanism and product for the given reactions, when they are catalyzed by the base, are to be drawn.

Concept introduction:

The carbonyl carbon is electrophilic in nature due to the electronegativity difference. The nucleophile adds the electrophilic carbon of the carbonyl group. The reaction follows a different mechanism for an acid and a base catalyst. When the reaction is catalyzed by the acid, the first step of the reaction is always protonation of the carbonyl oxygen, which increases the electrophilicity of the carbonyl carbon. For a base-catalyzed reaction, the first step is the deprotonation of the nucleophilic species to form the anion, which helps to enhance the nucleophilic character.

Answer to Problem 18.46P

1. The mechanism for the first reaction, when it catalyzed by the base:

   

2. The product for the first reaction, when it catalyzed by the base:

   

3. The mechanism for the second reaction, when it catalyzed by the base:

   

4. The product for the second reaction, when it catalyzed by the base:

   

Explanation of Solution

The first given reaction is

Under the basic condition, the base (${\text{HO}}^{\text{-}}$) deprotonates the nucleophile (${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{SH}$) to form the anion (${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{S}}^{\text{-}}$), which acts as a strong nucleophile.

The nucleophile adds the electrophilic carbon of the given substrate, leading to form an alkoxide ion. Alkoxide ion then abstracts proton from the solvent, water, to form the uncharged product.

Thus, the product of the first reaction, when it catalyzed by the base, is

The second given reaction is

Under the basic condition, the base (${\text{HO}}^{\text{-}}$) deprotonates the nucleophile (${\text{PhNH}}_{\text{2}}$) to form anion (${\text{PhNH}}^{\text{-}}$), which acts as a strong nucleophile.

In the second step, the attack of the nucleophile (${\text{PhNH}}^{\text{-}}$) takes place at the $\text{β}$ carbon of the substrate, leading to the formation of an alkoxide ion. An alkoxide ion is protonated by water to form the uncharged enol. An enol is then converted to a more stable keto form via tautomerization.

Thus, the product of the given reaction is

Conclusion

The mechanism and product for the given reaction are drawn on the basis of the reaction condition.

Interpretation Introduction

(b)

Interpretation:

The mechanism and product for the given reactions, when they are catalyzed by the acid, are to be drawn.

Concept introduction:

The carbonyl carbon is electrophilic in nature due to the electronegativity difference. The nucleophile adds the electrophilic carbon of the carbonyl group. The reaction follows a different mechanism for an acid and a base catalyst. When the reaction is catalyzed by the acid, the first step of the reaction is always protonation of the carbonyl carbon, which increases the electrophilicity of the carbonyl oxygen. For a base-catalyzed reaction, the first step is the deprotonation of the nucleophilic species to enhance the nucleophilic character.

Answer to Problem 18.46P

1. The mechanism for the first given reaction, when it catalyzed by the acid:

   

2. The product for the first reaction, when it catalyzed by the acid:

   

3. The mechanism for the second reaction, when it catalyzed by the acid:

   

4. The product for the second reaction, when it catalyzed by the acid:

   

Explanation of Solution

The first given reaction is

Under the acidic condition, the carbonyl oxygen of the given reactant (an aldehyde) is protonated by the acid to form the oxonium ion, which increases the electrophilicity of the carbonyl carbon.

Then the nucleophile attack of (${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{SH}$) takes place at the more electrophilic carbonyl carbon. The positive charge on the sulfur is deprotonated by the water to form the uncharged product.

Thus, the product of the given reaction is

The second given reaction is

Under the acidic condition, the carbonyl oxygen of the given reactant ($\text{α,β}$ unsaturated ketone) is protonated by the acid to form the oxonium ion, which increases the electrophilicity of the carbonyl carbon. The formed oxonium ion is also stabilized by resonance.

In the second step, the attack of the nucleophile (${\text{PhNH}}_{\text{2}}$) takes place at the $\text{β}$ carbon of the substrate. The positive charge on the nitrogen is deprotonated by the water to form the uncharged enol. An enol is then converted to a more stable keto form via tautomerization.

Thus, the product of the given reaction is

Conclusion

The mechanism and product for the given reaction are drawn on the basis of the reaction condition.