Q39E - Organic Chemistry (Mcmurry)

Question

21-39 Succinic anhydride yields the cyclic imide succinimide when heated with ammonium chloride at 200°C. Propose a mechanism for this reaction. Why do you suppose such a high reaction temperature is required?
 

Step-by-Step Solution

Answer

Here the Mechanism is illustrated below,

1.$\downarrow {\text{ NH}}_4\mathrm{CL}\underset{\mathrm{Sublimation}}{\overset{\begin{array}{l}\mathrm{gentle}\\ \mathrm{\Delta }\end{array}}{\to }}{\text{ NH}}_3\uparrow \text{ + HCl}\uparrow$

2.

Step 1: Introduction to the Concept

A nucleophilic substitution reaction occurs when an electron-rich nucleophile contacts a positively charged electrophile in order to replace a leaving group.

Step 2: Solution Explanation

The given reaction, which reflects the mechanistic aspects of both a nucleophilic acyl substitution and the effect of heat to cause an intramolecular nucleophilic attack by a nitrogen nucleophile, is an imide type (-NH) followed by a concerted lop of water ( ${\mathrm{H}}_2\mathrm{O}$), which is a better leaving group and weaker base than OH:

1.Ammonium chloride (white solid) sublimes to gaseous ammonic and hydrogen chloride gas potential nucleophile when heated gently.

 ${\text{NH}}_4\mathrm{CL}\underset{}{\overset{\mathrm{\Delta }}{\to }}\underset{\mathrm{Potential}\text{ nucleophile}}{\underset{\downarrow }{{\text{NH}}_3\uparrow }}\text{ + HCl}\uparrow$

2.For step 1, no heavy heating is required at first.

3. The reactant, succinic anhydride, has now been added to the reaction system.

Ammonia attacks the acyl carbon of one of the carbonyl groups of acetic anhydride immediately after formation (both carbonyl groups are equivalent). This is the initial step in nucleophilic addition: the nucleophilic targets the electron deficient resulting polarised carbonyl carbon.

It forms the typical tetrahedral ( ${\mathrm{sp}}^3$ carbon) intermediate. The solution is heated to a high temperature of >200°C; at the lowest level, a proton is lost and the carbon oxygen clearable bond is repaired; at the same time, the C-O bond breaks and the proton adds to the oxygen.

The correct completion of the reaction necessitates the production of OH as ${\mathrm{OH}}^{+}$. However, because OH is a poor leaving group, the excessively high temperature was used. Last, a second proton is added to the amide nitrogen ${\mathrm{NH}}_2$, and the same is done to OH, which is a better leaving group than  ${\mathrm{H}}_2\mathrm{O}$. The end product succinimide is made up of intramolecular bonds formed by the imide nitrogen. As a result of the dehydration, there is a nucleophilic acyl substitution.

In acyl substitution, the reaction exhibits the inversion of possible leaving group obesity, as well as internal nucleophilic attack to generate the desired product.