Animated Solution for Chemistry - Organic Chemistry: The increasing order of the reactivity of the following with ${\text{LiAlH}}_4$ is

A Tale of Leaving Groups

When dealing with carboxylic acid derivatives, understanding their relative reactivity is a fundamental pillar of organic chemistry. In this problem, we are asked to determine the increasing order of reactivity of four propanoic acid derivatives towards lithium aluminum hydride (${\text{LiAlH}}_4$).

The four compounds are: (A) Propanamide (an amide) (B) Methyl propanoate (an ester) (C) Propanoyl chloride (an acid chloride) (D) Propanoic anhydride (an acid anhydride)

Nucleophilic Acyl Substitution

Lithium aluminum hydride is a powerful reducing agent that acts as a source of nucleophilic hydride ions ($H^{-}$). When it reacts with an acid derivative, it undergoes a nucleophilic acyl substitution reaction.

The mechanism proceeds in two main steps. First, the nucleophile attacks the electrophilic carbonyl carbon, pushing the pi electrons onto the oxygen atom and forming a tetrahedral intermediate. Second, the carbonyl double bond reforms, and the leaving group is expelled.

The rate-determining step in this overall transformation is heavily influenced by the ability of the leaving group to depart. A fundamental rule in organic chemistry is that weaker bases make better leaving groups. This is because weaker bases are more stable on their own and are more willing to accept the electron pair from the breaking bond.

Let's evaluate the leaving groups for our four compounds: - For the acid chloride (C), the leaving group is the chloride ion (${\text{Cl}}^{-}$). - For the acid anhydride (D), the leaving group is the carboxylate ion (${\text{RCOO}}^{-}$). - For the ester (B), the leaving group is the alkoxide ion (${\text{RO}}^{-}$). - For the amide (A), the leaving group is the amide ion (${\text{NH}}_2^{-}$).

Now, we rank these leaving groups based on their basicity. Hydrochloric acid is a very strong acid, meaning its conjugate base, ${\text{Cl}}^{-}$, is extremely weak. Therefore, ${\text{Cl}}^{-}$ is an excellent leaving group. Carboxylic acids are moderately weak acids, making ${\text{RCOO}}^{-}$ a moderate base. Alcohols are weaker acids than carboxylic acids, so ${\text{RO}}^{-}$ is a stronger base than ${\text{RCOO}}^{-}$. Finally, ammonia is a very weak acid, making the ${\text{NH}}_2^{-}$ ion an exceptionally strong base and a terrible leaving group.

The leaving group ability follows the order:

Because the reactivity of the acid derivative is directly proportional to the leaving group ability, the reactivity order perfectly mirrors the leaving group ability. The acid chloride is the most reactive, and the amide is the least reactive.

The increasing order of reactivity is: Amide (A) < Ester (B) < Anhydride (D) < Acid Chloride (C)

This corresponds to option (a).