Question

The following reactivity order has been found for the saponification of alkyl acetates by aqueous \(\mathrm{NaOH}\). Explain. $$ \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3}>\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}>\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}>\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{3} $$

Short answer

Smaller alkyl groups with less steric hindrance react faster in saponification.

Step-by-step solution

Understand Saponification

Saponification is the hydrolysis of an ester under basic conditions to produce a carboxylate salt and an alcohol. In the case of alkyl acetates, aqueous \( \mathrm{NaOH} \) breaks the ester bond, forming sodium acetate and an alcohol.

Analyze Alkyl Groups

Consider the alkyl groups attached to the ester. Smaller alkyl groups allow better access to the reactive site, leading to faster reactions. Therefore, \( \mathrm{CH}_3 \mathrm{CO}_2 \mathrm{CH}_3 \) (methyl acetate) should be more reactive than larger groups.

Examine Steric Effects

As alkyl groups become bulkier, steric hindrance increases, slowing down the reaction rate. \( \mathrm{CH}_3 \mathrm{CO}_2 \mathrm{CH}_2 \mathrm{CH}_3 \) is less hindered than \( \mathrm{CH}_3 \mathrm{CO}_2 \mathrm{CH(CH}_3)_2 \), which is less hindered than \( \mathrm{CH}_3 \mathrm{CO}_2 \mathrm{C(CH}_3)_3 \).

Match Orders to Reactivity

Methanol produced from methyl acetate creates minimal hindrance, allowing for rapid reaction. Ethanol, isopropanol, and tert-butanol each contribute to increasing steric hindrance, correlating with the observed decreasing order of reactivity.

Key concepts

Reactivity order

When talking about the reactivity order in the context of saponification, it refers to the sequence in which different esters react under similar conditions. Reactivity is impacted by the size and shape of attached alkyl groups, which influences how easily a nucleophile can attack the ester bond.

In the given problem, methyl acetate ($$\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{CH}_3$$) is the most reactive among the listed esters, indicating a faster reaction rate. On the other hand, drowsy giants like tert-butyl acetate ($$\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{C(CH}_3)_3$$) lag due to the physical barrier created by bulky groups.

• Methyl acetate: Most reactive
• Ethyl acetate: Less reactive
• Isopropyl acetate: Even less reactive
• Tert-butyl acetate: Least reactive

This order shows that smaller, simpler structures facilitate a quicker saponification process by reducing obstacles to nucleophilic attack.

Steric hindrance

Steric hindrance involves the physical obstruction that prevents molecules from interacting as efficiently as they might prefer. In saponification, bulky alkyl groups attached to the ester hinder the approach of the nucleophile, which in this case is the hydroxide ion ($$\mathrm{OH}^-$$), to the carbonyl carbon.

Imagine trying to open a door when big boxes block your path. In chemistry, steric hindrance works similarly by obstructing reactant access. Small groups, like a methyl group ($$\mathrm{CH}_3$$), create less hindrance compared to isopropyl ($$\mathrm{CH(CH}_3)_2$$) or tert-butyl ($$\mathrm{C(CH}_3)_3$$).

• Minimal hindrance with small alkyl groups
• Moderate hindrance with isopropyl groups
• High hindrance with tert-butyl groups

Steric hindrance is a critical factor in slowing down the saponification rate.

Alkyl groups

Alkyl groups are chains or branches of carbon and hydrogen atoms. They play a significant role in determining the characteristics of compounds in reactions like saponification.

Consider them as the crowd at a concert: too many people, and it's hard to reach the stage. In esters, smaller alkyl groups allow better exposure of the reactive site, aiding the nucleophile's direct approach to the carbonyl carbon of the ester.

• Smaller alkyl groups, like methyl, speed up reactions
• Larger alkyl groups, like tert-butyl, slow it down due to increased bulk

Alkyl groups greatly influence the speed and ease of saponification by affecting steric hindrance and thus the overall reactivity.

Ester hydrolysis

Ester hydrolysis is the chemical process where an ester molecule is split into an alcohol and a carboxylic acid or their salts. Saponification is a special case of ester hydrolysis that occurs in basic conditions, producing soap as one of the products.

In this process, the ester bond is attacked by a nucleophile, breaking it down into its components. For example, when methyl acetate is saponified with sodium hydroxide, the bond is cleaved, leading to the formation of sodium acetate and methanol.

• Basic conditions favor the reaction due to hydroxide ions
• Alcohol chains dictate the reaction specifics

Understanding ester hydrolysis is crucial because it forms the basis of many industrial and biological processes.