Question

Barbiturates are prepared by treating a derivative of diethyl malonate with urea in the presence of sodium ethoxide as a catalyst. Following is an equation for the preparation of barbital, a long-duration hypnotic and sedative, from diethyl diethylmalonate and urea. Barbital is prescribed under one of a dozen or more trade names. (a) Propose a mechanism for this reaction. (b) The \(\mathrm{p} K_{\mathrm{a}}\) of barbital is 7.4. Which is the most acidic hydrogen in this molecule? How do you account for its acidity?

Short answer

Question: Provide a step-by-step mechanism for the preparation of barbital from diethyl diethylmalonate and urea, using sodium ethoxide as a catalyst. Identify the most acidic hydrogen in barbital and explain its acidity. Answer: 1. Formation of a nucleophilic intermediate: Sodium ethoxide abstracts a proton from the diethyl diethylmalonate, generating a nucleophilic enolate ion intermediate. 2. Nucleophilic attack: The enolate ion attacks the carbonyl carbon in the urea, forming a new C-N bond and a negative oxygen. 3. Proton transfer: The negative oxygen abstracts a proton from one of the amide nitrogens, creating a neutral intermediate. 4. Intramolecular nucleophilic attack: The amide nitrogen attacks the carbonyl carbon, forming a five-member ring and releasing an ethanol molecule. The most acidic hydrogen in barbital is attached to the nitrogen atom in the imidazole ring. Its acidity is due to the stabilization of the resulting conjugate base through resonance structures, delocalizing the negative charge across both nitrogen atoms and the carbonyl group.

Step-by-step solution

Write down the reaction equation

First, we need to write down the equation of the reaction used to prepare barbital from diethyl diethylmalonate and urea. Diethyl diethylmalonate (1): \(\displaystyle \mathrm{C}\mathrm{H}_{3}\mathrm{CH}_{2}\mathrm{OOC}\left(\mathrm{COOCH}_{2}\mathrm{CH}_{3}\right)_{2}\) Urea (2): \(\displaystyle \mathrm{NH}_{2}\left(\mathrm{CONH}_{2}\right)\) Sodium ethoxide: \(\displaystyle \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{O} \mathrm{Na}\) Barbital (product): \(\displaystyle \mathrm{NH}\left(\mathrm{CONHC}\mathrm{O}\mathrm{C}\mathrm{H}_{2}\mathrm{CH}_{2}\mathrm{N}\mathrm{H}\left(\mathrm{COOCH}_{2}\mathrm{CH}_{3}\right)_{2}\right)\) The overall reaction is: \(\displaystyle \mathrm{C}\mathrm{H}_{3}\mathrm{CH}_{2}\mathrm{OOC}\left(\mathrm{COOCH}_{2}\mathrm{CH}_{3}\right)_{2}+\mathrm{NH}_{2}\left(\mathrm{CONH}_{2}\right)\overset{\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{O} \mathrm{Na}}{\longrightarrow }\mathrm{NH}\left(\mathrm{CONHC}\mathrm{O}\mathrm{C}\mathrm{H}_{2}\mathrm{CH}_{2}\mathrm{N}\mathrm{H}\left(\mathrm{COOCH}_{2}\mathrm{CH}_{3}\right)_{2}\right)\)

Propose the reaction mechanism

The mechanism for this reaction can be proposed as follows: 1. Formation of a nucleophilic intermediate: Sodium ethoxide abstracts a proton from the diethyl diethylmalonate, generating a nucleophilic enolate ion intermediate: \(\displaystyle \mathrm{C}\mathrm{H}_{3}\mathrm{CH}_{2}\mathrm{OOC}\left(\mathrm{COOCH}_{2}\mathrm{CH}_{2}\mathrm{C}^{-}\mathrm{H}\right)_{2}+\mathrm{Na}^{+}\) 2. Nucleophilic attack: The enolate ion attacks the carbonyl carbon in the urea, forming a new C-N bond and a negative oxygen: \(\displaystyle \mathrm{C}\mathrm{H}_{3}\mathrm{CH}_{2}\mathrm{OOC}\left(\mathrm{COOCH}_{2}\mathrm{CH}_{2}\mathrm{C}\mathrm{H}\mathrm{CON}^{-}\mathrm{H}_{2}\right)_{2}+\mathrm{Na}^{+}\) 3. Proton transfer: The negative oxygen abstracts a proton from one of the amide nitrogens, creating a neutral intermediate: \(\displaystyle \mathrm{C}\mathrm{H}_{3}\mathrm{CH}_{2}\mathrm{OOC}\left(\mathrm{COOCH}_{2}\mathrm{CH}_{2}\mathrm{C}\mathrm{H}\mathrm{CON} \mathrm{H}\left(\mathrm{NH}_{2}\right)\right)_{2}\) 4. Intramolecular nucleophilic attack: The amide nitrogen attacks the carbonyl carbon, forming a five-member ring and releasing an ethanol molecule: \(\displaystyle \mathrm{NH}\left(\mathrm{CONHC}\mathrm{O}\mathrm{C}\mathrm{H}_{2}\mathrm{CH}_{2}\mathrm{N}\mathrm{H}\left(\mathrm{COOCH}_{2}\mathrm{CH}_{3}\right)_{2}\right)+\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\)

Determine the most acidic hydrogen in barbital

The barbital molecule has the structure shown below: \(\displaystyle \mathrm{NH}\left(\mathrm{CONHC}\mathrm{O}\mathrm{C}\mathrm{H}_{2}\mathrm{CH}_{2}\mathrm{N}\mathrm{H}\left(\mathrm{COOCH}_{2}\mathrm{CH}_{3}\right)_{2}\right)\) According to the given pKa value, let's analyze the most acidic hydrogen atom in the molecule. The most acidic hydrogen is the one that is most easily removed (donated as a proton) and results in the most stable conjugate base. In barbital, the most acidic hydrogen is the one attached to the nitrogen atom in the imidazole ring, \(\displaystyle \mathrm{NH}\left(\mathrm{CONHC}\mathrm{O}\mathrm{C}\mathrm{H}_{2}\mathrm{CH}_{2}\mathrm{N}\mathrm{H}\left(\mathrm{COOCH}_{2}\mathrm{CH}_{3}\right)_{2}\right)\).

Explain the acidity of the most acidic hydrogen in barbital

The most acidic hydrogen in barbital is that attached to the nitrogen atom in the imidazole ring because, when this hydrogen is removed as a proton (H+), the resulting conjugate base is stabilized by resonance structures in which the negative charge will be delocalized across both nitrogen atoms and the carbonyl group. This increased stabilization lowers the energy of the conjugate base, making it more likely for the hydrogen removal to occur, increasing its acidity.