Animated Solution for Chemistry - Organic Chemistry: Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in the above reaction is

Organic chemistry is often described as a molecular dance, where electrons flow from areas of high density to areas of low density, breaking old bonds and forging new ones. In this problem, we are presented with a classic reaction: the interaction between sodium ethoxide and ethanoyl chloride. To predict the product, we must understand the nature of the reactants and the fundamental mechanism that governs their interaction.

Let's break down our starting materials. We have sodium ethoxide ($C_2{H}_5{O}^{-}N{a}^{+}$), which is an ionic compound. In solution, it dissociates to give the ethoxide ion ($C_2{H}_5{O}^{-}$), a strong nucleophile and a strong base.

On the other hand, we have ethanoyl chloride ($C{H}_{3}COCl$), also known as acetyl chloride. This molecule belongs to the family of acid chlorides, which are the most reactive of all carboxylic acid derivatives. The carbonyl carbon in ethanoyl chloride is highly electrophilic. Why? Because it is bonded to two highly electronegative atoms: oxygen and chlorine. Both atoms pull electron density away from the carbon, leaving it with a significant partial positive charge ($delt{a}^{+}$), making it a prime target for nucleophilic attack.

Nucleophilic Acyl Substitution

When a strong nucleophile like the ethoxide ion encounters an acid chloride, it undergoes a reaction known as Nucleophilic Acyl Substitution. Unlike $S_{N}2$ reactions in alkyl halides, which happen in a single concerted step, nucleophilic acyl substitution occurs in two distinct stages: Addition followed by Elimination.

Addition

The negatively charged oxygen of the ethoxide ion attacks the electrophilic carbonyl carbon of ethanoyl chloride. Carbon can only form four bonds, so as the new carbon-oxygen bond forms, the $pi$-electrons of the carbon-oxygen double bond are pushed up onto the carbonyl oxygen.

This step destroys the $s{p}^2$ hybridization of the carbonyl carbon, converting it into an $s{p}^3$ hybridized state. The resulting structure is called a tetrahedral intermediate. In this intermediate, the central carbon is bonded to a methyl group, an ethoxy group, a chloride ion, and an oxygen atom bearing a full negative charge ($O^{-}$).

Elimination

The tetrahedral intermediate is highly unstable. The negatively charged oxygen strongly desires to reform the stable carbon-oxygen double bond. As the lone pair on the oxygen collapses back down to recreate the $pi$-bond, the carbon must expel one of its attached groups to maintain its tetravalency.

This is where the concept of leaving group ability becomes crucial. The intermediate has two potential leaving groups: the ethoxide ion ($C_2{H}_5{O}^{-}$) and the chloride ion ($C{l}^{-}$). Which one will leave?

A good leaving group is a weak base. Chloride is the conjugate base of hydrochloric acid ($HCl$), a very strong acid, making chloride an exceptionally weak base and a fantastic leaving group. Ethoxide, conversely, is the conjugate base of ethanol, a weak acid, making it a strong base and a poor leaving group. Therefore, the chloride ion is expelled.

With the expulsion of the chloride ion, the carbon-oxygen double bond is restored. The resulting organic molecule is $C{H}_{3}COO{C}_2{H}_5$, which is ethyl ethanoate (commonly known as ethyl acetate), an ester.

The expelled chloride ion ($C{l}^{-}$) pairs up with the spectator sodium ion ($N{a}^{+}$) that was present from the beginning, forming sodium chloride ($NaCl$) as a byproduct.

This reaction perfectly illustrates how understanding the flow of electrons and the relative stability of leaving groups allows us to confidently predict the outcome of complex organic transformations.